Find an equation of the tangent line to the curve at the given point
y=sin(sin(x)), (π,0)

The monthly payment for an $8,720 loan at 12.5 percent for 15 years is $312.84

The formula for calculating the compound interest is expressed as:

A = P(1+r/n)^nt

Given

P = $8720

r = 12.5% = 0.125
n = 12
t = 15

Substitute

A = 8720(1+0.125/12)^180
A = 1.8622 *  8720

A = $56,312.74

Monthly payment = 56,312.74/180
Monthly payment = 312.84

Hence the monthly payment for an $8,720 loan at 12.5 percent for 15 years is $312.84

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The correct option is:d.9/2 pie units cubbed. The given region is obtained by rotating the curve y = x2 – 3 about the x-axis, which can be represented as:y = x² - 3

We have to rotate this curve around the y-axis. The required volume can be obtained by integrating the area of the circular cross-sections formed after rotating the curve.The radius of each circular cross-section at any value of x is equal to the perpendicular distance between the curve and the y-axis, which is x.

Thus, the area of each circular cross-section can be represented as:A = π(x²)The limits of integration can be found by equating y = x² - 3 to 0, which is:0 = x² - 3x = ± √3

Thus, the volume of the solid obtained by rotating the region bounded by y = x² - 3 and the x-axis about the y-axis is given by:V = ∫-√3^√3 π(x²) dxV = π ∫-√3^√3 (x²) dx.

By using the formula for integration, we get:V = π [(x³)/3] -√3^√3V = π [(√3³ - (-√3)³)/3]V = π (18/3)V = 6π.Therefore, the volume of the solid obtained by rotating the region bounded by y = x² - 3 and the x-axis about the y-axis is 6π units cubed. Hence, the correct option is:d.9/2 [tex]\pi[/tex]units cubbed.

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Since the r-value is quite close to -1 (only 0.0562 away), we can conclude that the two variables are strongly negatively associated. This implies that when one variable increases, the other variable decreases.

In statistics, the correlation coefficient is used to quantify the linear relationship between two variables. The r-value ranges from -1 to 1, indicating the degree to which the variables are positively or negatively associated.

The r-value measures the strength and direction of the linear relationship between two variables. It ranges from -1 to 1.

The closer the r-value is to -1, the stronger and more negative the correlation is.

The closer the r-value is to 1, the stronger and more positive the correlation is.

A value of 0 indicates that there is no linear relationship between the variables.In this case, the r-value is -0.9438.

Because the r-value is negative, the two variables are negatively associated. The closer the r-value is to -1, the stronger the negative association is.

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The score on the second test that is comparable to a score of 127 on the first test is 131.

Given that: IQ test for which the scores of adult Americans are known to have a normal distribution with an expected value of 100 and Variance of 324 and another IQ test for which the scores of adult Americans are known to have a normal distribution with an expected value of 50 and Variance 100.

Let x denote the first test score that is comparable to a second test score of 127. Then, we have; (127 − 100) / 18 = (x − 50) / 10 (the z-scores corresponding to the scores)

Simplifying the above equation gives;

(127 − 100) × 10 = (x − 50) × 18

Solve for x as follows;

(127 − 100) × 10 + 50 = (x) × 18

Hence, x = 131

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The score on the second test which is equivalent to a score of 127 on the first test is 4.332, or approximately 4.33 (rounded to two decimal places).

If X1, X2 are the IQ scores of the first and second tests respectively, we are given that X1 and X2 are both normally distributed. This means that their expected values and variances are known.

E(X1) = 100, Var(X1) = σ1² = 324E(X2) = 50, Var(X2) = σ2² = 100

We need to find the score, x on the second test which is equivalent to a score of 127 on the first test.

Therefore, we need to find

P(X1 ≤ 127) and then solve for x so that

P(X2 ≤ x) = P(X1 ≤ 127).

We know that

Z1 = (X1 - μ1) / σ1 and

Z2 = (X2 - μ2) / σ2 where μ1, μ2 are the respective expected values.

For a given probability P(Z ≤ z), we can find z using a standard normal table. Therefore, we can rewrite P(X1 ≤ 127) in terms of Z1 as follows;

P(X1 ≤ 127) = P((X1 - μ1) / σ1 ≤ (127 - μ1) / σ1) = P(Z1 ≤ (127 - μ1) / σ1) = P(Z1 ≤ (127 - 100) / 18) = P(Z1 ≤ 1.5)

The corresponding score on the second test, X2 can be found using;

Z2 = (X2 - μ2) / σ2 ⇒ X2 = σ2 Z2 + μ2

From the given values, μ2 = 50 and σ2 = 10.

Therefore,

X2 = σ2 Z2 + μ2 = 10 Z2 + 50

We need to find the value of x such that

P(Z2 ≤ x) = P(Z1 ≤ 1.5).

From a standard normal table,

we have P(Z ≤ 1.5) = 0.9332 and therefore,

P(Z2 ≤ x) = 0.9332 implies that

x = (0.9332 - 0.5) / 0.1 = 4.332.

Therefore, the score on the second test which is equivalent to a score of 127 on the first test is 4.332, or approximately 4.33 (rounded to two decimal places).

We are given

E(X1) = 100, Var(X1) = σ1² = 324E(X2) = 50, Var(X2) = σ2² = 100

We need to find x such that P(X1 ≤ 127) = P(X2 ≤ x).

Therefore, we find P(X1 ≤ 127) in terms of Z1 as follows;

P(X1 ≤ 127) = P(Z1 ≤ (127 - 100) / 18) = P(Z1 ≤ 1.5) = 0.9332

The corresponding value of X2, denoted by x can be found from

P(Z2 ≤ x) = P(Z1 ≤ 1.5).

From a standard normal table, we have

P(Z ≤ 1.5) = 0.9332 and therefore,

P(Z2 ≤ x) = 0.9332 implies that

x = (0.9332 - 0.5) / 0.1 = 4.332.

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(a) The possible remainders when 12 + 16 + 20 is divided by 11 are 8 and 19.

(b) For every integer n, the expression 121 + n^2 + 16n + 20 is always divisible by 11.

In the first statement, we are asked to prove that a - c divides ab + cd if and only if a - c divides ad + bc. To prove this, we can use the property of divisibility. If a - c divides ab + cd, then there exists an integer k such that ab + cd = (a - c)k. Similarly, if a - c divides ad + bc, there exists an integer m such that ad + bc = (a - c)m. By rearranging the terms, we can express k and m in terms of a, b, c, and d. By substituting these expressions into the equation ab + cd = (a - c)k, we can show that ab + cd = (a - c)(ad + bc)m. Thus, proving the equivalence.

In the second statement, we are asked about the possible remainders when 12 + 16 + 20 is divided by 11. To find the remainder, we can calculate the sum as 48 and then divide it by 11. The quotient is 4 with a remainder of 4. Hence, 12 + 16 + 20 leaves a remainder of 4 when divided by 11.

To prove that 121 + n^2 + 16n + 20 is divisible by 11 for every integer n, we can use algebraic manipulation. By factoring the expression as (n + 4)(n + 4) + 16(n + 4), we can rewrite it as (n + 4)^2 + 16(n + 4). Now, we notice that both terms have a common factor of (n + 4). Factoring it out, we get (n + 4)(n + 4 + 16). This simplifies to (n + 4)(n + 20). Since both factors contain n + 4, we can conclude that the expression is divisible by 11 for any integer n.

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A performance task is a learning activity that asks students to perform to demonstrate their knowledge, understanding and proficiency.

Your information is incomplete. Therefore, an overview will be given. A performance task measure whether a student can apply his or her knowledge to make sense of a new phenomenon.

It should be noted that performance tasks yield a tangible product and performance that serve as evidence of learning.

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Answer:

g(f(x)) = 2x - 6

f(g(x)) = 2x - 4

Step-by-step explanation:

g(f(x)) = 2x - 8 + 2

g(f(x)) = 2x - 6

f(g(x)) = 2(x + 2) - 8

f(g(x)) = 2x - 4

Answer: g ( 2 x − 8 ) = 2 x− 6

Step-by-step explanation: MAKE ME BRAINLIEST!!!!!

Step-by-step explanation:

3 coffees + 2 muffins = $7

therefore,

[tex]3x + 2y = 7[/tex]

2 coffees + 4 muffins = $8

therefore,

[tex]2x + 4y = 8[/tex]

Answer:

i think in every coast we're adding two shillings in the first sentence the total was 5 and we paid $7 and in the second sentence the total was 8 dollars meaning we added to shillings.

Step-by-step explanation:

in my school we are taughtthat we shouldn't mix goats with sheeps so we this the same way we can't mix x and y so in my opinion I guess you can say x + y =to that amount of money at the end you paid if it's 8 dollars or 7 dollars. we can say x represents the coffees and y represents the muffins that's how I get it I hope you want to be angry with me. for example 3 + 2 equals to 7 .3 + 2 is 5 so we'll carry five to the other side of equals where we will make it be minus so we'll say 7 - 5 where will get 2

Answer :

AB = 48.88

Step-by-step explanation:

We will use the sin fulmera

[tex] \frac{ \sin(134°) }{ab} = \frac{ \sin(18°) }{21} \\ \\ ab = 48.88[/tex]

I hope that is useful for you :)

Answer:

48.9

Step-by-step explanation:

[tex]\frac{A}{sinA}=\frac{B}{sinB} =\frac{C}{sinC}[/tex]   (you can flip the equation around)

[tex]\frac{21}{sin18} = \frac{AB}{sin134}[/tex]

[tex]AB=\frac{21*134}{sin18}= 48.88448235[/tex]

Sorry for the late answer I was distracted

Answer:

the answer will be 14,3 ok but check it

Answer:

Triangle JKLM ~ Triangle STUV

Given: The mass of a species of mouse commonly found in houses is normally distributed with a mean of 20.9 grams with a standard deviation of 0.2 grams.

a) The probability that a randomly chosen mouse has a mass of less than 20.74 grams is 0.2119.

b) The probability that a randomly chosen mouse has a mass of more than 21.14 grams is 0.1151.

c) The proportion of mice have a mass between 20.8 and 21.04 grams is 0.4495.

d) 10% of all mice have a mass of less than 21.15632 grams.

(a) The probability that a randomly chosen mouse has a mass of less than 20.74 grams.

Z = (X - μ)/σ, where X = 20.74, μ = 20.9, σ = 0.2.

Z = (20.74 - 20.9)/0.2Z

= -0.8P(X < 20.74)

= P(Z < -0.8)

Using standard normal distribution table, P(Z < -0.8) = 0.2119.

Thus, the required probability is 0.2119.

(b) The probability that a randomly chosen mouse has a mass of more than 21.14 grams.

Z = (X - μ)/σ, where X = 21.14, μ = 20.9, σ = 0.2.

Z = (21.14 - 20.9)/0.2

Z= 1.2

P(X > 21.14) = P(Z > 1.2)

Using standard normal distribution table, P(Z > 1.2) = 0.1151.

Thus, the required probability is 0.1151.

(c) To find the proportion of mice have a mass between 20.8 and 21.04 grams.

Z_1 = (X1 - μ)/σ

= (20.8 - 20.9)/0.2

= -0.5

Z_2 = (X2 - μ)/σ

= (21.04 - 20.9)/0.2

= 0.7

P(20.8 < X < 21.04) = P(-0.5 < Z < 0.7)

= P(Z < 0.7) - P(Z < -0.5)

Using standard normal distribution table,

P(Z < 0.7) = 0.7580

P(Z < -0.5) = 0.3085

Therefore, P(-0.5 < Z < 0.7) = P(Z < 0.7) - P(Z < -0.5)

= 0.7580 - 0.3085

= 0.4495

Thus, the required probability is 0.4495.

(d) 10% of all mice have a mass of less than grams.

Let the mass be X.

We have to find X such that P(X < k) = 0.1

Z = (X - μ)/σ, where μ = 20.9, σ = 0.2.

P(X < k) = P(Z < (k - μ)/σ)

0.1 = P(Z < (k - 20.9)/0.2)

0.1 = P(Z < 5k - 104.5)

Using standard normal distribution table, we get

5k - 104.5 = -1.2816k

k= 21.15632

Thus, 10% of all mice have a mass of less than 21.15632 grams.

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Answer:

x = 8

Step-by-step explanation:

You can set up the equation 6/4 = 12/x and then cross multiply to solve for x

Answer:

Ok y does not = 0 (They say this because is y=0 it would be undefined anything divided by 0 is undefined)

[tex](3\sqrt{12x^2})/16y\\[/tex]

Answer: [tex](3x\sqrt{3})/8y[/tex]

To find the confidence interval for p1 - p2 at a given level of confidence 95%, with x1 = 354, n1 = 545, x2 = 406, n2 = 596, we need to first calculate the point estimate for the difference in proportions:

$$\hat{p_1} = \frac{x_1} {n_1} = \frac {354}{545} \approx. 0.6495$$$$\hat{p_2} = \frac{x_2} {n_2} = \frac {406}{596} \approx. 0.6822$$

Therefore, the point estimate of the difference in proportions is: $$\hat{p_1} - \hat{p_2} = 0.6495 - 0.6822 \approx. -0.0327$$. Now, we can use the formula for the confidence interval for the difference in proportions:

$$\text{Confidence interval} = (\hat{p_1} - \hat{p_2}) \pm z_{\alpha/2} \sqrt{\frac{\hat{p_1}(1 - \hat{p_1})}{n_1} + \frac{\hat{p_2}(1 - \hat{p_2})}{n_2}}$$where z_{\alpha/2} is the z-score for the level of confidence 95% (or 0.95), which is approximately 1.96

Using this information, the confidence interval for p1 - p2 at a 95% level of confidence is: $$(0.6495 - 0.6822) \pm 1.96 \sqrt {\frac {0.6495(1 - 0.6495)}{545} + \frac {0.6822(1 - 0.6822)}{596}}$$$$\approx. -0.0327 \pm 0.0472$$

Therefore, we can conclude that the researchers are 95% confident the difference between the two population proportions, P1 - P2, is between -0.080 and -0.004 (using ascending order and rounding to three decimal places as needed).

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Answer:

big shaq

Step-by-step explanation:

Answer:

Yes, 700 yards bigger than 2000 feet because 700 yards = 2100 feet

Step-by-step explanation:

There are 3 feet in a yard, so that means we need to multiply 700 yards by 3 to get the number of feet there are in 700 yards.

700 yards × 3 feet = 2100 feet

700 yards = 2100 feet

2100 feet > 2000 feet

Answer:

Yes!

Step-by-step explanation:

700 yards is equal to 2100 feet, making it 100 feet longer.

Answer:

They are the same thing!

Step-by-step explanation:

8*3=24 (Because 1 yard is 3 feet)

24 = 24!

Answer:

8yd 24ft= 48ft 0.000000in

Step-by-step explanation:

Answer:

x = 0

y = 2

Step-by-step explanation:

2x + y = 2 (Multiply by 2)

x – 2y = -4

4x + 2y = 4

x – 2y = -4

5x = 0

x = 0

2x + y = 2

2(0) + y = 2

0 + y = 2

y = 2

Answer:24 or 4

Step-by-step explanation:

2x2=4 4x6=24 24 or 4

The solution to the given initial value problem is u(x,t) = [e−(z−¹)² + e¯(x+t)²]. [infinity] < x

Given Differential equation is J²u(x, t) Ət² = dx²I.

C is u(x,0) = e-x² and ut (x,0) = 0

We need to find the solution to the given initial value problem. The general solution to the given Differential equation is

u(x,t) = f1(x − t) + f2(x + t)

where f1 and f2 are arbitrary functions. We need to find the particular solution that satisfies the given initial conditions. Given intial condition,

u(x,0) = e-x² …………(1)

ut (x,0) = 0…………(2)

Let us find the value of f1(x) and f2(x) using the given intial condition,

From equation (1)u(x,0) = f1(x) + f2(x) = e-x²

Now, Differentiating f1(x) + f2(x) = e-x² w.r.t x, we getf1'(x) + f2'(x) = -2x ………..(3)

From equation (2)

ut (x,0) = f1'(x) + f2'(x) = 0

On solving the above two equations, we get

f1'(x) = x, f1(x) = ½x² + C1

and

f2'(x) = -x, f2(x) = ½x² + C2

Now, the particular solution is

u(x,t) = ½(x − t)² + C1 + ½(x + t)² + C2

u(x,t) = (x² + t² + C1 + C2)………….(4)

Using the initial condition u(x,0) = e-x² in equation (4)

e-x² = x² + C1 + C2

Now, we know that for large value of x, the term e-x² becomes negligible. So, C1 + C2 = 0 ⇒ C2 = -C1

Substituting C2 = -C1 in equation (4), we get

u(x,t) = [e−(z−¹)² + e¯(x+t)²]. [infinity] < x

The solution to the given initial value problem is u(x,t) = [e−(z−¹)² + e¯(x+t)²]. [infinity] < x.

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A 98% confidence interval for the variance of all student's grades is [9.41, 31.41].

The degrees of freedom (df) are (n - 1) = (20 - 1) = 19.

A confidence interval is given by the formula: $[{\chi }_{\frac{\alpha }{2},(n-1)}^{2}\frac{s^{2}}{d_f},\ { \chi }_{1-\frac{\alpha }{2},(n-1)}^{2}\frac{s^{2}}{d_f}]$ where ${\chi }_{\frac{\alpha }{2},(n-1)}^{2}$ is the lower percentile of the chi-square distribution with (n - 1) degrees of freedom, ${ \chi }_{1-\frac{\alpha }{2},(n-1)}^{2}$ is the upper percentile of the chi-square distribution with (n - 1) degrees of freedom, and s² is the sample variance.

Lower percentile (L.P) = ${\chi }_{\frac{\alpha }{2},(n-1)}^{2}\frac{s^{2}}{d_f}$

Upper percentile (U.P) = ${ \chi }_{1-\frac{\alpha }{2},(n-1)}^{2}\frac{s^{2}}{d_f}$

Now, α = 0.02, n = 20, s² = 16, and df = 19.

Lower percentile:L.P = ${\chi }_{\frac{\alpha }{2},(n-1)}^{2}\frac{s^{2}}{d_f} = { \chi }_{\frac{0.02}{2},19}^{2}\frac{16}{19} = 9.410$

Upper percentile:U.P = ${ \chi }_{1-\frac{\alpha }{2},(n-1)}^{2}\frac{s^{2}}{d_f} = { \chi }_{1-\frac{0.02}{2},19}^{2}\frac{16}{19} = 31.410$

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The 98% confidence interval for the variance of all student's grades is (10.12, 30.29). Hence, option A is the correct answer.

Given data are: The sample size, n is 20, the mean of the sample is 72 and the variance of the sample, s² is 16.

Option A is the correct answer.

Step-by-step explanation: Given, Sample size, n = 20

Mean of the sample, [tex]\bar x = 72[/tex]

Variance of the sample, [tex]s^{2} =16[/tex].

We have to find a 98% confidence interval for the variance of all student's grades.

Assumption: Grades are distributed normally.

So, the formula for the confidence interval for the variance is: [tex]((n-1)s^{2} / \chi^{2} \alpha /2, (n-1)s^{2} / \chi^{2} 1-\alpha /2)[/tex]

Where, [tex]\alpha = 1-0.98[/tex]

= 0.02

n-1 = 19

Degrees of freedom [tex]\chi^{2}\alpha /2[/tex] and [tex]\chi^{2}²1-\alpha /2[/tex] are the critical values of chi-square distribution with (n-1) degrees of freedom.

From the chi-square distribution table, we can find the values of [tex]\chi^{2}\alpha /2[/tex] and [tex]\chi^{2}²1-\alpha /2[/tex] such that the area between these values is 0.98.

Here, [tex]\alpha = 0.02[/tex] (as 98% confidence interval)

[tex]\alpha /2 = 0.02/2[/tex]

= 0.01

Using the above values, we get,

[tex]\chi^{2}0.01/2,19 = 8.907[/tex] and

[tex]\chi^{2}1-0.01/2,19 = 32.852[/tex]

Now, using the formula of confidence interval for the variance, we get,

[tex]((n-1)s^{2} / \chi^{2} \alpha /2, (n-1)s^{2} / \chi^{2} 1-\alpha /2)=((20-1) \times 16 / 32.852, (20-1) \times 16 / 8.907)[/tex]

We get, (10.12, 30.29).

Therefore, the 98% confidence interval for the variance of all student's grades is (10.12, 30.29). Hence, option A is the correct answer.

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Answer:

1.4

Step-by-step explanation:

Probability of selecting a red counter :

Number of red counter / total number of counters

5 / (5 + 3) = 5 /8 = 0.625

Probability of selecting a blue counter = 0.375

Let number of red counters :

0, 1 or 2

0 red counters :

Expected value = x * p(x)

0 + (0.625*1) + (0.390625 * 2)

0 + 0.625 + 0.78125 = 1.40625

Expected number of red counters = 1.4

Answer:

x=0

Step-by-step explanation:

Answer:

Short answer: Empty set ∅

if x is less than -1 then it can't be greater than 1

Answer: Yes it is

Step-by-step explanation:

Answer:

Yes it is. Succese to The homework

Answer:

eight.

Step-by-step explanation:

2(6) ÷ 3

12÷3

4+4

8

There are 3150 ways the Kickers soccer team can field a 2/4/4 team.

To determine the number of ways the Kickers soccer team can field a 2/4/4 team, we need to consider the different positions and the number of players available in each position.

For a 2/4/4 team, we have:

2 strikers

4 midfielders

4 full-backs

2 goalies

To calculate the number of ways, we can multiply the number of choices for each position:

Number of ways = Number of choices for strikers [tex]\times[/tex] Number of choices for midfielders [tex]\times[/tex] Number of choices for full-backs [tex]\times[/tex] Number of choices for goalies.

For the strikers, there are 4 available players, and we need to choose 2 of them.

This can be calculated using the combination formula, denoted as C(n, r):

Number of choices for strikers = C(4, 2) = 4! / (2! [tex]\times[/tex] (4-2)!) = 6

For the midfielders, there are 6 available players, and we need to choose 4 of them:

Number of choices for midfielders = C(6, 4) = 6! / (4! [tex]\times[/tex] (6-4)!) = 15

Similarly, for the full-backs, there are 7 available players, and we need to choose 4 of them:

Number of choices for full-backs = C(7, 4) = 7! / (4! [tex]\times[/tex] (7-4)!) = 35

Finally, for the goalies, there are 2 available players, and we need to choose 2 of them:

Number of choices for goalies = C(2, 2) = 1

Now, we can multiply all the choices together:

Number of ways [tex]= 6 \times15 \times 35 \times 1 = 3150[/tex]

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a) Dependent variable: Systolic blood pressure

Independent variable: Age

b) Calculations: xi (age) yi (Systolic blood pressure)xiyi (age × systolic blood pressure)xi^2 (age²)yi^2 (systolic blood pressure²)16 109 1,744 256 11,88125 122 3,050 625 14,88439 143 5,577 1,521 20,44949 199 9,751 2,401 39,60122 118 2,596 484 13,92457 175 9,975 3,249 30,62522 118 2,596 484 13,92464 185 11,840 4,096 34,22570 199 13,930 4,900 39,601∑x = 561∑y = 1,450∑xy = 61,059∑x² = 19,200∑y² = 249,141r = [9 (∑xy) - (∑x)(∑y)] / sqrt([9∑x² - (∑x)²][9∑y² - (∑y)²]) = 0.664- The correlation is moderate positive. As age increases, systolic blood pressure also increases.

c) R² = r² = 0.4416- 44.16% of the variability in systolic blood pressure is explained by age.

d) Regression Equation: y = a + bx where a = (y mean) - b (x mean) Using the means x mean = 62.3 and y mean = 161.1a = (161.1) - b (62.3)a = 33.62y = 33.62 + 1.083x

The regression equation is y = 33.62 + 1.083x.

e) Predicting the blood pressure of a 50-year-old man:y = 33.62 + 1.083xy = 33.62 + 1.083(50)y = 33.62 + 54.15y = 87.77

The predicted blood pressure is 87.77.

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Answer:

21 1/2 yd

Step-by-step explanation:

43/4 + 43/4 = 21 1/2 yd