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2 years ago

9

A fair die is rolled 360 times. Find the probability that: a) less than50 results are a 6. B) between 55 and 65 results (inclusi

ve) are a 6

1 answer:

Karolina [17]2 years ago

8 0

For this problem, you haven = 360p = 1/6q = 5/6 E[x] = np = 360*1/6 = 60SD[x] = √(npq) = √(360*1/6*5/6) = √(50) ≈ 7.07

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The reason being is that we need to find it out of 100. So, we divide 400 by 4 and because we did that to the denominator we have to do it to the numerator so 16 divided by 4 is well 4. So we found out that is 4/100 we just divide that by 650 and we get 26.

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2 years ago

If Ethan takes out a $5,000 loan to buy a motorcycle and his payments are $168.45 a month for 36 months, how much money will he

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2 years ago

Which is the additive inverse of the complex number -8+3i?

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The additive inverse of a complex z is a complex number $-z,$ so that

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Tags: <em>complex number additive inverse opposite algebra</em>

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2 years ago

Eric is studying people's typing habits. He surveyed 525 people and asked whether they leave one space or two spaces after a per

hichkok12 [17]

Answer:

The confidence interval is (0.81, 0.87).

Step-by-step explanation:

There's 90% confidence that population proportion is within the interval obtained from the following formula:

$\hat{p}\pm z_{\alpha/2}\sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}$

Knowing that the sample size, $n=525$ we obtain the proportion of people from the sample who leave one space after a period as $\hat{p}=\frac{440}{525}=0.8381\approx 0.84$ .

We then look for the critical value:

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Now we can replace in the formula to obtain the confidence interval:

$0.84\pm 1.645\sqrt{\frac{0.84*(1-0.84)}{525}}= (0.8137; 0.8663)$

Therefore we can say that there's 90% probability that the population proportion of people who leave one space after a period lies between the values (0.8137; 0.8663).

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