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scoray [572]
2 years ago
6

5. Kamala worked for 7 1/2 h. She spent 2/3 of the time on her computer. How long was she on her computer?​

Mathematics
2 answers:
GREYUIT [131]2 years ago
6 0

She was 5 hours on her computer

Number of hours worked = 7\frac{1}{2}

This is an improper fraction so first we convert it

7\frac{1}{2} = > \frac{15}{2}

Time spent on her computer = \frac{2}{3} of working hours

∴ time spent on computer = (15/2)*(2/3)

= 5 hours

Lina20 [59]2 years ago
4 0

Answer:

5 hrs

Step-by-step explanation:

7 1/2 hours equates to 450 minutes total

2/3 of 450 = 300 total minutes

turning the minutes back into hours, 300 / 60 = <u>5</u>

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Solve for R. 3 + 3r &gt; 9
Nostrana [21]

Answer:

r > 2

Step-by-step explanation:

9-3 = 6

6/3 = 2

Have a good day :)

7 0
3 years ago
If the length of a rectangle is 4 more than the width and if the perimeter of the rectangle is 52, what is the width of the rect
Ymorist [56]

Answer:

11

Step-by-step explanation:

Perimeter is figured out by adding all 4 sides.

Width of 11 + width 11 + length 15 + length 15 = 52

And the length of 15 is 4 more than the width of 11

5 0
2 years ago
Write in simplest radical form: √208
Arte-miy333 [17]

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6 0
3 years ago
Use the exponential decay​ model, Upper A equals Upper A 0 e Superscript kt​, to solve the following. The​ half-life of a certai
Akimi4 [234]

Answer:

It will take 7 years ( approx )

Step-by-step explanation:

Given equation that shows the amount of the substance after t years,

A=A_0 e^{kt}

Where,

A_0 = Initial amount of the substance,

If the half life of the substance is 19 years,

Then if t = 19, amount of the substance = \frac{A_0}{2},

i.e.

\frac{A_0}{2}=A_0 e^{19k}

\frac{1}{2} = e^{19k}

0.5 = e^{19k}

Taking ln both sides,

\ln(0.5) = \ln(e^{19k})

\ln(0.5) = 19k

\implies k = \frac{\ln(0.5)}{19}\approx -0.03648

Now, if the substance to decay to 78​% of its original​ amount,

Then A=78\% \text{ of }A_0 =\frac{78A_0}{100}=0.78 A_0

0.78 A_0=A_0 e^{-0.03648t}

0.78 = e^{-0.03648t}

Again taking ln both sides,

\ln(0.78) = -0.03648t

-0.24846=-0.03648t

\implies t = \frac{0.24846}{0.03648}=6.81085\approx 7

Hence, approximately the substance would be 78% of its initial value after 7 years.

5 0
3 years ago
What is the area of the obtuse triangle below?
tekilochka [14]
D cause that’s the answer
4 0
2 years ago
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