Answer:
The amount in the account in the beginning of the 6th year is $6375 .
Step-by-step explanation:
Formula

As given
You invest $5000 in an account at 5.5% per year simple interest.
Principle = $ 5000
Rate = 5.5 %
Time = 5 years
(As calculate amount in the account for beginning of 6th year.)
Putting all the values in the formula


Simple interest = $ 1375
Amount = Principle + Simple interest
Putting values in the above formula
Amount = $5000 + $1375
Amount = $ 6375
Therefore the amount in the account in the beginning of the 6th year is $ 6375 .
Answer:
Step-by-step explanation:
![\sqrt[3]{125y^9z^6}\\ \\ \sqrt[3]{5^3(y^3)^3(z^2)^3}\\ \\ 5y^3z^2](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B125y%5E9z%5E6%7D%5C%5C%20%5C%5C%20%5Csqrt%5B3%5D%7B5%5E3%28y%5E3%29%5E3%28z%5E2%29%5E3%7D%5C%5C%20%5C%5C%205y%5E3z%5E2)
Answer:
Yes, we reject the auto maker's claim.
Step-by-step explanation:
H0 : μ ≥ 20
H1 : μ < 20
Sample mean, xbar = 18 ;
Sample size, n = 36
Standard deviation, s = 5
At α = 0.01
The test statistic :
(xbar - μ) ÷ s /sqrt(n)
(18 - 20) ÷ 5/sqrt(36)
-2 /0.8333333
= - 2.4
Pvalue from test statistic : Pvalue = 0.00819
Pvalue < α
0.00819 < 0.01
Hence, we reject the Null
9 I think, soo sorry if i'm wrong...