Answer:
10.20% probability that a randomly chosen book is more than 20.2 mm thick
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:
250 sheets, each sheet has mean 0.08 mm and standard deviation 0.01 mm.
So for the book.
What is the probability that a randomly chosen book is more than 20.2 mm thick (not including the covers)
This is 1 subtracted by the pvalue of Z when X = 20.2. So
has a pvalue of 0.8980
1 - 0.8980 = 0.1020
10.20% probability that a randomly chosen book is more than 20.2 mm thick
Answer:
y=1/6x -1
Step-by-step explanation:
Use point-slope form to convert it to y=mx+b.
y-y1=m(x-x1)
y+2=1/6(x+6)
Distribute and simplify.
y+2=1/6x+1
y=1/6x-1
Answer:
7.19 a pair
Step-by-step explanation:
To find unit price divide the total by the amount
for ex
5 pairs cost 35.95 so 35.95 is the price of 5 pairs. To find out the price of 1 pair divide the price by the items or the 5 pairs of socks. 35.95 / 5 is 7.19
im sorry for my lack of knowing how to explain :(
Answer:
U= 2i - 1i and V= 3i + ji
Step-by-step explanation:
I hope that is the correct answer
Answer:
Step-by-step explanation:
From the information given,
Mean, μ = (10.31 + 17.22 + 26.62 + 22.84)/4 = 19.2475
Standard deviation, σ = √summation(x - mean)/n
Summation(x - mean) = (10.31 - 19.2475)^2+ (17.22 - 19.2475)^2 + (26.62 - 19.2475)^2 + (22.84 - 19.2475)^2 = 151.249475
σ = √(151.249475/4)
σ = 6.15
number of sample, n = 4
The z score for 98% confidence interval is 2.33
We will apply the formula
Confidence interval
= mean ± z ×standard deviation/√n
It becomes
19.2475 ± 2.33 × 6.15/√4
= 19.2475 ± 2.33 × 3.075
= 19.2475 ± 7.16
The lower end of the confidence interval is 19.2475 - 7.16 = 12.09
The upper end of the confidence interval is 19.2475 + 7.16 = 26.41
