Answer:
0.1425 = 14.25% probability that the individual's pressure will be between 119.4 and 121.4mmHg.
Step-by-step explanation:
When the distribution is normal, we use 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 question, we have that:

Find the probability that the individual's pressure will be between 119.4 and 121.4mmHg
This is the pvalue of Z when X = 121.4 subtracted by the pvalue of Z when X = 119.4. So
X = 121.4



has a pvalue of 0.5987
X = 119.4



has a pvalue of 0.4562
0.5987 - 0.4562 = 0.1425
0.1425 = 14.25% probability that the individual's pressure will be between 119.4 and 121.4mmHg.
H + 4 < 6
h < 6 - 4
h<2
Therefore the answer is h< 2
You have to go past the decimal point in since the first number in the tenths place is a zero you can't really round here so you go to the hundreds place and not so three you round down because 3 is closer to 0 then it is 10 so it would be 1. 4
Answer:
Originally there was 324 nuts in the bag.
Phillip took 108, Joy took 54, Brent took 81 and Preston took 10.
Step-by-step explanation:
Let's call the total amount of nuts N, and the number of nuts each child took by the initial of their name (Phillip will be P1 and Preston will be P2).
So we can write the following equations:
P1 = N/3
After removing N/3, the remaining nuts is (N - N/3):
J = (N - N/3)/4 = (2N/3)/4 = N/6
After removing N/6 from (N - N/3), the remaining nuts is (N - N/3 - N/6):
B = (N - N/3 - N/6)/2 = (N/2)/2 = N/4
P2 = 10
In the final there were 71 nuts remaining, so we have that:
N - N/3 - N/6 - N/4 - 10 = 71
N - N/3 - N/6 - N/4 = 81
N/4 = 81
N = 324 nuts
The amount of nuts took by each child is:
P1 = N/3 = 108 nuts
J = N/6 = 54 nuts
B = N/4 = 81 nuts
P2 = 10 nuts