Let
x-----------> <span>this week's allowance
we know that
(3/4)x-----------> </span><span>was spent on candy
and
(5/6)[</span>(3/4)x]-------> <span>was spent on gummy bears
(5/6)*(3/4)x=(15/24)x------> (5/8)x
the answer is
5/8</span>
Answer:
The probability that a randomly selected chocolate bar will have a. Between 200 and 220 calories will be 0.3023
Step-by-step explanation:
Given:
True mean=225
S.D=10
X1=200 and X2=220
To Find:
P(X1<x<X2)
Solution:
<em>BY using Z-table for probability and Z-score we can proceed.</em>
So
For 200 calories Z will be ,
Z=(sample mean -true mean)/S.D
=200-225/10
=-25/10
=-2.5
For 220 calories Z will be ,
Z=220-225/10
=-5/10
=-0.5
So required probability will be




=0.3023
Answer:
I think the answer is go die and pls don’t forget to like this post
Answer:
Can you reformat?
So far, 0 is a solution to none of these.
Answer: 5
Step-by-step explanation:
If you do -8-3 you’d get 5