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Alik [6]
3 years ago
14

A librarian has 883 books to place on shelves. Each shelf holds 98 books.

Mathematics
2 answers:
egoroff_w [7]3 years ago
3 0

Answer:

She will be able to fill 9 shelves with 1 book left over.

Let me know if you need more help! :)

umka21 [38]3 years ago
3 0

Step-by-step explanation:

883/98

=9

9 books will be left

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An automatic machine is set to fill bags of dog food with exactly 25 pounds on average with a standard deviation of 0.5 pounds.
Rasek [7]

Answer:

Step-by-step explanation:

Given that an automatic machine is set to fill bags of dog food with exactly 25 pounds on average with a standard deviation of 0.5 pounds.

Sample size = n =100

x bar = 24.85 pounds

Sample std error = \frac{\sigma}{\sqrt{n} } =0.05

Hypotheses:

H_0: x bar = 25\\H_a: x bar \neq 25

(Two tailed test at 5% level)

Mean difference =0.15

Test statistic = mean diff/se = 3

Since population std dev is known z test can be used

p value =0.0000

Since p <0.05 we find that the bag does not contain exactly 25 pounds.

5 0
3 years ago
Ms. Curtis studies 25 hours for 2 tests, and studies 100 hours for how many tests?
enot [183]

Answer:

8

Step-by-step explanation:

Set up a ratio problem:

25 hours/2 tests = 100 hours/x

25x = 200

x = 8

4 0
3 years ago
Read 2 more answers
I need help simplifying this: √289
leonid [27]
The answer is 17 hoped this helped 
6 0
3 years ago
Becky wants to fence in a rectangular play area for her dog. If she wants the length of the play area to be 9 feet longer than h
Firdavs [7]

Answer:

The dimension of the play area written algebraically is y²/2 + 9y

Step-by-step explanation:

Let the be x and the width be y

If the length is to be 9 feet longer than half the width,

hence y/2 + 9 = x

hence dimension of the area = (y/2 + 9) × y = y²/2 + 9y

4 0
3 years ago
Read 2 more answers
A small lawnmower company produced 1,500 lawnmowers in 2008. In an effort to determine how maintenance-free these units were, th
ikadub [295]

Answer:

The 95% confidence interval for the average number of years until the first major repair is (3.1, 3.5).

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for the average using the finite correction factor is:

CI=\bar x\pm z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}\cdot\sqrt{\frac{N-n}{N-1}}

The information provided is:

N=1500\\n=183\\\sigma=1.47\\\bar x=3.3

The critical value of <em>z</em> for 95% confidence level is,

<em>z</em> = 1.96

Compute the 95% confidence interval for the average number of years until the first major repair as follows:

CI=\bar x\pm z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}\cdot\sqrt{\frac{N-n}{N-1}}

     =3.3\pm 1.96\times\frac{1.47}{\sqrt{183}}\times\sqrt{\frac{1500-183}{1500-1}}\\\\=3.3\pm 0.19964\\\\=(3.10036, 3.49964)\\\\\approx (3.1, 3.5)

Thus, the 95% confidence interval for the average number of years until the first major repair is (3.1, 3.5).

7 0
3 years ago
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