Suppose that 76% of americans prefer coke to pepsi. A sample of 27 was taken. What is the probability that less than sixty perce
1 answer:
Answer:
0.0403
Step-by-step explanation:
Given that 76% of americans prefer coke to pepsi.
Let x be the number of people who prefers coke to pepsi.
X is binomial as each trial is independent, and there are only two outcomes.
A sample of 27 was taken.
We have to find the probability that less than sixty percent of the sample prefers coke to pepsi
60% of 27 =
Required probability =
Hence prob=0.0403
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<h2>Answer:</h2>
A.
Let Annie's weight be = a
Let Benjie's weighs = b
Let Carmen's weight be = c
One day Annie weighed 24 ounces more than Benjie, equation forms:
......(1)
Benjie weighed 3 1/4 pounds less than Carmen.
In ounces:
1 pound = 16 ounces
pounds =
ounces
or
......(2)
Now adding (1) and (2), we get
a+b=b+24+c-52
=>
This gives Annie weighs 28 ounces less than Carmen.
B.
We cannot know anyone's actual weight, as we only know their relative weights.