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2 years ago

10 to the power of 2 divided by 4 divided by 3 plus 2

Mathematics

1 answer:

mylen [45]2 years ago

5 0

The answer would be = <span>10.3333333
</span>

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max2010maxim [7]

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2 years ago

Is 66,296,193 divisible by 3?

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James has organised a game to raise money for charity at a local fair. He rolls a fair six sided dice and flips a fair coin. If

Scorpion4ik [409]

The total amount that was raised for charity is £75.

<h3>What is the total amount raised?</h3>

Here is the sample space

1 2 3 4 5 6

Heads 1 4 9 16 25 36

Tails 1 8 27 64 125 216

Probability of getting a number greater than 30 = 4/12 = 1/3

1/3 x 450 = 150

Total value of people getting more than 30 : 150 x £1 = £150

Total amount paid = 50 x 450 = £225

Amount raised = £225 - 150 = £75

To learn more about probability, please check: brainly.com/question/13234031

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3 0

2 years ago

Suppose that each time Giannis Antetokounmpo shoots a free throw, he has a 3/4 probability of success. If Giannis shoots three f

GREYUIT [131]

Answer:

84.38% probability that he succeeds on at least two of them

Step-by-step explanation:

For each free throw, there are only two possible outcomes. Either Giannis makes it, or he does not. The free throws are independent. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

He has a 3/4 probability of success.

This means that $p = \frac{3}{4} = 0.75$

Giannis shoots three free throws

This means that $n = 3$

What is the probability that he succeeds on at least two of them

$P(X \geq 2) = P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{3,2}.(0.75)^{2}.(0.25)^{1} = 0.4219$

$P(X = 3) = C_{3,3}.(0.75)^{3}.(0.25)^{0} = 0.4219$

$P(X \geq 2) = P(X = 2) + P(X = 3) = 0.4219 + 0.4219 = 0.8438$

84.38% probability that he succeeds on at least two of them

6 0

2 years ago