Answer:
The minimum score required for admission is 21.9.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean 
and standard deviation 
, the zscore of a measure X is given by:
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:
A university plans to admit students whose scores are in the top 40%. What is the minimum score required for admission?
Top 40%, so at least 100-40 = 60th percentile. The 60th percentile is the value of X when Z has a pvalue of 0.6. So it is X when Z = 0.255. So
The minimum score required for admission is 21.9.
Answer:
y-intercept: (0, 2.5)
x-intercept: (1.67, 0)
Step-by-step explanation:
Right now, you have 3x + 2y = 5. You want to change it into slope-intercept form to make the problem easier to solve.
2y = -3x + 5
y = -3/2x + 5/2
From here, you can see that the y-intercept is (0, 5/2). You can also say that the y-intercept is (0, 2.5).
To get the x-intercept, substitute 0 for y.
0 = -3/2x + 5/2
3/2x = 5/2
3x = 5
x = 5/3
So, the x-intercept is (5/3, 0). You can also say that the x-intercept is (1.67, 0).
Hope this helps!
2n+1=3
2n=3-1
2n=2
n=1 is the answer
Answer:
Maximum revenue = $8000
The price that will guarantee the maximum revenue is $40
Step-by-step explanation:
Given that:
Price of product = $35
Total sale of items = 225
For every dollar increase in the price, the number of items sold will decrease by 5.
The total cost of item sold = 225 ×35
The total cost of item sold = 7875
If c should be the dollar unit in price increment;
Therefore; the cost function is : ![[35+c(1)][225-5(c)]](https://tex.z-dn.net/?f=%5B35%2Bc%281%29%5D%5B225-5%28c%29%5D)
For maximum revenue;
![\dfrac{d}{dc}[[35+c(1)][225-5(c)]]=0](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%7D%7Bdc%7D%5B%5B35%2Bc%281%29%5D%5B225-5%28c%29%5D%5D%3D0)
0+225-35× 5 -10c = 0
225 - 175 =10c
50 = 10c
c = 50/10
c = 5
Maximum revenue =
Maximum revenue = ![[35+5(1)][225-5(5)]](https://tex.z-dn.net/?f=%5B35%2B5%281%29%5D%5B225-5%285%29%5D)
Maximum revenue = (35 + 5)(225-25)
Maximum revenue = (40 )(200)
Maximum revenue = $8000
The price that will guarantee the maximum revenue is :
=(35 +c)
= 35 + 5
= $40
