Answer:
1
Step-by-step explanation:
Answer:
2² • 89
The prime factors are: 2 x 2 x 89
or also written as { 2, 2, 89 }
Written in exponential form: 2² x 891
Step-by-step explanation:
56 ÷ 2 = 178 - No remainder! 2 is one of the factors!
178 ÷ 2 = 89 - No remainder! 2 is one of the factors!
89 ÷ 2 = 44.5 - There is a remainder. We can't divide by 2 evenly anymore. Let's try the next prime number
89 ÷ 3 = 29.6667 - This has a remainder. 3 is not a factor.
89 ÷ 5 = 17.8 - This has a remainder. 5 is not a factor.
89 ÷ 7 = 12.7143 - This has a remainder. 7 is not a factor.
Answer:
(7^9)/4 = 40,353,607/4
Step-by-step explanation:
Assuming each digit is used once and exponentiation is allowed, the largest numerator and smallest denominator will result in the largest fraction.
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If other functions, such as factorial are allowed, then there might need to be a limit on the number of times they are applied. For example,
(7!)^(9!)/4 has about 1 million digits
something like ...
((7!)^(9!))!/4 has many more digits than that
and you can keep piling on the factorial symbols to any desired depth.
It’s the first answer. The function is y = x*2
Step-by-step explanation:
The solution to this problem is very much similar to your previous ones, already answered by Sqdancefan.
Given:
mean, mu = 3550 lbs (hope I read the first five correctly, and it's not a six)
standard deviation, sigma = 870 lbs
weights are normally distributed, and assume large samples.
Probability to be estimated between W1=2800 and W2=4500 lbs.
Solution:
We calculate Z-scores for each of the limits in order to estimate probabilities from tables.
For W1 (lower limit),
Z1=(W1-mu)/sigma = (2800 - 3550)/870 = -.862069
From tables, P(Z<Z1) = 0.194325
For W2 (upper limit):
Z2=(W2-mu)/sigma = (4500-3550)/879 = 1.091954
From tables, P(Z<Z2) = 0.862573
Therefore probability that weight is between W1 and W2 is
P( W1 < W < W2 )
= P(Z1 < Z < Z2)
= P(Z<Z2) - P(Z<Z1)
= 0.862573 - 0.194325
= 0.668248
= 0.67 (to the hundredth)
