Ask question

Ask question

All categories

4 years ago

10

Joshua wrote 13 articles for the school newspaper this year. Paulette wrote 7 more articles than Joshua. Jeff wrote as many arti

cles as Paulette. How many articles did they write in all?

1 answer:

Keith_Richards [23]4 years ago

6 0

53 articles? Because josh wrote 13 plus 20 that Paulette wrote then Jeff wrote as many as Paulette so another 20.

You might be interested in

Assume that foot lengths of women are normally distributed with a mean of 9.6 in and a standard deviation of 0.5 in.a. Find the

Makovka662 [10]

Answer:

a) 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.

b) 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

c) 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 9.6, \sigma = 0.5$ .

a. Find the probability that a randomly selected woman has a foot length less than 10.0 in

This probability is the pvalue of Z when $X = 10$ .

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{10 - 9.6}{0.5}$

$Z = 0.8$

$Z = 0.8$ has a pvalue of 0.7881.

So there is a 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.

b. Find the probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

This is the pvalue of Z when X = 10 subtracted by the pvalue of Z when X = 8.

When X = 10, Z has a pvalue of 0.7881.

For X = 8:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{8 - 9.6}{0.5}$

$Z = -3.2$

$Z = -3.2$ has a pvalue of 0.0007.

So there is a 0.7881 - 0.0007 = 0.7874 = 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

c. Find the probability that 25 women have foot lengths with a mean greater than 9.8 in.

Now we have $n = 25, s = \frac{0.5}{\sqrt{25}} = 0.1$ .

This probability is 1 subtracted by the pvalue of Z when $X = 9.8$ . So:

$Z = \frac{X - \mu}{s}$

$Z = \frac{9.8 - 9.6}{0.1}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772.

There is a 1-0.9772 = 0.0228 = 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.

5 0

4 years ago

Solve (round to two decimal places) <br><br> can someone please help me!

Degger [83]

Cos(40)= 21/ (hypotenuse)
hypotenuse(cos(40))= 21
hypotenuse = 21/ cos(40)
21/ cos40 = 27.41

4 0

4 years ago

Screenshot down below math question

Volgvan

I believe the answer is 1.35 x10^14

8 0

3 years ago

Find bc. Round the answer to the nearest tenth.

Nadusha1986 [10]

Is BC=2.3 one of the options?

For that question, you can use the cosine rule, where;
${a}^{2} = {b}^{2} + {c}^{2} - 2(b)(c) \cos( \alpha )$
In this case, we have..
${bc}^{2} = {1.8}^{2} + {2.1}^{2} - 2(1.8)(2.1) \cos(70)$
Solving for BC with this will get you 2.250406, rounded off to the tenth is 2.3.

6 0

4 years ago

Read 2 more answers

Theo bought these items shoes that cost $19.99 socks that cost $ 4.19, a shirt that costs $ 8.50 and pants that cost $27.75. how

Alina [70]

you must add all together 60. 43 .

3 0

3 years ago

Read 2 more answers