Answer with explanation:
Mean of the sample(m) = $ 5474
Standard deviation of the sample (S)=764
Number of observation(n)=36
So, Mean Monthly Expenses of Population =$ 5405.76, which is 90% upper confidence bound for the company's mean monthly expenses.
Answer:
A
Step-by-step explanation:
Time Distance
0 0
2 2*25 = 50
4 4*25 = 100
6 6*25= 150
8 8 * 25 = 200
Answer:
x as a fraction: 11
Step-by-step explanation:
Variable x cannot be equal to −2 since division by zero is not defined. Multiply both sides of the equation by 9(x+2), the least common multiple of 9,x+2.
2 ( x+ 2) = 9 x 3
Use the distributive property to multiply 2 by x + 2.
2x + 4 = 9 x 3
Multiply 9 and 3 to get 27.
2x + 4 = 27
Subtract 4 from both sides.
2x = 27 - 4
Subtract 4 from 27 to get 23.
2x = 23
Divide both sides by 2.
x =
Simplify
11
Hope it helps and have a great day! =D
~sunshine~
Answer:
1 / 24
Step-by-step explanation:
= > 3/8 - 2/6
=> {(3 x 3)/(8 x 3)} - {(2x4)/(6x4)}
=> 9/24 - 8/24
=> ( 9 - 8 ) / 24
= 1 / 24 ANS
Answer:
the dimension of the poster = 90 cm length and 60 cm width i.e 90 cm by 60 cm.
Step-by-step explanation:
From the given question.
Let p be the length of the of the printed material
Let q be the width of the of the printed material
Therefore pq = 2400 cm ²
q =
To find the dimensions of the poster; we have:
the length of the poster to be p+30 and the width to be
The area of the printed material can now be:
=
Let differentiate with respect to p; we have
Also;
For the smallest area
p² = 3600
p =√3600
p = 60
Since p = 60 ; replace p = 60 in the expression q = to solve for q;
q =
q =
q = 40
Thus; the printed material has the length of 60 cm and the width of 40cm
the length of the poster = p+30 = 60 +30 = 90 cm
the width of the poster = = = 40 + 20 = 60
Hence; the dimension of the poster = 90 cm length and 60 cm width i.e 90 cm by 60 cm.