Use the Pythagorean Theorem to find the length of the hypotenuse of a

Hundred chart: I am thinking of a number that is greater than 70 but less than 100.

Katyanochek1 [597]

The numbers which is greater than 70 but less than 100, not divisible by 10, not divisible by 4, not an odd number are 74, 78, 82, 86, 94, 98

<h3>Even and odd numbers</h3>

A number greater than 70 but less than 100
Not divisible by 10
Not divisible by 4
Not odd number

A number greater than 70 but less than 100

71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 98

Not odd number

72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98

Not divisible by 10

72, 74, 76, 78, 82, 84, 86, 88, 92, 94, 96, 98

Not divisible by 4

74, 78, 82, 86, 94, 98

Therefore, the numbers which is greater than 70 but less than 100, not divisible by 10, not divisible by 4, not an odd number are 74, 78, 82, 86, 94, 98.

Learn more about even and odd numbers:

brainly.com/question/2263958

#SPJ1

7 0

2 years ago

What is the number name?<br> 5.789

My name is Ann [436]

Answer:

B

Step-by-step explanation:

5.7 is five and 7 tenths and you keep moving up

8 0

3 years ago

Line segment CD is shown on a coordinate grid:

tekilochka [14]

Answer A) is the correct option because the rest are wrong. :)

6 0

3 years ago

Read 2 more answers

The freezing point of water is 32º F. Tim added potassium and found

HACTEHA [7]

Answer:

The freezing point of the water with the added potassium is 24 degrees Fahrenheit.

Step-by-step explanation:

To find the freezing point of the water with the added potassium, you need to subtract 8 degrees from the initial freezing point of water, This is because you know that after adding potassium to water, the freezing point went down by 8 degrees Fahrenheit. So you need to subtract 8 degrees from the initial freezing point of water (32 degrees Fahrenheit). 32º F - 8º = 24º F. So the freezing point of the water with the added potassium is 24 degrees Fahrenheit.

5 0

2 years ago

The Scholastic Aptitude Test (SAT) is a standardized test for college admissions in the U.S. Scores on the SAT can range from 60

kodGreya [7K]

Answer:

A. The PrepIt! claim of statistically significant differences is valid. PrepIt! classes produce improvements in SAT scores that are 3% to 13% higher than improvements seen in the comparison group.

False, We conduct a confidence interval associated to the difference of scores with additional preparation and without preparation. And we can't conclude that the results are related to a % of higher improvements.

B. Compared to the control group, the PrepIt! course produces statistically significant improvements in SAT scores. But the gains are too small to be of practical importance in college admissions.

Correct, since we net gain is between 3.0 and 13 with 90% of confidence and if we see tha range for the SAT exam is between 600 to 2400 and this gain is lower compared to this range of values.

C. We are 90% confident that between 3% and 13% of students will improve their SAT scores after taking PrepIt! This is not very impressive, as we can see by looking at the small p-value.

False, we not conduct a confidence interval for the difference of proportions. So we can't conclude in terms of a proportion of a percentage.

Step-by-step explanation:

Notation and previous concepts

$n_1$ represent the sample after the preparation

$n_2$ represent the sample without preparation

$\bar x_1 =678$ represent the mean sample after preparation

$\bar x_2 =1837$ represent the mean sample without preparation

$s_1 =197$ represent the sample deviation after preparation

$s_2 =328$ represent the sample deviation without preparation

$\alpha=0.1$ represent the significance level

Confidence =90% or 0.90

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{(\frac{s^2_1}{n_s}+\frac{s^2_2}{n_s})}$ (1)

The point of estimate for $\mu_1 -\mu_2$

The appropiate degrees of freedom are $df=n_1+ n_2 -2$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,df)

The standard error is given by the following formula:

$SE=\sqrt{(\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2})}$

After replace in the formula for the confidence interval we got this:

$3.0 < \mu_1 -\mu_2$

And we need to interpret this result:

A. The PrepIt! claim of statistically significant differences is valid. PrepIt! classes produce improvements in SAT scores that are 3% to 13% higher than improvements seen in the comparison group.

False, We conduct a confidence interval associated to the difference of scores with additional preparation and without preparation. And we can't conclude that the results are related to a % of higher improvements.

B. Compared to the control group, the PrepIt! course produces statistically significant improvements in SAT scores. But the gains are too small to be of practical importance in college admissions.

Correct, since we net gain is between 3.0 and 13 with 90% of confidence and if we see tha range for the SAT exam is between 600 to 2400 and this gain is lower compared to this range of values.

C. We are 90% confident that between 3% and 13% of students will improve their SAT scores after taking PrepIt! This is not very impressive, as we can see by looking at the small p-value.

False, we not conduct a confidence interval for the difference of proportions. So we can't conclude in terms of a proportion of a percentage.

5 0

3 years ago