Quick

Lila uses craft sticks to make stars for an art project. She uses 5 craft sticks for each star,and she has 19 craft sticks. Lila

Savatey [412]

Answer:

3

Step-by-step explanation:

5×3=15 she would have 4 left but can't make a full star

7 0

3 years ago

A statistics teacher started class one day by drawing the names of 10 students out of a hat and asked them to do as many pushups

mash [69]

Using the t-distribution, we have that the 95% confidence interval for the true mean number of pushups that can be done is (9, 21).

For this problem, we have the <u>standard deviation for the sample</u>, thus, the t-distribution is used.

The sample mean is of 15, thus $\overline{x} = 15$ .
The sample standard deviation is of 9, thus $s = 9$ .
The sample size is of 10, thus $n = 10$ .

First, we find the number of degrees of freedom, which is the one less than the sample size, thus df = 9.

Then, looking at the t-table or using a calculator, we find the critical value for a 95% confidence interval, with 9 df, thus t = 2.2622.

The margin of error is of:

$M = t\frac{s}{n}$

Then:

$M = 2.2622\frac{9}{\sqrt{10}} = 6$

The confidence interval is:

$\overline{x} \pm M$

Then

$\overline{x} - M = 15 - 6 = 9$

$\overline{x} + M = 15 + 6 = 21$

The confidence interval is (9, 21).

A similar problem is given at brainly.com/question/25157574

6 0

3 years ago

According to a study in a medical journal, 202 of a sample of 5,990 middle-aged men had developed diabetes. It also found that m

tekilochka [14]

Answer:

0.0588 = 5.88% probability that a middle-aged man with diabetes is very active

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Has diabetes.

Event B: Is very active.

Probability of having diabetes:

To find this probability, we take in consideration that:

It also found that men who were very active (burning about 3,500 calories daily) were a fourth as likely to develop diabetes compared with men who were sedentary. Assume that one-fifth of all middle-aged men are very active, and the rest are classified as sedentary.

So the probability of developing diabetes is:

x of 4/5 = x of 0.8(not active)

x/4 = 0.25x of 1/5 = 0.2(very active). So

$P(A) = 0.8x + 0.25*0.2x = 0.85x$