The ? Is 3
As 4^3 = 64
4x4x4= 64
Explanation:
Pair 1 is true if Jeff's monthly income is $600/20% = $3,000.
Pair 2 is true if Jeff's monthly income is $1200/10% = $12,000.
Both pairs can be true if Jeff's monthly income increased by a factor of 4 in the 20 years from 1990 to 2010.
Obviously, Jeff spent more on housing in 2010. (Fortunately for Jeff, that larger expenditure was a smaller fraction of his income.)
Your answer is incorrect. You forgot to get the square root of 25 and 4. Answer should be 16√2
we can only subtract radicals that are the same. At first glance, 4√50 - 2√8 are not the same, so they are not likely to be subtracted. However, each radical can still be simplified.
4 √50 = 4 √25 * 2 = 4 * 5 √2 = 20 √2
2 √8 = 2 √4 * 2 = 2 * 2 √2 = 4 √2
Now that the radicals are the same. then you can subtract the numbers.
20 √2 - 4 √2 = 16√2
Answer:
$6750 in the bank account and $20,250 in the stock fund
Step-by-step explanation:
If B is the money they put in the bank and S is the amount they put in the stock fund, then:
B + S = 27000
1.024 B + 1.072 S = 1.06 × 27000
Solving the system of equations:
1.024 (27000 − S) + 1.072 S = 28620
27648 − 1.024 S + 1.072 S = 28620
0.048 S = 972
S = 20250
B = 27000 − S
B = 6750
They should put $6750 in the bank account and $20,250 in the stock fund.
Answer:
Step-by-step explanation:
Given that a small manufacturing firm has 250 employees. Fifty have been employed for less than 5 years and 125 have been with the company for over 10 years. So remaining 75 are between 5 and 10 years.
Suppose that one employee is selected at random from a list of the employees
A) Probability that the selected employee has been with the firm less than 5 years = 
B) Probability that the selected employee has been with the firm between 5 and 10 years
= 
C) Probability that the selected employee has been with the firm more than 10 years
= 
a) P(A) = 0.2
P(C) = 0.5
P(A or B) = 0.2+0.3 = 0.5
P(A and C) = 0 (since A and C are disjoint)
