Let's say the number of days library book is late is X, and the total fee is Y.
Liability charges $0.30 dollars as a fee for being 1 day late,
For being 1 day late, fee charge is: 1* $0.30
So, for X days the charge would be: X*$0.30.
Total charge for being X days late is Y, Which means: Y= 0.30 * X.
Now We would have to check all the viable solutions in the answer to see if they satisfy the equation Y= 0.30 * X
Option
one(-3, -0.9) and two (-2.5, -0.75) Would not be a viable solution
because the value of number of days can not be negative and in option
one and two, value of days -3 and -2.5 is negative.
Option
three(4.5, 1.35) can not be correct because library charges fee for a
full day so the number for days would be a whole number. Library would
not charge for 4.5 days, they would either charge of 4 days or 5 days
because 4.5 is not an whole number.
Option four(8, 2.40) is the correct answer because it satisfies our equation;
Y= 0.30 * X
2.40= 0.30 * 8
2.40 = 2.40.
Fourth option (8, 2.40) is the only viable solution to this question.
First we need to calculate annual withdrawal of each investment
The formula of the present value of an annuity ordinary is
Pv=pmt [(1-(1+r)^(-n))÷(r)]
Pv present value 28000
PMT annual withdrawal. ?
R interest rate
N time in years
Solve the formula for PMT
PMT=pv÷[(1-(1+r)^(-n))÷(r)]
Now solve for the first investment
PMT=28,000÷((1−(1+0.058)^(−4))
÷(0.058))=8,043.59
The return of this investment is
8,043.59×4years=32,174.36
Solve for the second investment
PMT=28,000÷((1−(1+0.07083)^(
−3))÷(0.07083))=10,685.63
The return of this investment is
10,685.63×3years=32,056.89
So from the return of the first investment and the second investment as you can see the first offer is the yield the highest return with the amount of 32,174.36
Answer d
Hope it helps!
Answer:
a. 
b. 6.1 c. 0.6842 d. 0.4166 e. 0.1194 f. 8.5349
Step-by-step explanation:
a. The distribution of X is normal with mean 6.1 kg. and standard deviation 1.9 kg. this because X is the weight of a randomly selected seedless watermelon and we know that the set of weights of seedless watermelons is normally distributed.
b. Because for the normal distribution the mean and the median are the same, we have that the median seedless watermelong weight is 6.1 kg.
c. The z-score for a seedless watermelon weighing 7.4 kg is (7.4-6.1)/1.9 = 0.6842
d. The z-score for 6.5 kg is (6.5-6.1)/1.9 = 0.2105, and the probability we are seeking is P(Z > 0.2105) = 0.4166
e. The z-score related to 6.4 kg is 
and the z-score related to 7 kg is 
, we are seeking P(0.1579 < Z < 0.4737) = P(Z < 0.4737) - P(Z < 0.1579) = 0.6821 - 0.5627 = 0.1194
f. The 90th percentile for the standard normal distribution is 1.2815, therefore, the 90th percentile for the given distribution is 6.1 + (1.2815)(1.9) = 8.5349
Answer:
a) 15 house with 3 bedrooms
b) 40 ÷ 5 = 8
Step-by-step explanation:
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