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3 years ago

PLZZZZZZZZZZ HELP ASAP THIS QUIZ IS 5MIN FROM BEING OVERDUE HELPPPPPP AND THE FIRST ONE TO ANSWER WILL GET THE BRAIN!!!!!!!!!!!!

Mathematics

2 answers:

just olya [345]3 years ago

8 0

her ratio of local calls was 9:5

scZoUnD [109]3 years ago

4 0

36/20

36/4 = 9

20/4 = 5

C) 9:5

Hope this Helps!!

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The data list shows the scores of ten students in Mr. Smith's math class. 61, 67, 81, 83, 87, 88, 89, 90, 98, 100 What is the st

anzhelika [568]

Sample standard deviation

Mean of data = 84.4

Calculate deviations from the mean = 23.4, 17.8, 3.4, 1.4, -2.6, -3.6, -4.6, -5.6, -13.6 and -15.6

Squaring these deviations; we get 547.56, 316.84, 11.56, 1.96, 6.76, 12.96, 21.15, 31.36, 184.96, 243.36

Adding these we get 1378.47

As its a sample Std Dev we divide by 10 - 1 = 9. For the popklation Std Dev we divide by the number of scores (10).

Std Deviation for sample = sqrt ( 1378.47 / 9) = 12.4 to nearest tenth (answer)

Population Std Dev = sqrt (1378.47 / 10) = 11.7 answer

Theres a lot of arithmetic there so hope i haven't slipped up anywhere!

6 0

3 years ago

Original amount 90$ new amount 84.50 what's the percent of change?

diamong [38]

Percent change=100 times change/original

change=original-new
change=90-84.50=5.5

percent change=100 times 5.5/90
percent change=550/90
percent change=6.11111
percent change=6.1%

7 0

3 years ago

Luke saves $5 for every $2 he spends.

aalyn [17]

Answer:

5:2

10:4

15:6

20:8

Is this what format you wanted the answer?

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

A market surveyor wishes to know how many energy drinks teenagers drink each week. They want to construct a 98% confidence inter

frosja888 [35]

Answer:

The minimum sample size required to create the specified confidence interval is 1024.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.98}{2} = 0.01$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.01 = 0.99$ , so $z = 2.327$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

What is the minimum sample size required to create the specified confidence interval

This is n when $M = 0.08, \sigma = \sqrt{1.21} = 1.1$ .

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.08 = 2.327*\frac{1.1}{\sqrt{n}}$

$0.08\sqrt{n} = 2.327*1.1$

$\sqrt{n} = \frac{2.327*1.1}{0.08}$

$(\sqrt{n})^{2} = (\frac{2.327*1.1}{0.08})^{2}$

$n = 1024$

The minimum sample size required to create the specified confidence interval is 1024.

3 0

3 years ago

The LCM of 42, 63, and 84 is

frozen [14]

3| 42 , 63 , 84

7| 14 , 21 , 28

1| 2 , 3 , 4

2 × 3 × 4 × 7 × 3 = 504

5 0

2 years ago