Kira is choosing between two exercise routines.
In Routine #1, she burns 24 calories walking. She then runs at a rate that burns 15.5 calories per minute.
In Routine #2, she burns 52 calories walking. She then runs at a rate that burns 9.9 calories per minute.
For what amounts of time spent running will Routine #1 burn at most as many calories as Routine #2?
Use t for the number of minutes spent running, and solve your inequality for t.

Answer:<U=27.1

Step-by-step explanation:

undoing the law of sines to find <U:

[tex]\frac{sin161}{700} =\frac{sinU}{980}[/tex]

[tex]sin U 700=980sin161[/tex]

[tex]Sin U =\frac{980sin161}{700}[/tex]

[tex]< U=27.1[/tex]

Rounding the value to 2 decimal places, the geometric mean annual increase for the period is approximately 1.18.

Geometric mean annual increase for the period:

Let A be the initial value and B be the final value of the given data for the geometric mean annual increase for the period in the United States from 2000 to 2014.

A = 40,255,000 and B = 214,014,920.

To find the geometric mean annual increase for the period, we need to use the formula:

Geometric mean = (B/A)^(1/n), where n = the number of years elapsed.

Therefore, n = 2014 - 2000 = 14 years.

Substituting the values of A, B, and n in the above formula, we get:

Geometric mean = (214,014,920/40,255,000)^(1/14) ≈ 1.1802.

Rounding the value to 2 decimal places, the geometric mean annual increase for the period is approximately 1.18.

Answer: 1.18.

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The probability that a mother is a carrier is approximately 0.018.

The Hardy-Weinberg equilibrium formula can be used to determine the frequency of alleles in a population, as well as to make predictions about the probability of certain traits occurring in the next generation. In this case, we are interested in determining the probability that a mother is a carrier of a particular disease, given that the frequency of the disease is 1/4300.

To use the Hardy-Weinberg formula, we need to know the frequency of both the dominant and recessive alleles in the population. Assuming that the disease allele is recessive (i.e., carriers are heterozygous), we can represent the frequency of the disease allele (q) as

[tex]q = \sqrt{(1/4300)} = 0.004644.[/tex]

To determine the frequency of carriers (heterozygotes), we can use the formula p * q * 2,

where p represents the frequency of the dominant allele (which is equal to 1 - q in this case).

Therefore:

[tex]p = 1 - q \\ = 1 - 0.004644 = 0.995356\\q = \sqrt{(1/4300} = 0.004644\\p * q * 2 =0.018[/tex]

Approximately 1.8% of the population will be carriers of the disease allele. This means that the probability that a mother is a carrier is approximately 0.018.

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The linear measure of arc KML on OO is approximately 3.33 units, rounded to two decimal places.

Since KM is a diameter, angle KOM is a right angle. Therefore, angle KOL is a straight angle, which means that angle MOL is 180 - 145 = 35 degrees.

Now, we can use the fact that the measure of an arc is proportional to the measure of the angle it subtends. In particular, if the measure of an angle in degrees is θ and the radius of the circle is r, then the length of the arc it subtends is given by:

length of arc = (θ/360) * 2πr

In this case, the radius of the circle is half of the diameter KM, which is 36/2 = 18. So we have:

length of arc KML = (35/360) * 2 * 3.14 * 18

≈ 3.33

Therefore, the linear measure of arc KML on OO is approximately 3.33 units, rounded to two decimal places.

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Answer:

The answer to your question would be true.

The given statement 'full-time high school students spend more time per week on their courses than full-time college students' is true because full-time high school students spend approximately 30 hours per week in class, and full-time college students only spend about half of that in class.

Full-time high school students spend more time per week on their courses than full-time college students. It is because full-time high school students spend approximately 30 hours per week in class, and full-time college students only spend about half of that in class.

College courses usually last for one hour per day, whereas high school students attend classes for 5-6 hours per day. Hence, college students spend around 15-18 hours per week attending classes, while high school students spend around 30 hours per week in the classroom.

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The probability that at most 5 of these lights are red is 0.971.

To calculate this probability, we can use the binomial distribution with n = 7 and p = 0.4, since we want to know the probability of getting at most 5 red lights out of 7. We can use the formula:

P(X ≤ 5) = ΣP(X = k) for k = 0 to 5

Using this formula, we can calculate the probability as follows:

P(X = 0) = (0.5)^7 = 0.0078

P(X = 1) = 7(0.5)^7 = 0.0547

P(X = 2) = 21(0.4)(0.6)^6 = 0.2028

P(X = 3) = 35(0.4)^2(0.6)^5 = 0.322

P(X = 4) = 35(0.4)^3(0.6)^4 = 0.2458

P(X = 5) = 21(0.4)^4(0.6)^3 = 0.0907

Therefore, the probability of getting at most 5 red lights out of 7 is:

P(X ≤ 5) = 0.0078 + 0.0547 + 0.2028 + 0.322 + 0.2458 + 0.0907 = 0.971.

So the probability that at most 5 of these lights are red is 0.971.

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there are twο pοssible values οf x that make the length οf segment MP equal tο the length οf segment OP: x = 4 and x = 8.

In geοmetry, a line segment is a part οf a straight line that is bοunded by twο distinct end pοints, and cοntains every pοint οn the line that is between its endpοints.

Tο find the value οf x such that the length οf the segment MP is equal tο the length οf the segment OP, we can use the distance fοrmula tο find the lengths οf bοth segments, and then sοlve fοr x.

First, we can find the length οf segment MN using the distance fοrmula:

MN = √[(7-6)² + (-4-(-3))²] = √[1² + (-1)²] = √2

Next, we can find the length οf segment OP using the distance fοrmula:

OP = √[(x-(-3))² + (-5-(-7))²] = √[(x+3)² + 2²]

Nοw, we can set the lengths οf MP and OP equal tο each οther, and sοlve fοr x:

MN / OP = MP / OP

√2 / √[(x+3)² + 2²] = √[(6-x)² + (-3-(-5))²] / √[(x+3)² + 2²]

Squaring bοth sides οf the equatiοn, we get:

2 / [(x+3)²+ 2²] = [(6-x)² + 2²] / [(x+3)² + 2²]

Multiplying bοth sides οf the equatiοn by [(x+3)² + 2²], we get:

2 = [(6-x)² + 2²]

Expanding and simplifying, we get:

2 = x² - 12x + 40

Rearranging, we get:

x² - 12x + 38 = 0

We can use the quadratic fοrmula tο sοlve fοr x:

x = (12 ± √[(-12)² - 4(1)(38)]) / (2(1))

x = (12 ± √16) / 2

x = 6 ± 2

Therefοre, there are twο pοssible values οf x that make the length οf segment MP equal tο the length οf segment OP: x = 4 and x = 8.

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8. Spread = SD = sqrt(2*df)

9. The difference between the observed and expected frequencies of the outcomes of a set of events or variables.

10. Degrees of 1 freedom does the chi-square distribution.

11.  The rule of thumb for all expected counts in a chi-square goodness of fit test is expected counts must be at least 5.

8. Describe the center and spread of the chi-square distributions.

Shape - skewed right because can't be zero

µ = df = tipping point

mode = df - 2 = peak

spread = SD = sqrt(2*df)

9. Chi-square test is a non-parametric test (a non-parametric statistical test is a test whose model does not specify conditions about the parameter of the population from which the sample is drawn.). It is used for identifying the relationship between a categorical variable and denoted by χ2.

10. A chi-squared distribution constructed by squaring a single standard normal distribution is said to have 1 degree of freedom. Thus, as the sample size for a hypothesis test increases, the distribution of the test statistic approaches a normal distribution.

11. The rule of thumb for all expected counts in a chi-square goodness of fit test is expected counts must be at least 5.

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The volume of the rectangular prism is 60 inches cube.

The diagram above is a rectangular prism. The volume of a rectangular prism can be found as follows:

volume of a rectangular prism = lwh

where

Therefore,

l = 4 inches

w = 3 inches

h = 5 inches

volume of a rectangular prism = 4 × 3 × 5

volume of a rectangular prism = 60 inches cube

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According to question,

DEF+PRS=90

DEF=78

PRS=?

WE KNOW,

DEF+PRS=90

or,78+PRS=90

or,PRS=90-78

:.PRS=12,,

Answer:

331/2

Step-by-step explanation:

5 1/2= 11/2

3 1/2 = 7/2

11/2x7/2x6=115 1/2 = 331/2

Answer:

115.5 cm^3

Step-by-step explanation:

Volume=LengthxWidthxHeight

The values from the diagram are 3.5, 6, and 5.5

Multiply!

3.5x6x5.5=115.5

To perform regression on this longitudinal data, where air pollution readings and traffic flow data are collected at regular intervals over time, we would use time series regression.

The link between a dependent variable (in this case, air pollution readings) and one or more independent variables (in this case, traffic flow) that are measured over time is examined using the statistical technique known as time series regression.

It is assumed that the dependent variable and the independent variable have a linear relationship in time series regression, a form of linear regression (s). It also takes into account the data's temporal character, where each observation depends on earlier observations.

In time series regression, a time series model is used to take the temporal component of the data into account. There are various kinds of time series models, such as the popular technique for time series regression known as ARIMA (autoregressive integrated moving average). ARIMA models forecast the future using the dependent variable's previous values.

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Answer:

Step-by-step explanation:

One potential drawback of using a histogram is that it can be sensitive to the choice of bin width or bin size. If the bin size is too small, the histogram may appear too noisy or have too many empty bins, which can obscure patterns in the data. If the bin size is too large, important features of the distribution may be lost or smoothed out. Additionally, histograms do not always show the actual values of the data points, but rather a summary of the data. This means that some details about the data may be lost, such as the exact values of outliers or individual data points.

Regional identities have led to independence movements in the United Kingdom but not in Mexico because of the history of the regions' existence as separate states in the United Kingdom.

Reason of regional identities:

In the United Kingdom, there are four countries with distinct regional identities: England, Scotland, Wales, and Northern Ireland. Each of these countries has a unique culture, language, and history, which has led to a sense of national pride and identity.Regional identities in the United Kingdom have played a significant role in independence movements.

        For example, Scotland has a long history of wanting independence from the United Kingdom. The country was an independent state until it joined the United Kingdom in 1707.

              However, many Scottish people feel that their cultural and political identity is different from the rest of the United Kingdom and have called for a referendum on Scottish independence.Mexico, on the other hand, is a unitary state with a strong central government. The country has a long history of political and cultural unity, which has led to a sense of national identity.

      Therefore, the history of the regions' existence as separate states in the United Kingdom best explains why regional identities have led to independence movements in the United Kingdom but not in Mexico.

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The dimensions that will produce the maximum enclosed area for two adjacent rectangular safe play areas of equal size with 200 feet of fencing is 100 feet by 100 feet.

To solve this problem, use the area of a rectangle formula A = lw, where l is the length and w is the width.

Since the two play areas must be of equal size, let's call the length and width l and w, respectively.

We know that the total length of fencing is 200 feet, so l + w = 200 feet.

We can then solve for l by rearranging the equation as l = 200 - w.

We then substitute this into the area equation to get A = (200 - w)w.

To maximize the area, differentiate this equation to get A' = w - 200.

Set this to 0 and solve for w to get w = 100. Since l + w = 200, l = 100 as well.

Therefore, the maximum enclosed area is produced by dimensions of l = 100 feet and w = 100 feet.

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Option D is correct, the p-value is greater than 0.10, and H₀ should not be rejected.

The null hypothesis is denoted as H₀, is a statement or assumption that suggests there is no significant or meaningful relationship between two variables in a statistical hypothesis test.

If a 90 percent t-interval constructed from the sample data contains the value of 100, it means that the sample mean is within the confidence interval.

Since the null hypothesis H₀: μ = 100 is also assuming that the population mean is 100, this suggests that the sample data is consistent with the null hypothesis.

Given a significance level of α = 0.10, we compare the p-value with this significance level to make a conclusion.

If the 90 percent confidence interval contains the hypothesized value (100 in this case), it suggests that the p-value is likely to be greater than 0.10.

There is insufficient evidence to reject the null hypothesis at a significance level of 0.10.

Therefore, the p-value is greater than 0.10, and H₀ should not be rejected.

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The time necessary for the cell cycle for cancerous cells is significantly shorter compared to normal cells.

This implies that cancerous cells reproduce faster and therefore are able to spread more quickly. Doctors should be aware of this accelerated process so that they can provide more effective treatment options to their patients.

Explanation: In Model 1, the length of the arrow represents the time necessary for a cell to complete its cell cycle. For cancerous cells, the cell cycle is significantly shorter compared to normal cells.

This implies that cancerous cells can reproduce more quickly, and therefore spread more quickly. It is important for doctors to be aware of this so that they can make more informed decisions when it comes to treating cancer patients.

They should have an understanding of the accelerated process of cancerous cells, and use this to create better treatment plans for their patients.

This may include more aggressive methods of treatment such as chemotherapy and radiation, in order to try and stop the rapid spread of cancerous cells.

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Answer:

$57

Step-by-step explanation:

12*4.75=57

Percents of student failed in the test is 15.87%

Final grades are distributed normally.

Assuming mean, which is = 70

Provided that the standard deviation is 10,

Standard score or z-score refers to the normal random variable in a standard normal distribution. The following equation can convert each normal random variable X into a z score:

z=(X−μ)/σ

If X is a normal random variable, represents its mean, and represents its standard deviation.

For X=60 Z=(60−70)/10=−1

P(X<60)=P(Z<−1) \s =0.1587

Hence, 15.87% of pupils failed the test.

By dividing the sum of the given numbers by the entire number of numbers, the mean—the average of the given numbers—is determined.

Mean is equal to (Sum of All Observations/Total Observations).

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Theoretical probability of drawing the letter "o" is 1/4, or 0.25 as a decimal.

The probability of drawing the letter "o" from the bag of tiles depends on the number of tiles with the letter "o" and the total number of tiles in the bag.

Since the word "Colorado" has two "o's", there are two "o" tiles in the bag. The total number of tiles in the bag is the number of letters in the word "Colorado", which is 8.

Therefore, the theoretical probability of drawing the letter "o" is:

P("o") = number of "o" tiles / total number of tiles

= 2/8

= 1/4

So the theoretical probability of drawing the letter "o" is 1/4, or 0.25 as a decimal.

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Answer:  3

Step-by-ste3p explanation:

The probabilty of selecting two white balls in the two scenarios are 1/3 and 1/13

Scenario 1

The probability of drawing a white ball from the first urn is 5/9, since there are 5 white balls out of 9 total balls.

Similarly, the probability of drawing a white ball from the second urn is 3/5.

To find the probability that both balls are white, we need to multiply the probabilities of drawing a white ball from each urn, since the events are independent:

(5/9) × (3/5) = 15/45 = 1/3

Scenario 2

Using the steps in (1), we have

P = 1/2 * 2/12

Evaluate

P = 1/12

Therefore, the probability that both balls are white is 1/12.

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The length of the chord is approximately 7.62 inches.

Let's call the center of the circle point O, the radius of the circle 5 inches, the point where the chord intersects the radius point A, and the point where the chord intersects the circle point B.

Since the chord is perpendicular to the radius, we know that angle AOB is a right angle. Also, since OA is 5 inches and AB is 2 inches, we can use the Pythagorean theorem to find the length of OB

OB^2 = OA^2 + AB^2

OB^2 = 5^2 + 2^2

OB^2 = 25 + 4

OB^2 = 29

OB = sqrt(29) ≈ 5.39 inches

Now that we know the length of OB, we can use it to find the length of the chord. Let's call the length of the chord CD, where C and D are the points where the chord intersects the circle. Since OB is perpendicular to CD, we can use the Pythagorean theorem again to find the length of CD

CD^2 = 2OB^2

CD^2 = 2(29)

CD^2 = 58

CD = sqrt(58) ≈ 7.62 inches

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The equation that represents the total area of the pool and deck is: A(x) = 2x² + 24x + 64.

In algebra, a binomial is a polynomial that consists of two terms connected by either an addition or a subtraction operator. A binomial can be written in the form (ax + b) or (ax - b), where "a" and "b" are constants and "x" is a variable.

Binomials are used in a variety of algebraic operations, such as factoring, expanding, and solving equations. They also appear in binomial probability distributions and binomial theorem.

Let's start by drawing a diagram to visualize the problem:

___________      

|                       |    4 ft

|   _________|_________

|   |                   |                    |

|   |                   |                    |

|  |                    |                    |   2x

|  |                    |                    |

|  |_________|_________ |

|                       |    4 ft

|__________ |

      2w

The rectangular pool is twice as wide as it is long, so we can let the length be x, and the width be 2x.

The deck around the pool is 4 feet wide on each side, so the total width of the pool and deck is (2x + 8), and the total length is (x + 8).

Therefore, the total area of the pool and deck can be represented as:

(2x + 8)(x + 8)

Expanding the expression, we get:

2x² + 24x + 64

So the equation that represents the total area of the pool and deck is:

A(x) = 2x² + 24x + 64

where A(x) is the area of the pool and deck, and x is the length of the pool.

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a) The probability that during next second the number of alpha particales omitted from 1 gram is at most 6 is 0.998.

b) The probability that during next second the number of alpha particales omitted from 1 gram is at least 2 is 0.985.

c) The probability that during next second the number of alpha particales omitted from 1 gram is at least 3 and at most 6 is 0.679.

This problem is an application of the Poisson distribution, which is commonly used to model the number of events occurring in a fixed interval of time. In this case, we are interested in the number of alpha particles emitted by 1 gram of the radioactive element in one second.

The average rate of emission is given as 3.9 alpha particles per second. This means that the mean or expected number of alpha particles emitted by 1 gram of the element in one second is also 3.9.

(a) To find the probability that at most 6 alpha particles are emitted from 1 gram of the element in one second, we can use the Poisson distribution with λ=3.9 and x=0, 1, 2, 3, 4, 5, or 6.

The probability is given by the cumulative distribution function (CDF) of the Poisson distribution, which can be calculated using software or a table. For example, using a Poisson table or calculator, the probability is approximately 0.998.

(b) To find the probability that at least 2 alpha particles are emitted, we can use the complementary probability: P(at least 2) = 1 - P(none or 1). Using the Poisson distribution with λ=3.9 and x=0 or 1, we find P(none or 1) = 0.015. Therefore, P(at least 2) = 1 - 0.015 = 0.985.

(c) To find the probability that at least 3 and at most 6 alpha particles are emitted, we need to add the probabilities for x=3, 4, 5, and 6. Using the Poisson distribution with λ=3.9, we can calculate P(x=3) = 0.089, P(x=4) = 0.172, P(x=5) = 0.212, and P(x=6) = 0.206. Therefore, the probability is approximately 0.679.

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Answer: y-98

Step-by-step explanation:

the abundance expressed as cells per ml would be: 7.70*10^8

Here,

Initial population = 9019

Growth rate = 0.9502 hr-1

Time = 17 hrs

We know that,

Final population = Initial population ( 1 + Growth rate)Time

= 9019 ( 1 + 0.9502 )17

= 9019 ( 1.9502 )17

= 9019 x 85378

= 770,024,182

= 7.70*10^8

A bacteria culture is a test to distinguish whether you have a bacterial disease. It very well may be performed on an example of blood, stool, pee, skin, bodily fluid, or spinal liquid. Utilizing this kind of test, a medical services supplier can distinguish what caused a disease and decide the best therapy.

Exponential growth is an example of information that shows more keen increments after some time.

In finance, compounding makes exponential returns.

Bank accounts with a building financing cost can show exponential growth.

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the complete question is:

7 + 35 2 points A culture of bacteria enters the exponential growth phase with 9,019 cells per ml. The culture has a growth rate constant of 0.9502 hours 1. After 17 hours, what is the abundance expressed as cells per ml? 1 • Report your answer in scientific notation to two decimal places. Remember to include trailing zeros!! o For example: • 42,000,000 would be expressed as 4.20*10^7 42,367,000 would be expressed as 4.24*10^7 2 3 0.54 4 Previous Next 5 6 7 00 8 9 10 11 12

Answer:

[tex]\dfrac{2x^3-x^2+5x}{(x^2+1)^2}\equiv \dfrac{2x-1}{(x^2+1)}+\dfrac{3x+1}{(x^2+1)^2}[/tex]

Step-by-step explanation:

As the denominator has a repeated irreducible quadratic factor, and the degree of the denominator is greater than the degree of the numerator, the partial fraction form is:

[tex]\boxed{\dfrac{N(x)}{(x^2+c)^2} \equiv\dfrac{Ax+B}{(x^2+c)}+\dfrac{Cx+D}{(x^2+c)^2}}[/tex]

Therefore, the given algebraic fraction can be written as partial fractions of the form:

[tex]\dfrac{2x^3-x^2+5x}{(x^2+1)^2}\equiv \dfrac{Ax+B}{(x^2+1)}+\dfrac{Cx+D}{(x^2+1)^2}[/tex]

Add the partial fractions:

[tex]\dfrac{2x^3-x^2+5x}{(x^2+1)^2}\equiv \dfrac{(Ax+B)(x^2+1)+Cx+D}{(x^2+1)^2}[/tex]

Cancel the denominators from both sides of the original identity, so the numerators are equal:

[tex]2x^3-x^2+5x = (Ax+B)(x^2+1)+Cx+D[/tex]

Expand the right side of the equation:

[tex]2x^3-x^2+5x=Ax^3+Ax+Bx^2+B+Cx+D[/tex]

Group elements according to the powers of x:

[tex]2x^3-x^2+5x=Ax^3+Bx^2+(A+C)x+B+D[/tex]

Equate the coefficients of the terms in x³ and x² to solve for A and B:

[tex]\implies A=2[/tex]

[tex]\implies B=-1[/tex]

Substitute the found values of A and B into the equation:

[tex]2x^3-x^2+5x=2x^3-x^2+(2+C)x-1+D[/tex]

Equate the coefficients of the terms in x and the constant to solve for C and D:

[tex]\implies 5=2+C \implies C=3[/tex]

[tex]\implies 0=-1+D \implies D=1[/tex]

Replace A, B, C and D in the original identity:

[tex]\dfrac{2x^3-x^2+5x}{(x^2+1)^2}\equiv \dfrac{2x-1}{(x^2+1)}+\dfrac{3x+1}{(x^2+1)^2}[/tex]

If one card is randomly selected from a well-shuffled standard deck of 52 cards, the probability that the card selected is a queen is 1/13. This is because there are 4 queens in a deck of 52 cards, therefore, the probability of getting one queen out of the 52 cards is 4/52, which can be simplified to 1/13.

The probability that the card selected is a queen is calculated using the formula:

P(E) = n(E) / n(S)

where:

P(E) = probability of an event

n(E) = number of ways an event can happen

n(S) = total number of possible outcomes

In this case, n(E) is 4 (since there are 4 queens in a deck of 52 cards), and n(S) is 52 (since there are 52 cards in a deck), so:

P(E) = 4 / 52

P(E) = 1/13

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The missing side of the right triangle after using Pythagorean theorem we get the answer as  approximately 10.908 feet.

To find the missing side of a right triangle, we can use the Pythagorean theorem, which states that in a right triangle, the sum of the squares of the lengths of the two shorter sides is equal to the square of the length of the longest side (the hypotenuse).

Let's label the sides of the right triangle as follows:

According to the problem, we have:

We can use the Pythagorean theorem to find side b:

a² + b² = c²

5² + b² = 12²

25 + b² = 144

b² = 144 - 25

b² = 119

b = √(119) ≈ 10.908

Therefore, the missing side of the right triangle is approximately 10.908 feet.

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