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alex41 [277]
2 years ago
13

What number : Increased by 130% is 69 ?

Mathematics
1 answer:
e-lub [12.9K]2 years ago
4 0
<h2>Answer:<u> 158.7 is the answer</u></h2><h2><u>or it can be</u> <u>89.7</u></h2>

Step-by-step explanation:

<h2><u>69 + Percentage increase = </u></h2><h2><u>69 + (130% × 69) = </u></h2><h2><u>69 + 130% × 69 = </u></h2><h2><u>(1 + 130%) × 69 = </u></h2><h2><u>(100% + 130%) × 69 = </u></h2><h2><u>230% × 69 = </u></h2><h2><u>230 ÷ 100 × 69 = </u></h2><h2><u>230 × 69 ÷ 100 = </u></h2><h2><u>15,870 ÷ 100 = </u></h2><h2><u>158.7</u></h2><h2><u /></h2><h2><u /></h2><h2><u>69 increased by 130% = 158.7 </u></h2><h2><u>Absolute change (actual difference): </u></h2><h2><u>158.7 - 69 = 89.7</u></h2>
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5 0
2 years ago
Set Hill 1 to 75 cm and the other hills to 0 cm. What is the height in meters?
yanalaym [24]
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2 years ago
5 markers cost \$6.55$6.55dollar sign, 6, point, 55. Which equation would help determine the cost of 444 markers? Choose 1 answe
Ivenika [448]

Answer:

  E  None of the above

Step-by-step explanation:

An appropriate proportion is ...

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or any of the alternate ways this proportion can be written. None of the offered choices matches this, so it is appropriate to choose ...

  None of the above.

3 0
2 years ago
A contaminant is leaking into a lake at a rate of R(t) = 1400e0.06t gallons/h. Enzymes have been added to the lake that neutrali
ruslelena [56]

Answer:

Step-by-step explanation:

Rate of leakage, R(t) = 1400 e^0.06t gallons/h

fraction remains , S(t) = e^(-0.32t)

initial contaminant = 1000 gallon

gallons contaminant present after t hour is S(t) R(t)

G(t) = S(t) R(t)

G(t) = 1400 e^{0.06t}\times e^{-0.32t}

G(t) = 1400 e^{- 0.26t}

Put t = 18 hours

G(t) = 1400 e^{- 0.26\times 18}

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5 0
2 years ago
Consider the midterm and final for a statistics class. Suppose 13% of students earned an A on the midterm. Of those students who
padilas [110]

Answer:

There is a 38.97% probability that this student earned an A on the midterm.

Step-by-step explanation:

The first step is that we have to find the percentage of students who got an A on the final exam.

Suppose 13% students earned an A on the midterm. Of those students who earned an A on the midterm, 47% received an A on the final, and 11% of the students who earned lower than an A on the midterm received an A on the final.

This means that

Of the 13% of students who earned an A on the midterm, 47% received an A on the final. Also, of the 87% who did not earn an A on the midterm, 11% received an A on the final.

So, the percentage of students who got an A on the final exam is

P_{A} = 0.13(0.47) + 0.87(0.11) = 0.1568

To find the probability that this student earned an A on the final test also earned on the midterm, we divide the percentage of students who got an A on both tests by the percentage of students who got an A on the final test.

The percentage of students who got an A on both tests is:

P_{AA} = 0.13(0.47) = 0.0611

The probability that the student also earned an A on the midterm is

P = \frac{P_{AA}}{P_{A}} = \frac{0.0611}{0.1568} = 0.3897

There is a 38.97% probability that this student earned an A on the midterm.

5 0
2 years ago
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