Answer:
0.36878
Step-by-step explanation:
Given that:
Mean number of miles (m) = 2135 miles
Variance = 145924
Sample size (n) = 40
Standard deviation (s) = √variance = √145924 = 382
probability that the mean of a sample of 40 cars would differ from the population mean by less than 29 miles
P( 2135 - 29 < z < 2135 + 29)
Z = (x - m) / s /√n
Z = [(2106 - 2135) / 382 / √40] < z < [(2164 - 2135) / 382 / √40]
Z = (- 29 / 60.399503) < z < (29 / 60.399503)
Z = - 0.4801364 < z < 0.4801364
P(Z < - 0.48) = 0.31561
P(Z < - 0.48) = 0.68439
P(- 0.480 < z < 0.480) = 0.68439 - 0.31561 = 0.36878
= 0.36878
Assuming that the two investments are X & Y
X + Y = 6300
X = 6300 - Y (1)
9/100X + 4/100 Y = 372 (2)
replacing X from (1) into (2)
9/100(6300-Y) + 4/100 Y = 372
567 - 9/100Y +4/100Y = 372
(-9+4)/100Y = 372 - 567
5/100Y = -195
Y = 100*195/5 = 3900
From (1) we can get X
X = 6300 - 3900 = 2400
I hope this is helpful
Answer:
There isn't a specific answer in my knowledge but the answer that's matches with most of them is 149
Answer:
all
Step-by-step explanation:
have the awnserbbbnnnnnnnnnnnn
Answer:
expression: (30/t) - 10
evaluation: 30/2 - 10 = 5
Step-by-step explanation:
