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1 answer:
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Emily earned $166 more in her account than Katie.
We use the compound interest formula for both of these:
For Katie's deposit:
This gives Katie 6083.26-5000 = 1083.26 in interest (1083, to the nearest dollar).
For Emily's deposit:
This means she earned 11248.64-10000=1248.64 in interest (1249, to the nearest dollar).
The difference in interest is given by 1249-1083=166
We use the compound interest formula for both of these:
For Katie's deposit:
This gives Katie 6083.26-5000 = 1083.26 in interest (1083, to the nearest dollar).
For Emily's deposit:
This means she earned 11248.64-10000=1248.64 in interest (1249, to the nearest dollar).
The difference in interest is given by 1249-1083=166
Answer:
There will be 9897.47€ in the account.
La cantidad será de 9897.47€.
Step-by-step explanation:
Future value with annuity:
The future value formula, for an annuity, is:
An annuity means that a number of payments happen during the period(an year, for example.
P is the value of the deposit, r is the interest rate, as a decimal, and n is the number of deposits.
In this question:
1000 € each year, so
Interest rate of 6%, so
One application per year for 8 years, so
Total amount:
There will be 9897.47€ in the account.
La cantidad será de 9897.47€.
Answer:
Step-by-step explanation:
Hello!
The objective is to test if the two machines are filling the bottles with a net volume of 16.0 ounces or if at least one of them is different.
The parameter of interest is μ₁ - μ₂, if both machines are filling the same net volume this difference will be zero, if not it will be different.
You have one sample of ten bottles filled by each machine and the net volume of the bottles are measured, determining two variables of interest:
Sample 1 - Machine 1
X₁: Net volume of a plastic bottle filled by the machine 1.
n₁= 10
X[bar]₁= 16.02
S₁= 0.03
Sample 2 - Machine 2
X₂: Net volume of a plastic bottle filled by the machine 2
n₂= 10
X[bar]₂= 16.01
S₂= 0.03
With p-values 0.4209 (for X₁) and 0.6174 (for X₂) for the normality test and using α: 0.05, we can say that both variables of interest have a normal distribution:
X₁~N(μ₁;δ₁²)
X₂~N(μ₂;δ₂²)
Both population variances are unknown you have to conduct a homogeneity of variances test to see if they are equal (if they are you can conduct a pooled t-test) or different (if the variances are different you have to use the Welche's t-test)
H₀: δ₁²/δ₂²=1
H₁: δ₁²/δ₂²≠1
α: 0.05
Using a statistic software I've calculated the test
p-value 0.6168
The p-value is greater than the significance level, so the decision is to not reject the null hypothesis then you can conclude that both population variances for the net volume filled in the plastic bottles by machines 1 and 2 are equal.
To study the difference between the population means you can use
The hypotheses of interest are:
H₀: μ₁ - μ₂ = 0
H₁: μ₁ - μ₂ ≠ 0
α: 0.05
The p-value for this test is: 0.882433
The p-value is greater than the significance level, the decision is to not reject the null hypothesis.
Using a significance level of 5%, there is no significant evidence to reject the null hypothesis. You can conclude that the population average of the net volume of the plastic bottles filled by machine one and by machine 2 are equal.
I hope this helps!
