Eight hundred and eighty-six ten-thousandths in decimal

The perimeter of Raul's picture frame is 108 centimeters. The length of the picture frame is 18 centimeters. What is the width o

o-na [289]

Perimeter of a rectangle= 2(length)+2(width)

Data:
Perimeter=108 cm
length=18 cm

therefore:
108 cm=2(18 cm)+2(width)
108 cm=36 cm+2 (width)
2 width=108 cm -36 cm=
2 width=72 cm
width=72cm/2
width=36 cm

Answer: the width of the picture frame would be 36 cm.

7 0

4 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

Suppose aggressive behavior is the dependent variable in a regression model and the other variables are independent variables. I

vova2212 [387]

Answer and explanation:

Multicollinearity happens when multiple independent variables are correlated and therefore it is hard to determine which independent variable has the most impact or is most significantly affecting the dependent variable. This could be a source of confusion as the coefficients become less reliable.

Extreme multicollinearity occurs when there is exaggerated standard errors or increased variance of coefficient estimates.

6 0

3 years ago

Plssssssssss helpppppppppppp <br> 21 points

ziro4ka [17]

Answer:

196, obtuse

Step-by-step explanation:

To go around a point, you would go 360 degrees

Let x be the third angle

113+51+x=360

164+x = 360

x = 360-164

x=196

Since the angle is bigger than 180 the angle is obtuse

5 0

3 years ago

-5.73q = 97.41 q= please help me???

jek_recluse [69]

First Divide both sides by -5.73

-5.73q/-5.73=97.41/-5.73

Answer: q=-17

7 0

4 years ago