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wolverine [178]
3 years ago
5

Edmund purchased a computer for $1,800 on a payment plan (month zero). Three months after he purchased the computer, his balance

was $1,350. Five months after he purchased the computer, his balance was $1,050. Create a table from this information and find out his balance for two other months.​
Mathematics
1 answer:
trapecia [35]3 years ago
6 0

Answer:

Well I can't make a table but I can explain the answer.

Step-by-step explanation:

So you have $1,800 on a payment plan.  In three months, the balance is 1,350.  So 1,800-1,350=450.  This means he paid $450 in three months.  450/3 is 150.  He pays $150 a month.  Then 2 months later, (5 months after purchase) he has balance is $1,050.  1,350-1,050=300.  300/2=150.  Further evidence of the fact that he pays 150 dollars a month.  This means that the 6th month, he will have $900 dollars left on the payment.  7th month... $850, 8th month... $600, 9th month... $450, 10th month... 300 dollars, 11th month... 150, and 12 month he will have paid off all of his computer.  Hope it helps, sorry it is so long.  Have a great day!  :D

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The following information was obtained from independent random samples taken of two populations. Assume normally distributed pop
Volgvan

Answer:

1. The 95% confidence interval for the difference between means is (-5.34, 11.34).

2. The standard error of (x-bar1)-(x-bar2) is 4.

s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{9.2195^2}{10}+\dfrac{9.4868^2}{12}}\\\\\\s_{M_d}=\sqrt{8.5+7.5}=\sqrt{16}=4

Step-by-step explanation:

We have to calculate a 95% confidence interval for the difference between means.

The sample 1, of size n1=10 has a mean of 45 and a standard deviation of √85=9.2195.

The sample 2, of size n2=12 has a mean of 42 and a standard deviation of √90=9.4868.

The difference between sample means is Md=3.

M_d=M_1-M_2=45-42=3

The estimated standard error of the difference between means is computed using the formula:

s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{9.2195^2}{10}+\dfrac{9.4868^2}{12}}\\\\\\s_{M_d}=\sqrt{8.5+7.5}=\sqrt{16}=4

The critical t-value for a 95% confidence interval is t=2.086.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_{M_d}=2.086 \cdot 4=8.34

Then, the lower and upper bounds of the confidence interval are:

LL=M_d-t \cdot s_{M_d} = 3-8.34=-5.34\\\\UL=M_d+t \cdot s_{M_d} = 3+8.34=11.34

The 95% confidence interval for the difference between means is (-5.34, 11.34).

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4 years ago
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