Can anyone help with the last problem about The Difference Quotient of a Function

Add 13.7% to $45.50

Yuri [45]

i think first you must multiply $45.5 × 13.7/100

then you will get 6.2335

later you add 6.2335 to $45.5 and you will get $51.7335

i hope it is correct

8 0

3 years ago

Days before a presidential election, a nationwide random sample of registered voters was taken. Based on this random sample, i

dimaraw [331]

Answer:

D. No, because 50% is within the bounds of the confidence interval.

Step-by-step explanation:

"52% of registered voters plan on voting for Robert Smith with a margin of error of plus or minus±3%"

This means that the 95% confidence interval for the percentage of people that plan on voting for Robert Smith is

(52-3 = 49%, 52+3 = 55%).

We can't say that he will win the election, because at the 95% confidence level, the proportion is not guaranteed to be above 50%.

So the correct answer is:

D. No, because 50% is within the bounds of the confidence interval.

4 0

4 years ago

Read 2 more answers

Comparing observations from different populations: The heights of adult men in America are normally distributed, with a mean of

Schach [20]

Answer:

a. Z = 1.99

b. 97.67%

c. Z = 2.56

d. 0.52%

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

a. If a man is 6 feet 3 inches tall, what is his z-score (to two decimal places)?

The heights of adult men in America are normally distributed, with a mean of 69.7 inches and a standard deviation of 2.66 inches, and thus, we have $\mu = 69.7, \sigma = 2.66$

6 feet 3 inches = 6*12 + 3 = 75 inches, which means that we have to find z when X = 75. So

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

$Z = \frac{75 - 69.7}{2.66}$

$Z = 1.99$

b. What percentage of men are SHORTER than 6 feet 3 inches?

The proportion is the p-value of Z = 1.99.

Looking at the z-table, Z = 1.99 has a p-value of 0.9767.

0.9767*100% = 97.67%, which is the answer.

c. If a woman is 5 feet 11 inches tall, what is her z-score (to two decimal places)?

The heights of adult women in America are also normally distributed, but with a mean of 64.5 inches and a standard deviation of 2.54 inches, and thus, we have $\mu = 64.5, \sigma = 2.54$ . We have to find Z when X = 5*12 + 11 = 71. So

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

$Z = \frac{71 - 64.5}{2.54}$

$Z = 2.56$

d. What percentage of women are TALLER than 5 feet 11 inches?

The proportion is 1 subtracted by the p-value of Z = 2.56.

Looking at the z-table, Z = 2.56 has a p-value of 0.9948.

1 - 0.9948 = 0.0052

0.0052*100% = 0.52%, which is the answer.

3 0

3 years ago

Mr. Snow bought 90 grams of Christmas candy for each of his 14 grandchildren. How many total kilograms of candy did he buy?

nignag [31]

Answer:

1.26 kg.

Step-by-step explanation:

We have been given that Mr. Snow bought 90 grams of Christmas candy for each of his 14 grandchildren. We are asked to find the amount of candy bought by Mr. Snow in kilograms.

First of all, we will find the amount of candy in grams by multiplying 90 grams by 14 as:

$\text{Amount of candy bought in grams}=14\times 90$