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2 years ago

Becky spent h hours studying

Mathematics

1 answer:

Otrada [13]2 years ago

7 0

Answer:

6.5 = h + 2

h = 4.5

Step-by-step explanation:

We are given h and since she did 2 additional hours of work, we add the 2 hours. and since we are trying to find h, we set the equation equal to 6.5

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D = 7 mm Calculate the radius of the circle.

Vikentia [17]

Answer:

r= d/2

7/2

3.5

3.5 is the answer

Step-by-step explanation:

5 0

3 years ago

A piece of string is 112 inches long. The student wants to cut it into 2 pieces so that one piece is 3 times as long as the othe

gogolik [260]

37.33333333333 so about 37" I'm thinking..

3 0

3 years ago

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In a standardized IQ test, scores are normally distributed with a mean score of 100 and a standard deviation of 15. Find the pro

Zanzabum

Answer:

0.0475

Step-by-step explanation:

First find the z-score.

z = (x − μ) / σ

z = (125 − 100) / 15

z = 1.67

Use a calculator or chart to find the probability. Using a chart:

P(Z > 1.67) = 1 − P(Z < 1.67)

P(Z > 1.67) = 1 − 0.9525

P(Z > 1.67) = 0.0475

8 0

3 years ago

Find the area. Simplify your answer

castortr0y [4]

Answer:

Solution given:

length[l]=3x+7

breadth [b]=2x+3

we have

area of rectangle=l×b=(3x+7)(2x+3)

=6x²+14x+9x+21=6x²+23x+21 units ²

3 0

2 years ago

A sample of a radioactive isotope had an initial mass of 360 mg in the year 1998 and

RUDIKE [14]

Answer:

193 mg

Step-by-step explanation:

Exponential decay formula:

$A_t = A_0e^r^t$
where Aₜ = mass at time t, A₀ = mass at time 0, r = decay constant (rate), t = time

Our known variables are:

1998 to the year 2004 is a total of t = 6 years.
The sample of radioactive isotope has an initial mass of A₀ = 360 mg at time 0 and a mass of Aₜ = 270 mg at time t.

Let's solve for the decay constant of this sample.

$270=360e^-^r^(^6^)$
$270=360e^-^6^r$
$\frac{3}{4} =e^-^6^r$
$\text{ln} (\frac{3}{4} )= \text{ln}(e^-^6^r)$
$\text{ln} (\frac{3}{4} )=-6r$
$r=-\frac{\text{ln}\frac{3}{4} }{6}$
$r=0.04794701$

Using our new variables, we can now solve for Aₜ at t = 7 years, since we go from 2004 to 2011.

Our new initial mass is A₀ = 270 mg. We solved for the decay constant, r = 0.04794701.

$A_t=270e^-^(^0^.^0^4^7^9^4^8^0^1^)^(^7^)$
$A_t=270e^-^0^.^3^3^5^6^2^9^0^7$
$A_t=193.01982213$

The expected mass of the sample in the year 2011 would be 193 mg.

3 0

2 years ago