The answer is 7, hope it’s right
6/8a+4,6-2/4a. I think that this is the answer.
Answer:
x=(3/2) OR 1.5
Step-by-step explanation:
2x+3=6
2x=6-3
2x=3
x=(3/2) OR 1.5
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 (
) follows a Normal distribution.
The mean of this distribution is:

The standard deviation of this distribution is: 
Compute the probability that 75% or more of the women in the sample have been on a diet as follows:

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

Thus, the probability that 75% or more of the women in the sample have been on a diet is 0.037.
Answer:
left 7
Step-by-step explanation: