Answer: No, they would not get the same amount.
Step-by-step explanation:
Let the number of coins be x
Eliot counts a group of coins staring with the quarters.
As we know that
So, Number of amount Eliot counts with the quarters is given by
Similarly, His sister counts the same coins she counting the pennies .
As we know that
So, Total amount his sister counts with the pennies is given by
But we can see that
Hence, No, they would not get the same amount.
Answer:
Population standard deviation, = 3683.063 .
Step-by-step explanation:
We are given that the width of the estimated confidence interval i.e. 99% is 600 and the sample size used in estimating the mean is 1000 which means ; n = 1000 and width = 600
We know that Width of confidence interval = 2 * Margin of error
<em> Margin of error</em><em> </em>= = 2.5758 *
{because at 1% significance level
z table has value of 2.5758 .}
Therefore, 600 = 2 * 2.5758 *
⇒ =
= 3683.063 .
Hence, the Population standard deviation = 3683.063 .
Answer: a) 83, b) 28, c) 14, d) 28.
Step-by-step explanation:
Since we have given that
n(B) = 69
n(Br)=90
n(C)=59
n(B∩Br)=28
n(B∩C)=20
n(Br∩C)=24
n(B∩Br∩C)=10
a) How many of the 269 college students do not like any of these three vegetables?
n(B∪Br∪C)=n(B)+n(Br)+n(C)-n(B∩Br)-n(B∩C)-n(Br∩C)+n(B∩Br∩C)
n(B∪Br∪C)=
So, n(B∪Br∪C)'=269-n(B∪Br∪C)=269-156=83
b) How many like broccoli only?
n(only Br)=n(Br) -(n(B∩Br)+n(Br∩C)+n(B∩Br∩C))
n(only Br)=
c) How many like broccoli AND cauliflower but not Brussels sprouts?
n(Br∩C-B)=n(Br∩C)-n(B∩Br∩C)
n(Br∩C-B)=
d) How many like neither Brussels sprouts nor cauliflower?
n(B'∪C')=n(only Br)= 28
Hence, a) 83, b) 28, c) 14, d) 28.