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Helen [10]
3 years ago
5

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Mathematics
2 answers:
Aleonysh [2.5K]3 years ago
7 0

Answer:

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VLD [36.1K]3 years ago
6 0
J is insists sis’s sjshjssjens jssbjss
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What is 9 over 5 in a percent
Nataly_w [17]
So you convert the fraction into a decimal then multiply it by 100 so the percentage would be 180%
3 0
3 years ago
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A clothing store sells men's hats for $30 each. At this price, the store sells 100 men's hats per week. The owner estimates that
Mkey [24]

Answer:

Step-by-step explanation:

At this price, the store sells 100 men's hats per week. The owner estimates that for every $1 increase in price, one fewer men's hat is sold per week

6 0
2 years ago
I have no idea how to work this problem out and the answer
NeTakaya

Answer:

B, D, C, A

Step-by-step explanation:

The question is asking to put the photographers into least to greast.

For the best case, I suggest you change them into decimals and put them in order.

Photgrapher A: 2/5 = 0.4 = 40%

Photographer B: 4% = 0.04 = 4%

Photographer C: 0.29 = 29%

Photographer D: 27/100 = 0.27 = 27%

Now put them into least to greatest.

Photographer B, Photographer D, Photographer C, And Photographer A

3 0
3 years ago
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A game involves correctly choosing the 5 correct numbers from 1 through 18 that are randomly drawn. What is the probability that
bazaltina [42]

Answer:

=\frac{1}{8568}\ = .00011\\\ =\frac{7}{8568}  = .00081

Step-by-step explanation:

5/18\cdot \:4/17\cdot \:3/16\cdot \:2/15\cdot \:1/14=\frac{1}{8568}

4 0
2 years ago
A manufacturer of college textbooks is interested in estimating the strength of the bindings produced by a particular binding ma
IrinaVladis [17]

Answer:

538 books should be tested.

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{1-0.99}{2} = 0.005

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 1-0.005 = 0.995, so z = 2.575

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.

How many books should be tested to estimate the average force required to break the binding to within 0.08 lb with 99% confidence?

n books should be tested.

n is found when M = 0.08

We have that \sigma = 0.72

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

0.08 = 2.575*\frac{0.72}{\sqrt{n}}

0.08\sqrt{n} = 2.575*0.72

\sqrt{n} = \frac{2.575*0.72}{0.08}

(\sqrt{n})^{2} = (\frac{2.575*0.72}{0.08})^{2}

n = 537.1

Rounding up

538 books should be tested.

6 0
3 years ago
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