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2 years ago

The tile along the edge of a triangular community pool needs to be replaced.

Mathematics

2 answers:

Gre4nikov [31]2 years ago

6 0

Answer:

Third option: 12x^2+8x+25

Step-by-step explanation:

s1=8x^2

s2=4x^2+15

s3=8x+10

Total perimeter of the pool edge: P

P=s1+s2+s3

Replacing s1, s2 and s3 in the formula above:

P=(8x^2)+(4x^2+15)+(8x+10)

P=8x^2+4x^2+15+8x+10

Adding like terms:

P=12x^2+8x+25

suter [353]2 years ago

6 0

Answer:

$12x^2+8x+25$

Step-by-step explanation:

We are given that

Perpendicular length side of triangle = $4x^2+15$

Base of triangle= $8x+10$

Hypotenuse of right triangle = $8x^2$

We have to find the expression which represents the total perimeter of the pool edge.

We know that

Perimeter of triangle=Sum of three sides of triangle

Substitute the values then we get

Perimeter of right triangular pool edge= $4x^2+15+8x+10+8x^2$

Combine like terms

Perimeter of right triangular pool edge= $12x^2+8x+25$

Hence, the perimeter of the pool edge is given by

$12x^2+8x+25$

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Is a line segment a defined or undefined term?

Savatey [412]

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3 0

2 years ago

The average length of a field goal in the National Football League is 38.4 yards, and the s. d. is 5.4 yards. Suppose a typical

Tasya [4]

Answer:

a) The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 0.8538.

b) 0.25% probability that his average kicks is less than 36 yards

c) 0.11% probability that his average kicks is more than 41 yards

d-a) The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 1.08

d-b) 1.32% probability that his average kicks is less than 36 yards

d-c) 0.80% probability that his average kicks is more than 41 yards

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 38.4, \sigma = 5.4, n = 40, s = \frac{5.4}{\sqrt{40}} = 0.8538$

a. What is the distribution of the sample mean? Why?

The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 0.8538.

b. What is the probability that his average kicks is less than 36 yards?

This is the pvalue of Z when X = 36. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{36 - 38.4}{0.8538}$

$Z = -2.81$

$Z = -2.81$ has a pvalue of 0.0025

0.25% probability that his average kicks is less than 36 yards

c. What is the probability that his average kicks is more than 41 yards?

This is 1 subtracted by the pvalue of Z when X = 41. So

$Z = \frac{X - \mu}{s}$

$Z = \frac{41 - 38.4}{0.8538}$

$Z = 3.05$

$Z = 3.05$ has a pvalue of 0.9989

1 - 0.9989 = 0.0011

0.11% probability that his average kicks is more than 41 yards

d. If the sample size is 25 in the above problem, what will be your answer to part (a) , (b)and (c)?

Now $n = 25, s = \frac{5.4}{\sqrt{25}} = 1.08$

The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 1.08

$Z = \frac{X - \mu}{s}$

$Z = \frac{36 - 38.4}{1.08}$