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Sav [38]
3 years ago
8

HELP PLEASE I DONT HAVE MUCH TIME!!!

Mathematics
1 answer:
BaLLatris [955]3 years ago
8 0

Answer:

B

Step-by-step explanation:

The other answers don't divide into rational numbers

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How many degrees are on the celsius thermometer between the freezing and boiling points of water
andriy [413]
The freezing point is 0 degrees Celsius, and the boiling point is at 100 degrees.  The difference is thus 100-0=100.
3 0
3 years ago
What is 50+ (100x99 = what
tamaranim1 [39]
50+ (100x99)
50 + 9900

9950

(You have to multiply first because of pemdas

Parentheses
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4 0
3 years ago
1. Solve for p in: -2(3p-6)+4(5p-8)=2(2p+3)+4
skad [1K]

Answer:

p = 3

Step-by-step explanation:

Hey there!

In order to find what p equals, we must first simply the equation to the most, then solve

SIMPLIFY

----------------------------------------------------------------------------------------------------------

-2 (3p - 6) + 4(5p - 8) = 2(2p + 3) + 4

-6p + 12 + 20p - 32 = 4p + 6 + 4

Now that we expanded the equation, we can combine all like terms

14p - 20 = 4p + 10

10p = 30

Now that we simplified the equation to the most, now we solve the equation

SOLVE

----------------------------------------------------------------------------------------------------------

10p = 30

p = 30/10

p = 3/1

p = 3

8 0
1 year 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
Algebra 1. Questions 2-4. Answering any would be awesome!
boyakko [2]
----------2.
6pink 7purple. total count 13.

without replacement draws 1 and then another. (odds of drawing purple)
first draw 7/13
second draw 6/12

7/13•6/12 = 42/156 or 21/78 or 7/26

with replacement.
first draw 7/13
second draw 7/13

7/13•7/13 = 49/169

-------------3.
total $8.40, p=n-60

60•5¢= $3.00

$8.40-$3.00=$5.40
1¢+5¢= 6¢

$5.40/6¢=90¢

1¢•90=90¢
5¢•90=$4.50

150•5¢=$7.50
90•1¢= $0.90

$7.50+$0.90=$8.40


5 0
3 years ago
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