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11111nata11111 [884]
3 years ago
13

A vet borrows $12,000 for 4 years. The total interest paid was $875. What was the interest rate on the loan?

Mathematics
1 answer:
lina2011 [118]3 years ago
8 0
They paid $218.75 every year as an interest rate.
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Please help I’ll give brainliest
bekas [8.4K]

Answer:

Total area:75

Step-by-step explanation:

Rectangle area=6×5=30

triangle area=(18×5)/2=45

total area: 30+45=75

5 0
3 years ago
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Find the slope of the line using the ordered pairs (-1, -3/4) and (4, -3)
wel

Answer:

- 2.25/3

Step-by-step explanation:

5 0
3 years ago
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The seat of a bench is the shape of a Trapezoid with bases of 8 feet and 7 feet and a height on 3.5 feet whats the are of the se
Pepsi [2]

Answer:

26.25 square feet

Step-by-step explanation:

Calculation to determine the area of the seat

Using this formula

Area=1/2*(b1+b2)*h

Where,

b1=8 feet

b2=7 feet

h=3.5 feet

Let plug in the formula

Area=1/2*(8+7)*3.5

Area=1/2*15*3.5

Area=26.25

Therefore the area of the seat is 26.25 square feet

6 0
2 years ago
4 2/9 as a decimal work shown
Xelga [282]
The Answer Is:<span>4.222222</span>

4 2/9 = 4 + 2/9

= 4/1 + 2/9

= (4/1 * 9/9) + 2/9

= 36/9 + 2/9

= 38/9

= 38 ÷ 9 = 4.222222
4 0
3 years ago
A survey of 85 families showed that 36 owned at least one DVD player. Find the 99% confidence interval estimate of the true prop
Katyanochek1 [597]

Answer:   (0.367,\ 0.473)

Step-by-step explanation:

The confidence interval for population mean is given by :-

\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}

, where \hat{p} is the sample proportion, n is the sample size , z_{\alpha/2} is the critical z-value.

Given : Significance level : \alpha:1-0.99=0.01

Sample size : n= 85

Critical value : z_{\alpha/2}=2.576

Sample proportion: \hat{p}=\dfrac{36}{85}\approx0.42

Now, the  99% confidence level will be :

\hat{p}\pmz_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\=0.42\pm(2.576)\sqrt{\dfrac{0.42(1-0.42)}{85}}\\\\\approx0.42\pm0.053\\\\=(0.42-0.053,\ 0.42+0.053)=(0.367,\ 0.473)

Hence, the  99% confidence interval estimate of the true proportion of families who own at least one DVD player is  (0.367,\ 0.473)

3 0
3 years ago
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