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WARRIOR [948]
3 years ago
14

Need help asap please!!!

Mathematics
1 answer:
Tanya [424]3 years ago
8 0

Answer:

251.32741

Step-by-step explanation: Later Surface Area Formula: 2(pi)rh

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A bottle holds 5 milliliters of perfume. How many bottles can we fill with 2 milliliters of perfume?
Sonja [21]

Answer:

2/5 or 40% of a bottle.

Step-by-step explanation:

A bottle can hold 5 milliliters of perfume. You only have 2 milliliters. Divide 2 with 5:

(2/5)(20/20) = 40/100 = 0.40 = 40%

~

3 0
3 years ago
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Can a triangle be drawn with the following side lengths? Why or why not?
Ghella [55]
Yes, because it still have 3 sides.
4 0
3 years ago
Write the following decimal number in its equivalent fraction form. Show all work for full credit. 0.225
zalisa [80]
Rewrite the decimal number as a fraction with 1 in the denominator

0.225=0.22510.225=0.2251

Multiplying by 1 to eliminate 3 decimal places, we multiply top and bottom by 103 = 1000

0.2251×10001000=2251000


7 0
3 years ago
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An automobile manufacturer would like to know what proportion of its customers are not satisfied with the service provided by th
butalik [34]

Answer:

a) A sample size of 5615 is needed.

b) 0.012

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

99.5% confidence level

So \alpha = 0.005, z is the value of Z that has a pvalue of 1 - \frac{0.005}{2} = 0.9975, so Z = 2.81.

(a) Past studies suggest that this proportion will be about 0.2. Find the sample size needed if the margin of the error of the confidence interval is to be about 0.015.

This is n for which M = 0.015.

We have that \pi = 0.2

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.015 = 2.81\sqrt{\frac{0.2*0.8}{n}}

0.015\sqrt{n} = 2.81\sqrt{0.2*0.8}

\sqrt{n} = \frac{2.81\sqrt{0.2*0.8}}{0.015}

(\sqrt{n})^{2} = (\frac{2.81\sqrt{0.2*0.8}}{0.015})^{2}

n = 5615

A sample size of 5615 is needed.

(b) Using the sample size above, when the sample is actually contacted, 12% of the sample say they are not satisfied. What is the margin of the error of the confidence interval?

Now \pi = 0.12, n = 5615.

We have to find M.

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

M = 2.81\sqrt{\frac{0.12*0.88}{5615}}

M = 0.012

7 0
3 years ago
A bag contains 6 white marbles, 5 red
Artist 52 [7]

Answer:

1/30

Step-by-step explanation:

      6 white marbles

      5 red marbles

+    19 marbles of other colors

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

     30 marbles in total

First drawing:

p(white) = (number of white marbles)/(total number of marbles)

p(white) = 6/30 = 1/5

Second drawing:

Since the first marble is replaced, there are still 30 marbles in the bag.

p(red) = (number of red marbles)/(total number of marbles)

p(red) = 5/30 = 1/6

The two drawings are independent events, so the overall probability of the two events is the product of the individual probabilities.

p(white then red) = p(white) × p(red)

p(white then red) = 1/5 × 1/6

p(white then red) = 1/30

4 0
2 years ago
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