.98 and98% .................
24
8
31
17
absolute value with positive numbers are always positive if I remember correctly.
oooo and the opposite
24
-8
-31
and 17
To solve this we are going to use the future value of annuity ordinary formula:
![FV=P[ \frac{(1+ \frac{r}{n} )^{kt} -1}{ \frac{r}{n} } ]](https://tex.z-dn.net/?f=FV%3DP%5B%20%5Cfrac%7B%281%2B%20%5Cfrac%7Br%7D%7Bn%7D%20%29%5E%7Bkt%7D%20-1%7D%7B%20%5Cfrac%7Br%7D%7Bn%7D%20%7D%20%5D)
where

is the future value

is the periodic payment

is the interest rate in decimal form

is the number of times the interest is compounded per year

is the number of payments per year

is the number of years
We know for our problem that

and

. To convert the interest rate to decimal form, we are going to divide the rate by 100%:

Since the deposit is made semiannually, it is made 2 times per year, so

.
Since the type of the annuity is ordinary, payments are made at the end of each period, and we know that we have 2 periods, so

.
Lets replace the values in our formula:
![FV=P[ \frac{(1+ \frac{r}{n} )^{kt} -1}{ \frac{r}{n} } ]](https://tex.z-dn.net/?f=FV%3DP%5B%20%5Cfrac%7B%281%2B%20%5Cfrac%7Br%7D%7Bn%7D%20%29%5E%7Bkt%7D%20-1%7D%7B%20%5Cfrac%7Br%7D%7Bn%7D%20%7D%20%5D)
![FV=6200[ \frac{(1+ \frac{0.06}{2} )^{(2)(5)} -1}{ \frac{0.06}{2} } ]](https://tex.z-dn.net/?f=FV%3D6200%5B%20%5Cfrac%7B%281%2B%20%5Cfrac%7B0.06%7D%7B2%7D%20%29%5E%7B%282%29%285%29%7D%20-1%7D%7B%20%5Cfrac%7B0.06%7D%7B2%7D%20%7D%20%5D)
We can conclude that the correct answer is <span>
$71,076.06</span>
Draw two lines that connect at one point. label where they connect w and the other two ends j
Answer:
There is significant evidence that treatment will create effect on the subject. Hence We reject H0
Step-by-step explanation:
H0 : μ = 37
H1 : μ ≠ 37
μ = 37 ; σ = 7 ; sample size, x = 51 ; sample size, n = 1
Decision region :
If P value < α ;
Reject H0
Using the Z test statistic :
Zstatistic = (x - μ) ÷ (σ / √n)
Zstatistic = (51 - 37) ÷ (7 / 1)
Zstatistic = 14 ÷ 7
Zstatistic = 2
Obtaining p value from Zstatistic using the p value calculator ;
Zscore = 2 ; 2-tailed test, significance level = 0.05
P value = 0.0455
Zcritical at α = 0.05 for a 2 - tailed test = 1.96
0.0455 < 0.05
There is significant evidence that treatment will create effect on the subject. Hence We reject H0