Answer:
The perimeter is 52
Step-by-step explanation:
The area is 36x^2 -60x+25
We set this equal to 0 to find the length and width
A= 36x^2 -60x+25
We notice that this is a perfect square trinomial
(a^2 -2ab-b^2) = (a-b)^2
let a = 6x and b=5
A=(6x-5) (6x-5)
The length and width are the same since is it a square (we know it is a square so they have to be equal)
The perimeter of a square is
P =4s
P =4 (6x-5)
Distribute the 4
= 24x -20
Let x =3
P = 24(3) -20
=72 -20
= 52
The perimeter is 52
Answer: 524
Step-by-step explanation: The opposite number is the negative or positive number of the number given.
check a website called math-way without the dash it might help
Answer:
a) 40.13% probability that a laptop computer can be assembled at this plant in a period of time of less than 19.5 hours.
b) 34.13% probability that a laptop computer can be assembled at this plant in a period of time between 20 hours and 22 hours.
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:

a)Less than 19.5 hours?
This is the pvalue of Z when X = 19.5. So



has a pvalue of 0.4013.
40.13% probability that a laptop computer can be assembled at this plant in a period of time of less than 19.5 hours.
b)Between 20 hours and 22 hours?
This is the pvalue of Z when X = 22 subtracted by the pvalue of Z when X = 20. So
X = 22



has a pvalue of 0.8413
X = 20



has a pvalue of 0.5
0.8413 - 0.5 = 0.3413
34.13% probability that a laptop computer can be assembled at this plant in a period of time between 20 hours and 22 hours.