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2 years ago

A bank pays $15 in interest for every $100 deposited. The bank pays how much interest?

Mathematics

1 answer:

Veronika [31]2 years ago

8 0

Answer:

15% interest

Step-by-step explanation:

because it's 15 out of 100 so that would be 15%

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A flashlight has 7 batteries, 5 of which are defective. If 5 are selected at random without replacement, find the probability th

TEA [102]

71.4 % chance I think is your answer.

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3 years ago

What is the constant of proportionality between y/x

Y_Kistochka [10]

Answer:

3.5

Step-by-step explanation:

To find the constant of proportionality between y/x you investigate the trend in values of x with corresponding values of y

For this case, when x =1, y=3.5

when x=2, y=7

Taking in case 1, y/x = 3.5/1 = 3.5

In case 2, y/x= 7/2 = 3.5

But you know that the constant of proportionality k is given by;

y=kx------------------------making k subject of formula

y/x=k

Hence in this case, your constant of proportionality is 3.5

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3 years ago

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Romashka [77]

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3 years ago

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How do you solve 2h + 7 > 21

Mrrafil [7]

$2h + 7 > 21\\
2h>14\\
h>7$

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3 years ago

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Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

3 0

3 years ago