ΔACB and ΔMNB are similar. Therefore the corresponding sides are in proportion:
Substitute:
suppose marisol scored a 96% on an english test. There were 50 questions on the test. How many did marisol answer correctly?
2 answers:
Answer: 48
Step-by-step explanation:
Given : Marisol scored a 96% on an English test.
We can rewrite 96% as 0.96 [To convert percent into decimal we divide it by 100]
There were 50 questions on the test.
Then, the number of questions Marisol answered correctly:-
Hence, Marisol answered 48 questions correctly.
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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