2 years ago

A deposit of 8000 at 9.5% for 120 days

Mathematics

1 answer:

brilliants [131]2 years ago

5 0

So here is what I got for you

I=prt
I=8,000×0.095×(120÷360)
I=253.33

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Please help! I attached the question below.

kompoz [17]

Answer:

$\frac{2(c+2)}{c(c-2)}$

Step-by-step explanation:

$\frac{c^{2}-4 }{6c^{4}+15c^{3}}=\frac{(c-2)(c+2)}{c(6c^{3}+15c^{2}) }$

Identity used:

$a^{2}-b^{2}=(a-b)(a+b)$

$\frac{c^{2}-4c+4}{12c^{3}+30c^{2}}=\frac{(c-2)^{2}}{2(6c^{3}+15c^{2}) }$

Now let us divide the modified expressions:

$\frac{(c-2)(c+2)}{c(6c^{3}+15c^{2})}$ ÷ $\frac{(c-2)^2}{2(6c^{3}+15c^{2}) }$

we get:

$\frac{2(c+2)}{c(c-2)}$

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2 years ago

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notka56 [123]

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2 years ago

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nydimaria [60]

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2 years ago

Read 2 more answers

Which is the correct algebraic representation for the area of a circle?

fiasKO [112]

B because pie radius square = area of a circle

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2 years ago

The time to complete a standardized exam in the BYU-Idaho Testing Center is approximately normal with a mean of 70 minutes and a

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Answer:

Step-by-step explanation:

Given that X the time to complete a standardized exam in the BYU-Idaho Testing Center is approximately normal with a mean of 70 minutes and a standard deviation of 10 minutes.

We have 68 rule as 2/3 of total would lie within 1 std deviation, and 95 rule as nearly 95% lie within 2 std deviations from the mean.

We have std deviation = 10

Hence 2 std deviations from the mean

= Mean ±2 std deviations

= $70$ ±20

= $(50,90)$

Below 50, 0.25 or 2.5% would complete the exam.

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2 years ago