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3 years ago

If 100% of a number is 15, what is 50% of the number?

Mathematics

1 answer:

anzhelika [568]3 years ago

6 0

Answer:

Let n be the number

100% of a number is 15 is written as: 100% × n = 15

100% = 100 / 100 = 1

Hence

100% × n = n = 15

We now need to find 50% of the number

50% of 15 = (50 / 100) × 15 = 750 / 100 = 7.5

Step-by-step explanation:

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3 years ago

Randall has an account balance of $675.54 in a savings account that earns interest at a rate of 2.5% compounded twice a year. If

Alex_Xolod [135]

To calculate amount accrued after a given period of time we use the compound interest formula: A= P(1+r/100)∧n where A i the amount, P is the principal amount, r is the rate of interest and n is the interest period.
In the first part; A= $ 675.54, r= 1.25% (compounded semi-annually) and n =22 ( 11 years ), hence, 675.54 = P( 1.0125)∧22
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P= $ 514.109 , therefore the principal amount was $ 514 (to nearest dollar)
Part 2
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3 years ago

What value of x satisfies this equation, 4/5 x - 6 = -2 ? Group of answer choices

andrew11 [14]

Answer:

Step-by-step explanation:

4/5 x -6 = -2

4 x -30 = -10

-120 = -10

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2 years ago

How do you do this question?

Ksivusya [100]

Answer:

V = (About) 22.2, Graph = First graph/Graph in the attachment

Step-by-step explanation:

Remember that in all these cases, we have a specified method to use, the washer method, disk method, and the cylindrical shell method. Keep in mind that the washer and disk method are one in the same, but I feel that the disk method is better as it avoids splitting the integral into two, and rewriting the curves. Here we will go with the disk method.

$\mathrm{V\:=\:\pi \int _a^b\left(r\right)^2dy\:},\\\mathrm{V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy}$

The plus 1 in '1 + 2/x' is shifting this graph up from where it is rotating, but the negative 1 is subtracting the area between the y-axis and the shaded region, so that when it's flipped around, it becomes a washer.

$V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy,\\\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\=\pi \cdot \int _1^3\left(1+\frac{2}{y}\right)^2-1dy\\\\\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\= \pi \left(\int _1^3\left(1+\frac{2}{y}\right)^2dy-\int _1^31dy\right)\\\\$

$\int _1^3\left(1+\frac{2}{y}\right)^2dy=4\ln \left(3\right)+\frac{14}{3}, \int _1^31dy=2\\\\=> \pi \left(4\ln \left(3\right)+\frac{14}{3}-2\right)\\=> \pi \left(4\ln \left(3\right)+\frac{8}{3}\right)$

Our exact solution will be V = π(4In(3) + 8/3). In decimal form it will be about 22.2 however. Try both solution if you like, but it would be better to use 22.2. Your graph will just be a plot under the curve y = 2/x, the first graph.

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3 years ago