Answer:
She will donate P-50/3 each time
Step-by-step explanation:
Now assuming total donation needed to be in the silver category Is P
We can assume also the three additional and equal donations be Q
So
Then 50 + 3Q = P
So we still don't know P
We can get Q by saying
3Q= P - 50
Q = (P - 50) /3
Answer:
81.85% of the workers spend between 50 and 110 commuting to work
Step-by-step explanation:
We can assume that the distribution is Normal (or approximately Normal) because we know that it is symmetric and mound-shaped.
We call X the time spend from one worker; X has distribution N(μ = 70, σ = 20). In order to make computations, we take W, the standarization of X, whose distribution is N(0,1)
The values of the cummulative distribution function of the standard normal, which we denote 
, are tabulated. You can find those values in the attached file.
Using the symmetry of the Normal density function, we have that 
. Hece,
The probability for a worker to spend that time commuting is 0.8185. We conclude that 81.85% of the workers spend between 50 and 110 commuting to work.
Finding the value of the second variable when the first variable is equal to one and then multiplying the second variable by the first variable of each answer will always work in this situation (two variable).
There are three questions related to this problem.
First, the probability of the mail will arrive after 2:30
PM
<span>Find the z-score of 2:30 which is 30 minutes after 2:00.</span>
<span>
z(2:30) = (2:30 – 2:00)/15 = -30/15 = -2</span>
<span>
P(x < 2:30) = P(z<-2) = 0.0228</span>
<span>
</span>
Second, the probability of the mail will arrive at 1:36
PM
<span>Find the z-score of 1:36 which is 24 minutes before 2:00.</span>
<span>
z(1:36) = (1:36 – 2:00)/15 = -24/15 = -1.6</span>
<span>
P(x < 1:36) = P(z<-1.6) = 0.0548</span>
Lastly, the probability of the mail will arrive between 1:48
PM and 2:09 PM
Find the z-score of 1:46 and 2:09 PM which will result to
a z value of 0.034599
<span>P(1:48 < x < 2:09) = P(z<0.034599) = 0.5138</span>
GIVEN
the dimensions ofrectangular prism are
length = 54 in
breadth = 32.25in
height = 24.5 in
formula
volume of the rectangular prism = length×breadth×height
= 54 ×32.25×24.5
= 42997.5 in³
first statement is false because the volume of the rectangular prism in two significant digit.
Second statement is false because the volume of the rectangular prism in two significant digit.
FORMULA
surface area of a rectangular prism = 2(wl+hl+hw)
by putting the value of the length , breadth & height
we get
= 2 (54×32.25 + 32.25×24.5 + 24.5×54)
= 7709.25 in²
= 7709.3 in²(approx)
Third statement is true the surface area of the container is rounded to ones place.
Hence proved
