The primarily result of DDT in some animal populations would be birth defects in their offsprings. The DDT was used during the second world war as a government measure to control the population of pests most likely the mosquitos. In addition, the organisms regulated are fishes and shrimps.
Answer:
(A) P (D > 0) = 99.38%
(B) P (D > 15) = 10.56%
Step-by-step explanation:
The random variable D = difference, is defined as the difference between the reading test scores after and before the program.
The random variable <em>D </em>follows a normal distribution with mean,
and standard deviation,
.
(A)
Compute the probability that the children showed any improvement, i.e.
P (D > 0):
![P(D>0)=P(\frac{D-\mu_{D}}{\sigma_{D}} >\frac{0-10}{4} )=P(Z>-2.5)=P(Z](https://tex.z-dn.net/?f=P%28D%3E0%29%3DP%28%5Cfrac%7BD-%5Cmu_%7BD%7D%7D%7B%5Csigma_%7BD%7D%7D%20%3E%5Cfrac%7B0-10%7D%7B4%7D%20%29%3DP%28Z%3E-2.5%29%3DP%28Z%3C2.5%29)
Use the standard normal random variable to determine the probability.
![P(D>0)=P(Z](https://tex.z-dn.net/?f=P%28D%3E0%29%3DP%28Z%3C2.5%29%3D0.9938)
The percentage of children showed any improvement is:
0.9938 × 100 = 99.38%
Thus, 99.38% of children showed improvement.
(B)
Compute the probability that the children improved by more than 15 points, i.e. P (D > 15):
![P(D>15)=P(\frac{D-\mu_{D}}{\sigma_{D}} >\frac{15-10}{4} )=P(Z>1.25)=1-P(Z](https://tex.z-dn.net/?f=P%28D%3E15%29%3DP%28%5Cfrac%7BD-%5Cmu_%7BD%7D%7D%7B%5Csigma_%7BD%7D%7D%20%3E%5Cfrac%7B15-10%7D%7B4%7D%20%29%3DP%28Z%3E1.25%29%3D1-P%28Z%3C1.25%29)
Use the standard normal random variable to determine the probability.
![P(D>0)=1-P(Z](https://tex.z-dn.net/?f=P%28D%3E0%29%3D1-P%28Z%3C1.25%29%3D1-0.8944%3D0.1056)
The percentage of children improved by more than 15 points is:
0.1056 × 100 = 10.56%
Thus, 10.56% of children showed improvement by more than 15 points.
Answer:
Commision rate 13%
Explanation:
The commission rate is equal to the commission as a percent of sells.
Now in our case,
sells = $2700
commission = $351
Therefore, the commission rate is equal to
Area of a circle= pi*r^2
r^2= 10^2= 100
100*pi is the area of the whole circle. Since it is a semicircle, divide it in half.
100*pi/2= 50*pi
Final answer: 50*pi cm^2
Answer:
just take the whole number (1) times the denominator (18) plus the numerator (7)
therefore: 1*18 = 18
18+7 = 25/18