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3 years ago

11

A cooler contains 7 cans of soda; 2 colas, 4 orange, and 1 cherry. Two cans are selected at random without replacement. Find

the probability that at least one can is cherry

1 answer:

Mazyrski [523]3 years ago

7 0

The probability is 2/14

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Sam bought some toys. He bought a baseball for $5.88, and paid $8.41

myrzilka [38]

Answer:

$5.71

Step-by-step explanation:

20-5.88=14.12

14.12-8.41=5.71

6 0

3 years ago

Read 2 more answers

4,792÷8 show your work

Svetradugi [14.3K]

Answer:

<h2>4,792 ÷ 8 = 599</h2>

Step-by-step explanation:

Look at the picture.

Use the long division.

3 0

3 years ago

If the probability of an event is 0.7 repeating, what are the odds against the event

ivann1987 [24]

So the probability is the number of favorable outcomes divided by the number of total outcomes. This means that the favorable outcomes are 7/9, and the unfavorable outcomes are 2/9. The odds against the event are the unfavorable outcomes. Therefore the odds against the event is 2/9, or 0.2 repeating. Hope this helps. Feel free to ask more questions, and feel free to ask questions about my explanation.

7 0

3 years ago

What is 2+2+?+10+?-2+9=?

Dennis_Churaev [7]

$2+2+x+10+x-2+9=x\\2+2+x+10+x-2+9 -x=0\\(2+2+10-2+9)+(x+x-x)=0\\21+x=0\\x=-21$

ANSWER: ? = -21

P/s: ? = x < let ? is x >

ok done. Thank to me :>

3 0

2 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:

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

Therefore the half-life is between 600 and 700 years

5 0

3 years ago