He has 9 bowls...and each bowl holds ,at most, 12 fish...so he can house (12 * 9) = 108 fish at most.
so... no, he does not have enough bowls ....he only needs one more bowl
Answer:
81.85% of the workers spend between 50 and 110 commuting to work
Step-by-step explanation:
We can assume that the distribution is Normal (or approximately Normal) because we know that it is symmetric and mound-shaped.
We call X the time spend from one worker; X has distribution N(μ = 70, σ = 20). In order to make computations, we take W, the standarization of X, whose distribution is N(0,1)

The values of the cummulative distribution function of the standard normal, which we denote
, are tabulated. You can find those values in the attached file.

Using the symmetry of the Normal density function, we have that
. Hece,

The probability for a worker to spend that time commuting is 0.8185. We conclude that 81.85% of the workers spend between 50 and 110 commuting to work.
1 yd is equivalent to 3 ft, so 7 times 3 is 21. 21 is greater than 20, so your answer is 7 yd.
Answer:
The answer to your question is (17 + 8x)/24
Step-by-step explanation:
Data
First distance = 3/8 km
second distance = (x + 1)/3 km
Process
1.- Sum both distances
3/8 + (x + 1)/3
- Find the Least common factor
8 3 2
4 3 2
2 3 2
1 3 3
1
LCF = 2x 2 x 2 x 3 = 24
- Sum the fractions
3/8 + (x + 1)/3 = (9 + 8(x + 1)) / 24
-Simplification
= (9 + 8x + 8) / 24
-Result
= (17 + 8x)/24