<u>ΔACB</u> <u>ΔCDA</u>
AC² + BC² = AB² AD² + CD² = AC²
BC² = AB² - AC² BC² + CD² = AC² (AD=BC is given)
BC² = AC² - CD²
AB² - AC² = AC² - CD² (both sides were = to BC²)
AB² + CD² = 2AC²
(3)² + (√2)² = 2AC² (AB=3 and CD=√2 were given)
9 + 2 = 2AC²
11 = 2AC²
= AC²
= AC
= AC
Answer:
17,065 of those batteries can be expected to last between three years and one month and three years and seven months
Step-by-step explanation:
Problems of normally distributed samples can be 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:
I will calculate the time in months. Each year has twelve months.
Mean shelf life is three years and four months, with a standard deviation of three months. So
Proportion lasting between three years and one month and three years and seven months:
This is the pvalue of Z when X = 3*12 + 7 = 43 subtracted by the pvalue of Z when X = 3*12 + 1 = 37
X = 43
has a pvalue of 0.8413.
X = 37
has a pvalue of 0.1587.
0.8413 - 0.1587 = 0.6826
Out of 25,000 batteries:
68.26% of the batteries are expected to last between three years and one month and three years and seven months.
0.6826*25000 = 17,065
17,065 of those batteries can be expected to last between three years and one month and three years and seven months