The amount to be invested today so as to have $12,500 in 12 years is $6,480.37.
The amount that would be in my account in 13 years is $44,707.37.
The amount I need to deposit now is $546.64.
<h3>How much should be invested today?</h3>
The amount to be invested today = future value / (1 + r)^nm
Where:
- r = interest rate = 5.5 / 365 = 0.015%
- m = number of compounding = 365
- n = number of years = 12
12500 / (1.00015)^(12 x 365) = $6,480.37
<h3>What is the future value of the account at the end of 13 years?</h3>
Future value = monthly deposits x annuity factor
Annuity factor = {[(1+r)^n] - 1} / r
Where:
- r = interest rate = 5.3 / 12 = 0.44%
- n = 13 x 12 = 156
200 x [{(1.0044^156) - 1} / 0.0044] = $44,707.37
<h3>What should be the monthly deposit?</h3>
Monthly deposit = future value / annuity factor
Annuity factor = {[(1+r)^n] - 1} / r
Where:
- r = 6.7 / 12 = 0.56%
- n = 2 x 12 = 24
$14,000 / [{(1.0056^24) - 1} / 0.0056] = $546.64
To learn more about annuities, please check: brainly.com/question/24108530
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An outlier causes the range of the population to move slightly towards the outlier.
Answer:
Null Hypothesis: H_0: \mu_A =\mu _B or \mu_A -\mu _B=0
Alternate Hypothesis: H_1: \mu_A >\mu _B or \mu_A -\mu _B>0
Here to test Fertilizer A height is greater than Fertilizer B
Two Sample T Test:
t=\frac{X_1-X_2}{\sqrt{S_p^2(1/n_1+1/n_2)}}
Where S_p^2=\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}
S_p^2=\frac{(14)0.25^2+(12)0.2^2}{15+13-2}= 0.0521154
t=\frac{12.92-12.63}{\sqrt{0.0521154(1/15+1/13)}}= 3.3524
P value for Test Statistic of P(3.3524,26) = 0.0012
df = n1+n2-2 = 26
Critical value of P : t_{0.025,26}=2.05553
We can conclude that Test statistic is significant. Sufficient evidence to prove that we can Reject Null hypothesis and can say Fertilizer A is greater than Fertilizer B.
Answer:
5
Step-by-step explanation:
