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marusya05 [52]
3 years ago
7

A company decides to donate $50,000 to charity. It will select up to 20 charitable organizations, as nominated by its employees.

Each selected organization will receive an equal amount of donation.
Mathematics
1 answer:
motikmotik3 years ago
3 0
$50,000 / 20 = $2,500
each charity will receive $2,500
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20

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If (6^0)^x=1 what are the possible values of x
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A sequence of transformations maps ∆ABC to ∆A′B′C′. The sequence of transformations that maps ∆ABC to ∆A′B′C′ is a reflection ac
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Simplify the following expression 6X2 plus 8X -3 minus 3X2-6
Murrr4er [49]

Answer:

-24

Step-by-step explanation:

I´m guessing by the big x sign you want me to multiply so the first step in the equation is to multiply because of PEMDAS so(6*2)+(8*-3)-3*2-6,

so 6*2=12,8*-3=-24,-3*2=-6

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7 0
3 years ago
The probability that a randomly selected 2 2​-year-old male garter snake garter snake will live to be 3 3 years old is 0.98861 0
Mnenie [13.5K]

Answer:

a. Probability = 0.97735

b. Probability = 0.92294

c. P(At\ Least\ One) = 1

No, it is not unusual if at least 1 lives up to 3.

Step-by-step explanation:

Given

Represent the probability that a 2 year old snake will live to 3 with P(Live);

P(Live) = 0.98861

Solving (a): Probability that two selected will live to 3 years.

Both snakes have a chance of 0.98861 to live up to 3 years.

So, the required probability is:

Probability = P(Live)\ and\ P(Live)

Probability = 0.98861 * 0.98861

Probability = 0.9773497321

Probability = 0.97735 <em>--- Approximated</em>

Solving (b): Probability that seven selected will live to 3 years.

All 7 snakes have a chance of 0.98861 to live up to 3 years.

So, the required probability is:

Probability = P(Live)^n

Where n = 7

Probability = 0.98861^7

Probability = 0.92294324145

Probability = 0.92294 <em>--- Approximated</em>

Solving (c): Probability that at least one of seven selected will not live to 3 years.

In probabilities, the following relationship exist:

P(At\ Least\ One) = 1 - P(None).

So, first we need to calculate the probability that none of the 7 lived up to 3.

If the probability that one lived up to 3 years is 0.98861, then the probability than one do not live up to 3 years is 1 - 0.98861

This gives:

P(Not\ Live) = 0.01139

The probability that none of the 7 lives up to 3 is:

P(None) = P(Not\ Live)^7

P(None) = 0.01139^7

Substitute this value for P(None) in

P(At\ Least\ One) = 1 - P(None).

P(At\ Least\ One) = 1 - 0.01139^7

P(At\ Least\ One) = 0.99999999999997513055642436060443621

P(At\ Least\ One) = 1 ---- Approximated

No, it is not unusual if at least 1 lives up to 3.

This is so because the above results, which is 1 shows that it is very likely for at least one of the seven to live up to 3 years

7 0
3 years ago
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