If Danica ran 12 meters in 6 seconds and Colby ran 15 meters in 5 seconds, who was faster?

Mathematics

1 answer:

salantis [7]3 years ago

7 0

Answer:

colby is faster

Step-by-step explanation:

he ran more meters in less time he is faster

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Find the exact circumference of a circle with diameter equal to 8 ft

lina2011 [118]

The formula for circumference is C= 2(pi)(r)^2. Because diameter is double the radius, and the diameter is 8 ft, the radius is 4 ft. So, plug that into the equation, and you get: C=2(pi)(4)^2. The answer is 32 pi ft. If the question asks for the exact answer, keep the pi as it is, but if it says to round, multiply 32 by pi and round.

6 0

3 years ago

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A high school spends approximately $9,000 hosting a football games (e.G. The hourly pay for staff, the electricity, water usage,

Yanka [14]

Answer:

$4800

Step-by-step explanation:

The expected net income is the probability when rains multiplied by net income (profit) summed with the probability of not raining with that days net income multiplied.

First, Chance of rain = 12% = 12/100 = 0.12

Profit/Loss = Income - Cost

Income is 5000

Cost = 9000

So,

Loss = 5000 - 9000 = 4000

Or, Net Income (profit) = - 4000

Now, Change of not raining = 100 - 12 = 88% = 88/100 = 0.88

Profit = Income - Cost

Income = 15,000

Cost = 9000

Net Income (profit) = 15000 - 9000 = 6000

So, we can write the expected net income as:

Expected NI = (0.12)(-4000) + (0.88)(6000) = $4800

8 0

3 years ago

Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li

krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

$A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years$

First, let us calculate the decay constant (k)

$75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036$

Next, let us calculate the half-life as follows:

$\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years$