Answer:
15.87% probability that a randomly selected individual will be between 185 and 190 pounds
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:
What is the probability that a randomly selected individual will be between 185 and 190 pounds?
This probability is the pvalue of Z when X = 190 subtracted by the pvalue of Z when X = 185. So
X = 190
has a pvalue of 0.8944
X = 185
has a pvalue of 0.7357
0.8944 - 0.7357 = 0.1587
15.87% probability that a randomly selected individual will be between 185 and 190 pounds
Answer:
y=4/5+3 and 3/5
Step-by-step explanation:
Answer:
a type nut is 10 pounds
a different one is 14 pounds
Step-by-step explanation:
let a type of the nut be represented by t
Let a different one be represented by d
a type of nut cost $7 per pound
a different one cost $4.20 per pound
The cost of the mixture for 24 pounds = 5.37 * 24
= $128.88
t + d = 24 ........(1)
7t + 4.2d = 128.88 ..........(2)
From equation (1), t = 24 - d
Put t = 24 - d in equation 2
7(24 - d) + 4.2d = 128.88
168 - 7d + 4.2d = 128.88
168 - 2.8d = 128.88
-2.8d = 128.88 - 168
-2.8d = -39.12
d = -39.12 / -2.8
d= 13.97
d = 14 pounds
t = 24 - d
t = 24 - 14
t = 10 pounds
A type nut is 10 pounds. A different one is 14 pounds
5x^2y is the greatest common factor