All categories

3 years ago

If i have a 15 question test and the test is worth 25 points how many points is each question worth

Mathematics

1 answer:

marta [7]3 years ago

5 0

1 point because 15 goes into 25 one time.

You might be interested in

Answer this please tell me if I’m correct

jek_recluse [69]

Yes you are correct, :)

6 0

2 years ago

For each day that Sasha travels to work, the probability that she will experience a delay due to traffic is 0.2. Each day can be

lions [1.4K]

Answer:

0.8213

Step-by-step explanation:

-This is a binomial probability problem given by the function:

$P(X=x)={n\choose x}p^x(1-p)^{n-x}$

Given that n=21 and p=0.2, the probability that she experience a delay on at least 3 days is calculated as:

$P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X\geq 3)=1-P(X$

Hence, the probability that she experience delay on at least 3 days is 0.8213

6 0

3 years ago

Pls help thx , I’m behind ;(

sweet-ann [11.9K]

5 is the in its value bc its on the y intercept meaning the starting point

6 0

2 years ago

in 2006 ,the enrollment at jefferson high school was 1,200 students in 2012 the enrollment at jefferson high school was 1,650 st

Usimov [2.4K]

1650 - 1200 = 450
What percent is 450 of 1200?
y = 450/1200
y = 0.375
37.5% increase

3 0

2 years ago

Read 2 more answers

g A researcher wants to construct a 95% confidence interval for the proportion of children aged 8-10 living in Atlanta who own a

Helga [31]

Answer:

The sample size needed is 2401.

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

The margin of error of the interval is:

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

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 $Z = 1.96$ .

Estimate the sample size needed if no estimate of p is available so that the confidence interval will have a margin of error of 0.02.

We have to find n, for which M = 0.02.

There is no estimate of p available, so we use $\pi = 0.5$

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

$0.02 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.02\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.02}$

$(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.02})^{2}$

$n = 2401$

The sample size needed is 2401.

7 0

2 years ago

Read 2 more answers