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zavuch27 [327]
3 years ago
7

A random selection will be made from a bag containing different

Mathematics
1 answer:
Ganezh [65]3 years ago
8 0

Answer:B

Step-by-step explanation:because it will give you more information and the more information you have the more likely you would get it correct.

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Kayla wanted to buy a sweatshirt. The original price was $45. When she went to the store the sale price was $37. What is the per
crimeas [40]

Answer:

You need to cross multiply 37/45+x/100 which equals 17%

Step-by-step explanation:

The first part represents the 37 dollars out of the original 45 and the second part represents x= the unknown percent out of 100 percent. This would be 17%

7 0
3 years ago
Select the correct solution in each column of the table.
aalyn [17]

Answer:

Can you take a picture of the problem to see if i could help!?

Step-by-step explanation:

5 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
3y +(-2) +48=180 solve for y
GalinKa [24]

Answer:

The answer to the problem is

y = 134/3

Step-by-step explanation

7 0
3 years ago
Here's another one :)
jekas [21]
Area Of Triangle:


4/2 = 2
2x6 = 12

12x2 = 24ft^2

Remaining area:

150 - 24 = 126ft^2

Maximum length:

126/6 = 21

x = 21ft
8 0
3 years ago
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