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4 years ago

Plz answer will be marked BRAINLIEST

Mathematics

2 answers:

ivanzaharov [21]4 years ago

8 0

Answer:

1 duh

Step-by-step explanation:

zmey [24]4 years ago

3 0

Pretty sure it’s 3 i could be wrong

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

aleksley [76]

Answer:

Step 1: The estimate the proportion of tenth graders reading at or below the eighth grade level is 0.2.

Step 2: The 99% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.181, 0.219).

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 z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

Suppose a sample of 2845 tenth graders is drawn. Of the students sampled, 2276 read above the eighth grade level.

So 2845 - 2276 = 569 read below, and the estimate of the proportion of tenth graders reading at or below the eighth grade level is:

$\pi = \frac{569}{2845} = 0.2$ , and the answer to step 1 is 0.2.

The sample size is $n = 2845$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a p-value of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2 - 2.575\sqrt{\frac{0.2*0.8}{2845}} = 0.181$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2 + 2.575\sqrt{\frac{0.2*0.8}{2845}} = 0.219$

The 99% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.181, 0.219).

4 0

3 years ago

There is a 10% discount on all reusable water bottles this week. The regular price of a reusable water bottle is $20. Raul wants

Brilliant_brown [7]

Answer:$18

Step-by-step explanation:

6 0

3 years ago

Does anyone know the answer to this? i’ll mark brainliest

Ilia_Sergeevich [38]

Answer:

It is greater

Step-by-step explanation:

6 0

3 years ago

What is the solution to the system of liner equations shown on the graph?

Valentin [98]

Answer:

C The system has no solution

Step-by-step explanation:

They are parallel lines so there is no solution :).

5 0

3 years ago

A standard roulette wheel has 38 numbered slots for a small ball to land in: 36 are marked from 1 to 36, with half of those blac

Anastasy [175]

Answer:

The expected value of betting $500 on red is $463.7.

Step-by-step explanation:

There is not a fair game. This can be demostrated by the expected value of betting a sum of money on red, for example.

The expected value is calculated as:

$E(G)=\sum p_iG_i$

being G the profit of each possible result.

If we bet $500, the possible outcomes are:

- <em>Winning</em>. We get G_w=$1,000. This happens when the roulette's ball falls in a red place. The probability of this can be calculated dividing the red slots (half of 36) by the total slots (38) of the roulette:

$P(R)=R/T=18/38\approx0.4737$

- <em>Losing</em>. We get G_l=$0. This happens when the ball does not fall in a red place. The probability of this is the complementary of winning, so we have:

$P(not \,R)=1-P(R)=1-18/38=20/38\approx 0.5263$

Then, we can calculate the expected value as:

$E(G)=\sum p_iG_i=p_wG_w+p_lG_l=0.4637*1,000+0.5263*0=463.7$

We expect to win $463.7 for every $500 we bet on red, so we are losing in average $36.3 per $500 bet.

4 0

3 years ago