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mart [117]
3 years ago
8

A jet travels 430 miles in 5 hours. At this rate, how far could the jet fly in

Mathematics
2 answers:
erastovalidia [21]3 years ago
8 0

divide 430 by 9 and get the answer

Galina-37 [17]3 years ago
4 0

86 miles per hour and 86 times 9 is 774 miles in 9 hours

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Please help I am stuck on this
9966 [12]
Answer: B. y = 1/2x - 1

Explanation:
I know someone else already answered, but I wanted to explain why the answer is B.

Consider an equation of a line in slope-intercept form: y = mx + b

Since the line intersects the y-axis at -1, that’s the y intercept, or b. We know the slope of the line is positive because it slopes upward from left to right. From each whole number solution point, the line goes up one and two to the right. Slope, represented by m, is “rise” over “run”, or the change in y over the change in x. This makes the slope 1/2.

I hope this helps.
7 0
2 years ago
Subtract. Write your answer as a fraction in simplest form. -5/10 - 3/10
grin007 [14]

Answer:

7/60

Step-by-step explanation:

U find the common denominator, subtract how you would with negative and postive integers, and check if you can simplify

3 0
3 years ago
Read 2 more answers
BRAINLIEST WILL BE GIVEN TO CORRECT ANSWER (stu.pid answers or links will result in DE4TH)
Serhud [2]

Answer:

2.=3

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
*<br> #17) Michelle has 8 pints of milk. How many quarts of milk does she have?
pshichka [43]

Answer:

4 quarts

Step-by-step explanation:

1 quart = 2 pints

2 quart = 4 pints

3 quart = 6 pints

4 quarts = 8 pints,

therefore Michelle has 4 quarts of milk

hope i helped you!

6 0
3 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
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