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2 years ago

I’ll give brainliest!! please help and answer correctly! plsss answer quick

Mathematics

1 answer:

Troyanec [42]2 years ago

5 0

Answer:

Step-by-step explanation:

brainlest

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4 in.

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length times width = area

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3 years ago

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Integers contain whole numbers True or False

murzikaleks [220]

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Step-by-step explanation:

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2 years ago

Suppose that 5 cards are drawn from a well-shuffled deck of 52 cards. What is the probability that all 5 are not black?

lions [1.4K]

Here's the way I see it: 5 cards are drawn, one by one, without replacement. Half (or 26) of the original deck are black and half (26) are white.

P(5 are not black) = P(5 are red)

P(5 are not black) = P(first card is red) * P(second card is red) * P(third card is red)*P(fourth card is red)*P(fifth card is red) =

(26/52) * (25/51) * (24/50) * (23/49) * (22/48) = 0.025 (answer)

We start with 52 cards. We draw one, leaving 51 cards, 25 of which are red. And so on.

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2 years ago

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

Viefleur [7K]

Answer:

The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.2087, 0.2507).

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 $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

Suppose a sample of 1537 tenth graders is drawn. Of the students sampled, 1184 read above the eighth grade level. So 1537 - 1184 = 353 read at or below this level. Then

$n = 1537, \pi = \frac{353}{1537} = 0.2297$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2297 - 1.96\sqrt{\frac{0.2297*0.7703}{1537}} = 0.2087$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2297 + 1.96\sqrt{\frac{0.2297*0.7703}{1537}} = 0.2507$

The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.2087, 0.2507).

7 0

2 years ago