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3 years ago

Find the simply interest earned on an investment of $15,000 at a rate of 5.25% for two years.

2 answers:

4 0

Ok so you know that simple interest formula is
Simple interest= money x percent (expressed as a decimal) x time (years)
So 15,000(0.0525)2
Work from there :)

uysha [10]3 years ago

3 0

Ill show you step by step.

This is what the equation should look like

Y=Initial value(1+/- Increasing or Decreasing value)^2 Which stands for time

$So Y=15,000(1+.0525)^2$

So you need to do 1.0525 to the 2nd power Which equals 1.10775625

So now you need to do 15,000(1.10775625)
------------------------------------------------------------------------------------
Which equals <span>16616 rounding to the nearest Ones

16616.3 rounding to the nearest tenth</span>

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4(x-2+y) what is this for the distributive property

DochEvi [55]

Answer:

4x + 4y - 8

Step-by-step explanation:

4(x - 2 +y) = 4*x -4*2 + 4*y by the distributive property

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3 years ago

What is 9 times 9. (my calculator Broke)

boyakko [2]

Answer:81

Step-by-step explanation: 9*1= 9 9*2= 18 9*3 = 27 9*4= 36 9*5=45 9*6 = 54 9*7=63 9*8= 72 9*9 81 9*10= 90

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3 years ago

Read 2 more answers

If the ratio of red hair bands to green hair bands is 5 to 9 with a total of 70, how many are green

Eduardwww [97]

$5x-\ red\\9x-\ green\\\\5x+9x=70\\
14x=70\ \ \ | divide\ by\ 14\\\\
x=5\\\\
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5 0

3 years ago

The number of defective circuit boards coming off a soldering machine follows a Poisson distribution. During a specific ten-hour

Alexus [3.1K]

Answer:

a) the probability that the defective board was produced during the first hour of operation is $\frac{1}{10}$ or 0.1000

b) the probability that the defective board was produced during the last hour of operation is $\frac{1}{10}$ or 0.1000

c) the required probability is 0.2000

Step-by-step explanation:

Given the data in the question;

During a specific ten-hour period, one defective circuit board was found.

Lets X represent the number of defective circuit boards coming out of the machine , following Poisson distribution on a particular 10-hours workday which one defective board was found.

Also let Y represent the event of producing one defective circuit board, Y is uniformly distributed over ( 0, 10 ) intervals.

f(y) = $\left \{ {{\frac{1}{b-a} }\\\ }} \right _0$ ; ( a ≤ y ≤ b ) $_{elsewhere$

= $\left \{ {{\frac{1}{10-0} }\\\ }} \right _0$ ; ( 0 ≤ y ≤ 10 ) $_{elsewhere$

f(y) = $\left \{ {{\frac{1}{10} }\\\ }} \right _0$ ; ( 0 ≤ y ≤ 10 ) $_{elsewhere$

Now,

a) the probability that it was produced during the first hour of operation during that period;

P( Y < 1 ) = $\int\limits^1_0 {f(y)} \, dy$

we substitute

= $\int\limits^1_0 {\frac{1}{10} } \, dy$

= $\frac{1}{10} [y]^1_0$

= $\frac{1}{10} [ 1 - 0 ]$

= $\frac{1}{10}$ or 0.1000

Therefore, the probability that the defective board was produced during the first hour of operation is $\frac{1}{10}$ or 0.1000

b) The probability that it was produced during the last hour of operation during that period.

P( Y > 9 ) = $\int\limits^{10}_9 {f(y)} \, dy$

we substitute

= $\int\limits^{10}_9 {\frac{1}{10} } \, dy$

= $\frac{1}{10} [y]^{10}_9$

= $\frac{1}{10} [ 10 - 9 ]$

= $\frac{1}{10}$ or 0.1000

Therefore, the probability that the defective board was produced during the last hour of operation is $\frac{1}{10}$ or 0.1000

no defective circuit boards were produced during the first five hours of operation.

probability that the defective board was manufactured during the sixth hour will be;

P( 5 < Y < 6 | Y > 5 ) = P[ ( 5 < Y < 6 ) ∩ ( Y > 5 ) ] / P( Y > 5 )

= P( 5 < Y < 6 ) / P( Y > 5 )

we substitute

$= (\int\limits^{6}_5 {\frac{1}{10} } \, dy) / (\int\limits^{10}_5 {\frac{1}{10} } \, dy)$

$= (\frac{1}{10} [y]^{6}_5) / (\frac{1}{10} [y]^{10}_5)$

= ( 6-5 ) / ( 10 - 5 )

= 0.2000

Therefore, the required probability is 0.2000

4 0

3 years ago

Find the area of the following shape. Show all work

Pavel [41]

Best way to solve this is by using

$\sqrt{s(s - a)(s - b)(s - c)}$

$where \: s = \frac{a + b + c}{2}$

s=(12+8+17)/2

=18.5

using the formulae

area =43.5

4 0

3 years ago