Answer:
Step-by-step explanation:
Since the number of pages that this new toner can print is normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = the number of pages.
µ = mean
σ = standard deviation
From the information given,
µ = 2300 pages
σ = 150 pages
1)
the probability that this toner can print more than 2100 pages is expressed as
P(x > 2100) = 1 - P(x ≤ 2100)
For x = 2100,
z = (2100 - 2300)/150 = - 1.33
Looking at the normal distribution table, the probability corresponding to the z score is 0.092
P(x > 2100) = 1 - 0.092 = 0.908
2) P(x < 2200)
z = (x - µ)/σ/√n
n = 10
z = (2200 - 2300)/150/√10
z = - 100/47.43 = - 2.12
Looking at the normal distribution table, the probability corresponding to the z score is 0.017
P(x < 2200) = 0.017
3) for underperforming toners, the z score corresponding to the probability value of 3%(0.03) is
- 1.88
Therefore,
- 1.88 = (x - 2300)/150
150 × - 1.88 = x - 2300
- 288 = x - 2300
x = - 288 + 2300
x = 2018
The threshold should be
x < 2018 pages
The volume of cube is 1000 cubic feet
<em><u>Solution:</u></em>
Given that,
<em><u>The volume of cube is given by formula:</u></em>
Where,
V is the volume of cube
"s" is the length of one side
<em><u>What is the volume of a cube if the length of a side s is 10 feet</u></em>
Given, s = 10 feet
<em><u>Substitute s = 10 in given formula,</u></em>
Thus volume of cube is 1000 cubic feet
Answer:
1. 10 in / 3 ft
2. 5 min / 1 h
3. 1 kg / 250 g
Step-by-step explanation:
Answers:
CB = 14
GF = 8
FB = 9
EF is parallel to CB
====================================
Explanations:
Points E and F are midpoints of their respective sides. They form the midsegment EF. Because EF is a midsegment, A midsegment is half the length of its parallel counterpart, so CB is two times longer than EF. If EF is 7 units long, then CB = 2*EF = 2*7 = 14
For similar reasons, GF is parallel to AC. If AC = 16, then half of that is GF = (1/2)*AC = 0.5*16 = 8.
FB = FA = 9 as these segments have the same single tickmark to indicate they are the same length
EF is parallel to CB because EF is a midsegment, and this is one of the properties of being a midsegment. We can show that quadrilateral EGBF is a parallelogram to help prove this.
