Answer:
Step-by-step explanation:
jjjjjjjjjjjjj
<span>I = 1832 </span>
<span>P = 16000 </span>
<span>T = 206/365 years </span>
<span>Rate of interest = [100*I] / [P*T] </span>
<span>= [100*1832] / [16000*206/365] </span>
<span>= 20.29% per annum</span>
Answer:
The answer to your question is: letter A
Step-by-step explanation:
From the graph we get the points,
P (2,1)
Q (6,8)
Formula
d = √((x2-x1)² + (y2-y1)²)
d = √((6-2)² + (8-1)²)
d = √ (4² + 7²)
d = √ (16 + 49
d = √65 letter A
Answer:
6x+2-1x+5=42
Step-by-step explanation:
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
