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3 years ago

CAN SOMEONE HELP ME PLEASE??

Mathematics

1 answer:

lapo4ka [179]3 years ago

8 0

I think you go from the top right right to the bottom right 8, sorry of this is wrong❤️

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Justin is ordering an ice cream dessert. He must order a size, a flavor of ice cream, and a topping. There are 2 sizes, 2 flavor

Soloha48 [4]

Answer:

Step-by-step explanation:

because thats the answer

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3 years ago

Look At The Picture. Thats The Question I Need Answered ASAP

Monica [59]

Its the letter G because 17.64 divided by 7 is 2.52

8 0

3 years ago

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Customers arrive at a movie theater at the advertised movie time only to find that they have to sit through several previews and

Naddika [18.5K]

Answer:

Using a 90% confidence level

A. A sample size of 68 should be used.

B. A sample size of 98 should be used.

Step-by-step explanation:

I think there was a small typing mistake and the confidence level was left out. I will use a 90% confidence level.

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

$\alpha = \frac{1-0.9}{2} = 0.05$

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 $1-0.05 = 0.95$ , so $z = 1.645$

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.

A. If we want to estimate the population mean time for previews at movie theaters with a margin of error of 72 seconds, what sample size should be used?

We have the standard deviation in minutes, so the margin of error should be in minutes.

72 seconds is 72/60 = 1.2 minutes.

So we need a sample size of n, and n is found when M = 1.2. We have that $\sigma = 6$ . So

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

$1.2 = 1.645*\frac{6}{\sqrt{n}}$

$1.2\sqrt{n} = 6*1.645$

$\sqrt{n} = \frac{6*1.645}{1.2}$

$(\sqrt{n})^{2} = (\frac{6*1.645}{1.2})^{2}$

$n = 67.65$

Rounding up.

A sample size of 68 should be used.

B. If we want to estimate the population mean time for previews at movie theaters with a margin of error of 1 minute, what sample size should be used?

Same logic as above, just use M = 1.

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

$1 = 1.645*\frac{6}{\sqrt{n}}$

$\sqrt{n} = 6*1.645$

$(\sqrt{n})^{2} = (6*1.645)^2$

$n = 97.42$

Rounding up

A sample size of 98 should be used.

3 0

2 years ago

Vicki would like to estimate the probability of tails coming up fewer than 2 times in 3 coin flips.

Ganezh [65]

Answer:

0.4

Step-by-step explanation:

Vicki has a computer randomly select 0 or 1 three times, with 0 representing <u>heads</u> and 1 representing <u>tails</u>. The results of 15 trials are shown in the table.

110 111 000 011 000

010 101 100 000 000

111 101 110 101 111

Vicki would like to estimate the probability of tails (1) coming up fewer than 2 times (must be 0 or 1 digit 1 in record). All such seuences are marked in bold int the above table. There are 6 such trails. Hence, Vicki's estimated probability, based on this simulation is

$\dfrac{6}{15}=\dfrac{2}{5}=0.4$

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3 years ago

An ice-cream shop uses the following ingredients to make one sundae.

vampirchik [111]

Answer:

Step-by-step explanation:

If each sundae requires 2 Ice-cream scoops, and you have 14, then you can find the number of sundaes by dividing 14/2

14/2 = 7

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2 years ago

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