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bulgar [2K]
2 years ago
7

April wants to borrow $200 from her father and is willing to pay $9 in Interest. Her father wants to charge an interest rate of

6%. How long can April keep the money?
Mathematics
1 answer:
Lunna [17]2 years ago
7 0

Answer:

principle (p)= $200

interest (I) =$9

Rate (R) = 6%

time (T)= ?

we know,

Time(T) = I×100÷P×R

= 9×100÷ 200×6

= 3÷4 years

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3/6 + 1/2 what is the answer.​
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2 years ago
A market research firm supplies manufacturers with estimates of the retail sales of their products from samples of retail stores
EastWind [94]

Answer:

(52-49) -4.300= -1.300

(52-49) +4.300= 7.300

And the 95% confidence would be :

-1.300 \leq \mu_1 -\mu_2 \leq 7.300

Step-by-step explanation:

We have the following info given from the problem

\bar X_1 = 52 sample mean for this year

s_1= 13 sample deviation for this year

n_1 = 75 random sample selected for this year

\bar X_2 = 49 sample mean for last year

s_2= 11 sample deviation for last year

n_1 = 53 random sample selected for last year

And we want to construct a 95% confidence interval estimate of the difference μ1−μ2, where μ1 is the mean of this year's sales and μ2 is the mean of last year's sales

For this case the formula that we need to use is:

(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}

The degrees of freedom are given by:

df= n_1 +n_2 -2 = 75+53-2= 126

The confidence level is 0.95 and the significance would be \alpha=0.05 and \alpha/2 =0.025 so then the critical value for this case is :

t_{\alpha/2}= 1.979

The margin of error would be:

ME = 1.979 \sqrt{\frac{13^2}{75} +\frac{11^2}{49}}= 4.300

And the confidence interval would be given by:

(52-49) -4.300= -1.300

(52-49) +4.300= 7.300

And the 95% confidence would be :

-1.300 \leq \mu_1 -\mu_2 \leq 7.300

6 0
3 years ago
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