Answer:
Step-by-step explanation:
Given that the Acme Company manufactures widgets, which have a mean of 60 ounces and a standard deviation of 7 ounces
We know that 95% of the area lie between -2 and 2 std deviations from the mean.
i.e. Probability for lying in the middle of 95%
Z score 
Between 46 and 74 oz.
b) Between 12 and 57
convert into Z score
P(-6.86<z<-0.43)
=0.5-0.1664=0.3336
c) X<30 gives Z<-4.83
i.e. P(X<30) =0.00
To answer problems like this you have to use binomial:
P (x > 1) = 1 – p (0
< x < 1) > .7
So:
1 – p (0) – p (1) >
.7
1 – (3/ 4) ^n – (3/ 4)
^n (n – 1 ) (1/ 4) > .7
Therefore n > 5.185,
and the smallest value of n so that we can satisfy the given condition is 6
(rounded up)
<span> </span>
Defined as the average of a set of numbers
If I understand your question correctly, you want to know how many pounds of vegetables the zoo feeds the animals a month if its a 30 day month?
In that case it would be 23 pounds times 30 days which is 690 pounds of vegetables. SO the zoo feeds the animals 690 pounds of vegetables to the animals each month.
Is that what your asking for?
Answer:
240%
Step-by-step explanation:
240 / 100 Answer: 240%
