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spayn [35]
3 years ago
14

GIVING Brainly TO CORRECT Answer

Mathematics
2 answers:
ikadub [295]3 years ago
8 0

Answer:

28

Step-by-step explanation:

23 + 24 + 27 + 38 = 112

112 ÷ 4 (total numbers) = 28

Mean = 28

nalin [4]3 years ago
3 0

Answer:

23

Step-by-step explanation:

25 + 17 + 38 + 11 + 24 = 115 / 5 = 23

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Svet_ta [14]
About 6 ;) ............
8 0
3 years ago
Sharon rented a car for a base price of $200 and has to pay $13 for each day she drives. If her total bill is $252, how many day
ollegr [7]

three thousand two handrend and seventy six

3 0
3 years ago
Please tell me what this is (correct answer)
Kamila [148]

Answer:

3.17

Step-by-step explanation:

0.45x + 0.72 + 5x = 18

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x = 3.17

6 0
3 years ago
Which is bigger:<br> 3/8 or 0.38.<br> Can you tell me how to work it out?please!
Usimov [2.4K]
You do 3 x 8 / 100 to work out 3/8 as a decimal. So its 0.24. 0.38 is bigger
7 0
3 years ago
Suppose that 70% of college women have been on a diet within the past 12 months. A sample survey interviews an SRS of 267 colleg
Fofino [41]

Answer:

The probability that 75% or more of the women in the sample have been on a diet is 0.037.

Step-by-step explanation:

Let <em>X</em> = number of college women on a diet.

The probability of a woman being on diet is, P (X) = <em>p</em> = 0.70.

The sample of women selected is, <em>n</em> = 267.

The random variable thus follows a Binomial distribution with parameters <em>n</em> = 267 and <em>p</em> = 0.70.

As the sample size is large (n > 30), according to the Central limit theorem the sampling distribution of sample proportions (\hat p) follows a Normal distribution.  

The mean of this distribution is:

\mu_{\hat p} = p = 0.70

The standard deviation of this distribution is: \sigma_{\hat p}=\sqrt{\frac{p(1-p}{n}} =\sqrt{\frac{0.70(1-0.70}{267}}=0.028

Compute the probability that 75% or more of the women in the sample have been on a diet as follows:

P(\hat p \geq 0.75)=P(\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}} \geq \frac{0.75-0.70}{0.028}) =P(Z\geq0.179)=1-P(Z

**Use the <em>z</em>-table for the probability.

P(\hat p \geq 0.75)=1-P(Z

Thus, the probability that 75% or more of the women in the sample have been on a diet is 0.037.

7 0
3 years ago
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