2. D
3. 2t^3 -4t^2 -3t-4
4. D
5. C
And the work I did to get the answers is in the picture
3. 2t^3 -4t^2 -3t-4
4. D
5. C
And the work I did to get the answers is in the picture
20
1 answer:
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Answer:
a. b. 6.1 c. 0.6842 d. 0.4166 e. 0.1194 f. 8.5349
Step-by-step explanation:
a. The distribution of X is normal with mean 6.1 kg. and standard deviation 1.9 kg. this because X is the weight of a randomly selected seedless watermelon and we know that the set of weights of seedless watermelons is normally distributed.
b. Because for the normal distribution the mean and the median are the same, we have that the median seedless watermelong weight is 6.1 kg.
c. The z-score for a seedless watermelon weighing 7.4 kg is (7.4-6.1)/1.9 = 0.6842
d. The z-score for 6.5 kg is (6.5-6.1)/1.9 = 0.2105, and the probability we are seeking is P(Z > 0.2105) = 0.4166
e. The z-score related to 6.4 kg is and the z-score related to 7 kg is
, we are seeking P(0.1579 < Z < 0.4737) = P(Z < 0.4737) - P(Z < 0.1579) = 0.6821 - 0.5627 = 0.1194
f. The 90th percentile for the standard normal distribution is 1.2815, therefore, the 90th percentile for the given distribution is 6.1 + (1.2815)(1.9) = 8.5349
I would convert to degrees first seems to make it easier.
×
= -468
by adding 360 twice we get the coterminal angle 252
252 is in the 3rd quadrant so 180 is the closest angle on the x-axis
ref angle = 252 - 180 = 72
If necessary convert back to radians...
= -468
by adding 360 twice we get the coterminal angle 252
252 is in the 3rd quadrant so 180 is the closest angle on the x-axis
ref angle = 252 - 180 = 72
If necessary convert back to radians...