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4 years ago

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Santos flipped a coin 300 times. The coin landed heads up 175 times. Find the ration of head to total number of coin flips. Expr

ess as a simplified ratio.

1 answer:

xxMikexx [17]4 years ago

5 0

Hi there!

Let's solve this problem step by step!
If a coin is flipped 300 times and it landed 175 times in heads, we can find out that it landed on coins for 125 times, since 300 - 175 = 125.

Therefore the ratio heads to coins is
175 : 125

Now we can search for the simplest form of this ratio. We can divide both sides by 25 to find our answer (we do this because we must divide by the largest possible integer). When we divide both sides by 25, we find the following ratio
7 : 5

Hence, the answer is D. 7 : 5
~ Hope this helps you!

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If 7x=3y and 5y=7z then X/Z?

Aleks04 [339]

Answer:

3/5=x/z

Step-by-step explanation:

7x=3y

7(3)=3(7)

21=21

5y=7z

5(7)=7z (plug in the seven from the first step)

35=7z

then divide both sides by 7

5=z

therefore, X/Z=3/5

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4 years ago

A programmer plans to develop a new software system. In planning for the operating system that he will use, he needs to estimat

Vedmedyk [2.9K]

Using the z-distribution, we have that:

a) A sample of 601 is needed.

b) A sample of 93 is needed.

c) A. Yes, using the additional survey information from part (b) dramatically reduces the sample size.

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

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

In which z is the z-score that has a p-value of $\frac{1+\alpha}{2}$ .

The margin of error is:

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

95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so $z = 1.96$ .

For this problem, we consider that we want it to be within 4%.

Item a:

The sample size is <u>n for which M = 0.04.</u>

There is no estimate, hence $\pi = 0.5$

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

$0.04 = 1.96\sqrt{\frac{0.5(0.5)}{n}}$

$0.04\sqrt{n} = 1.96\sqrt{0.5(0.5)}$

$\sqrt{n} = \frac{1.96\sqrt{0.5(0.5)}}{0.04}$

$(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.5(0.5)}}{0.04}\right)^2$

$n = 600.25$

Rounding up:

A sample of 601 is needed.

Item b:

The estimate is $\pi = 0.96$ , hence:

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

$0.04 = 1.96\sqrt{\frac{0.96(0.04)}{n}}$

$0.04\sqrt{n} = 1.96\sqrt{0.96(0.04)}$

$\sqrt{n} = \frac{1.96\sqrt{0.96(0.04)}}{0.04}$

$(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.96(0.04)}}{0.04}\right)^2$

$n = 92.2$

Rounding up:

A sample of 93 is needed.

Item c:

The closer the estimate is to $\pi = 0.5$ , the larger the sample size needed, hence, the correct option is A.

For more on the z-distribution, you can check brainly.com/question/25404151

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