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3 years ago

I have 23 hundreds,7 tens and 10 ones. what number am i ?

Mathematics

1 answer:

torisob [31]3 years ago

8 0

You would be 2,380 because there are 10 ones there would be 8 tens instead of 8.

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BRAINLIEST!!!

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Answer: $\frac{8}{17}$ or 8:17

Step-by-step explanation:

For any angle x (other than right angle) in a right triangle ,the trigonometric ratio of sin x is given by :-

$\sin x=\frac{\text{side opposite to x}}{\text{Hypotenuse}}$

Given: A right triangle with hypotenuse = 68 units

The side adjacent to S = 60

Let h be the side opposite to S, then using Pythagoras in the given right triangle, we get

$(68)^2=60^2+h^2\\\\\Rightarrow\ h^2=68^2-60^2\\\\\Rightarrow\ h^2=1024\\\\\Rightarrow\ h=\sqrt{1024}=32$

Thus, the side opposite to S = 32 units

Now, the trigonometric ratio for sin S is given by :-

$\sin S=\frac{\text{side opposite to S}}{\text{Hypotenuse}}\\\\\Rightarrow\sin S=\frac{32}{68}=\frac{8}{17}$

Hence, the trigonometric ratio for sin S = $\frac{8}{17}$ or 8:17

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Which expression is equivalent to the number 144?

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Korvikt [17]

Answer:

90.67% probability that John finds less than 7 golden sheets of paper

Step-by-step explanation:

For each container, there are only two possible outcomes. Either it contains a golden sheet of paper, or it does not. The probability of a container containing a golden sheet of paper is independent of other containers. 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.

At Munder Difflin Paper Company, the manager Mitchell Short randomly places golden sheets of paper inside of 30% of their paper containers.

This means that $p = 0.3$

14 of these containers of paper.

This means that $n = 14$

What is the probability that John finds less than 7 golden sheets of paper?

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

In which

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

$P(X = 0) = C_{14,0}.(0.3)^{0}.(0.7)^{14} = 0.0068$

$P(X = 1) = C_{14,1}.(0.3)^{1}.(0.7)^{13} = 0.0407$

$P(X = 2) = C_{14,2}.(0.3)^{2}.(0.7)^{12} = 0.1134$

$P(X = 3) = C_{14,3}.(0.3)^{3}.(0.7)^{11} = 0.1943$

$P(X = 4) = C_{14,4}.(0.3)^{4}.(0.7)^{10} = 0.2290$

$P(X = 5) = C_{14,5}.(0.3)^{5}.(0.7)^{9} = 0.1963$

$P(X = 6) = C_{14,6}.(0.3)^{6}.(0.7)^{8} = 0.1262$

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0068 + 0.0407 + 0.1134 + 0.1943 + 0.2290 + 0.1963 + 0.1262 = 0.9067$

90.67% probability that John finds less than 7 golden sheets of paper

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3 years ago