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larisa86 [58]
3 years ago
10

Hey guys! I am going to be posting multiple questions, but one by one... so be prepared!

Mathematics
1 answer:
Mashcka [7]3 years ago
8 0

Answer:

\frac{110}{3}    or     36\frac{2}{3}

Step-by-step explanation:

You might be interested in
The line that contains the point Q( 1, -2) and is parallel to the line whose equation is y - 4 = 2/3 (x - 3)
Musya8 [376]

Answer:

y + 2 = \frac{2}{3}(x - 1)

Step-by-step explanation:

The equation of a line in point- slope form is

y - b = m(x - a)

where m is the slope and (a, b) a point on the line

y - 4 = \frac{2}{3}(x - 3) ← is in point- slope form

with slope m = \frac{2}{3}

Parallel lines have equal slopes, thus equation of parallel line

with m = \frac{2}{3} and (a, b) = Q(1, - 2) is

y - (- 2) = \frac{2}{3}(x - 1) , that is

y + 2 = \frac{2}{3}(x - 1)

7 0
3 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

So

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

b)

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

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

Z = -2.22

Z = -2.22 has a pvalue of 0.0132

1.32% probability that his average kicks is less than 36 yards

c)

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

Z = \frac{41 - 38.4}{1.08}

Z = 2.41

Z = 2.41 has a pvalue of 0.9920

1 - 0.9920 = 0.0080

0.80% probability that his average kicks is more than 41 yards

4 0
3 years ago
Help please 10points
ikadub [295]

Answer:

Have you tried copying and pasting?

Step-by-step explanation:

ctrl c

ctrl v

6 0
3 years ago
Let f(x) = x2- 3x - 7. Find f(-3)
Alexxx [7]
F(x) = x^2 - 3x - 7
f(-3) = (-3)^ 2 - 3(-3) - 7
f(-3) = 9 + 9 - 7
f(-3) = 18 - 7
f(-3) = 11
6 0
3 years ago
Can someone please help asap thank you
Bumek [7]

Answer:

7%

Step-by-step explanation:

.07x1 is .07, therefore tax rate

please give brainliest im desperate

3 0
3 years ago
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