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denis23 [38]
3 years ago
11

Help me I know it is easy but i just cant think rn. :/​

Mathematics
1 answer:
Ipatiy [6.2K]3 years ago
3 0
A. The Answer Is 8
Multiplication: 1/4 • 8 = 2

B. The Answer Is 12
Multiplication: 1/2 • 12 = 6

C. The Answer Is 20
Multiplication: 1/4 • 20 = 5

D. The Answer Is 40
Multiplication: 1/8 • 40 = 5

E. The Answer Is 18
Multiplication: 1/3 • 18 = 6

F. The Answer Is 18
Multiplication: 1/6 • 18 = 3

G. The Answer Is 30
Multiplication: 1/5 • 30 = 6

H. The Answer Is 60
Multiplication: 1/10 • 60 = 6

Hope Im Not Too Late :)
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Recall that r^2 = x^2 + y^2, so that r = sqrt(x^2+y^2).

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r = 3 sin g becomes    sqrt(x^2+y^2) = 3*-----------------------
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Squaring both sides,

                    9y^2
x^2+y^2 = -----------------
                  x^2 + y^2

If this is correct (and I'm not convinced that it is), then (x^2+y^2)^2 = 9y^2
shows the relationship between x and y.  Can anyone improve on this result?

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2/5

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2 years ago
10 normal six sided dice are thrown.Find the probability of obtaining at least 8 failuresif a success is 5 or 6.
erastova [34]

Answer:

0.2992 = 29.92% probability of obtaining at least 8 failures.

Step-by-step explanation:

For each dice, there are only two possible outcomes. Either a failure is obtained, or a success is obtained. Trials are independent, which means that the binomial probability distribution is used 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.

A success is 5 or 6.

A dice has 6 sides, numbered 1 to 6. Since a success is 5 or 6, the other 4 numbers are failures, and the probability of failure is:

p = \frac{4}{6} = 0.6667

10 normal six sided dice are thrown.

This means that n = 10

Find the probability of obtaining at least 8 failures.

This is:

P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)

So

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

P(X = 8) = C_{10,8}.(0.6667)^{8}.(0.3333)^{2} = 0.1951

P(X = 9) = C_{10,9}.(0.6667)^{9}.(0.3333)^{1} = 0.0867

P(X = 10) = C_{10,10}.(0.6667)^{10}.(0.3333)^{0} = 0.0174

Then

P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.1951 + 0.0867 + 0.0174 = 0.2992

0.2992 = 29.92% probability of obtaining at least 8 failures.

8 0
3 years ago
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