The solution to the problem is as follows:
x = log (25) (125)
125 = 25^x
log 125 = x log 25
<span>
x = 3/2
</span>
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Answer: He will have $1974 in April.
Step-by-step explanation:
Given : Total money he earned = $2200
He spent $50 per month gym membership, a $68 monthly phone plan $48 monthly gas plan and $60 monthly fun plan.
So , Money he has left = Money he earned - Total expenditure
= $ ( 2200- (50+68+48+60))
= $ (2200- 226)
= $ 1974
Hence, He will have $1974 in April.
Complete question:
Assume that different groups of couples use a particular method of gender selection and each couple gives birth to one baby this mention is designed to increase the likelihood that each baby will be a girl, but assume that the method has no effect so the probability of a girl is 0.5. Assume that the group consists of 36 couples.
A) Find the mean and standard deviation for the number of girls in groups of 36 births.
B) Use the range rule of thumb to find the values separating results that are significantly low and significantly high.
C) Is the result of 33 girls significantly high? A result of 33 girls would suggest the method is effective or is not effective?
Answer:
a) mean = 18
Standard deviation =3
b) low range = 12
High range = 24
c) The result of 33 girls is significantly high. Yes, the method is effective.
Step-by-step explanation:
Given:
p = 0.5
n = 36
a) The mean is the product of n and p
Mean u = np
u = 36 * 0.5 = 18
The standard deviation is the square root of the product of n and p&q.
S.d ó = 


b) To find the range rule of thumb:
• For low range
Low range = u - 2ó
= 18 - (2 * 3)
= 12
• High range
= u + 2ó
= 18 + (2*3)
= 24
c) The result is significantly high, because 33 is greater than 24 girls.
A result of 33 girls would prove the method as effective.
28 divided by 3 is 9.33333333
3 divided by 28 is 0.10714286
I might be wrong but I think it's 504
I got this by multiplying 28 and 18 hope it helped : )