Let X be the number of female employee. Let n be the sample size, p be the probability that selected employee is female.
It is given that 45% employee are female it mean p=0.45
Sample size n=60
From given information X follows Binomial distribution with n=50 and p=0.45
For large value of n the Binomial distribution approximates to Normal distribution.
Let p be the proportion of female employee in the given sample.
Then distribution of proportion P is normal with parameters
mean =p and standard deviation = 
Here we have p=0.45
So mean = p = 0.45 and
standard deviation = 
standard deviation = 0.0642
Now probability that sample proportions of female lies between 0.40 and 0.55 is
P(0.40 < P < 0.45) = 
= P(-0.7788 < Z < 1.5576)
= P(Z < 1.5576) - P(Z < -0.7788)
= P(Z < 1.56) - P(Z < -0.78)
= 0.9406 - 0.2177
= 0.7229
The probability that the sample proportion of females is between 0.40 and 0.55 is 0.7229
Answer:
its b just toook the test
Step-by-step explanation:
If $36=75%,
That means we can divide both sides by three to get $12=25%,
And because 25x4 is 100, we multiply both sides of the equation by 4 to get $48=100%,
So the answer is $48 dollars.
This also worked for me but I’m not sure it’s the most reliable way:
36 ÷ 0.75 = 48
(amount after discount ÷ percentage that is remaining)
I believe the answer would be 4.2.
Add all the points together and divide by 10
Answer:
0.01364
Step-by-step explanation:
It is given that,
A store sells a 33-pound bag of oranges for $3.60 and a 55-pound bag of oranges for $5.25.
Price per pound of 33 pound bag is 3.60/33 = 0.10909 price per pound
Price per pound of 55 pound bag of oranges is 5.25/55 = 0.09545 price per pound
Difference between price per pound for the 33-pound bag of oranges and the price per pound for the 55-pound bag of oranges is :
D = 0.10909 - 0.09545
D = 0.01364
Therefore, this is the required solution.
