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3 years ago

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Ten pieces of paper numbered consecutively from 1 to 10 are put into a hat. if three pieces of paper are drawn at random from th

e hat one at a time without being returned to the hat, what is the probability that all three will be even numbers

1 answer:

DaniilM [7]3 years ago

7 0

There are 5 even numbers from 1 to 10.

The number of ways to draw all even numbers is 5!/(2!*3!) = 10

The number of ways to draw from the hat is 10!/(7!*3!) = 128

Therefore the probability is 10/128 = 5/64 or roughly 7.81%

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Algebra 2 question need help thanks.

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What set of reflections would carry parallelogram ABCD onto itself?

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In the given figure △ABC ≅△DEC. Which of the following relations can be proven using CPCTC ?

Serhud [2]

Option B:

$\overline{A B}=\overline{D E}$

Solution:

In the given figure $\triangle A B C \cong \triangle D E C$ .

If two triangles are similar, then their corresponding sides and angles are equal.

By CPCTC, in $\triangle A B C \ \text{and}\ \triangle D E C$ ,

$\overline{AB }=\overline{DE}$ – – – – (1)

$\overline{B C}=\overline{EC}$ – – – – (2)

$\overline{ CA}=\overline{CD}$ – – – – (3)

$\angle ACB=\angle DCE$ – – – – (4)

$\angle ABC=\angle DEC$ – – – – (5)

$\angle BAC=\angle EDC$ – – – – (6)

Option A: $\overline{B C}=\overline{D C}$

By CPCTC proved in equation (2) $\overline{B C}=\overline{EC}$ .

Therefore $\overline{B C}\neq \overline{D C}$ . Option A is false.

Option B: $\overline{A B}=\overline{D E}$

By CPCTC proved in equation (1) $\overline{AB }=\overline{DE}$ .

Therefore Option B is true.

Option C: $\angle A C B=\angle D E C$

By CPCTC proved in equation (4) $\angle ACB=\angle DCE$ .

Therefore $\angle A C B\neq \angle D E C$ . Option C is false.

Option D: $\angle A B C=\angle E D C$

By CPCTC proved in equation (5) $\angle ABC=\angle DEC$ .

Therefore $\angle A B C\neq \angle E D C$ . Option D is false.

Hence Option B is the correct answer.

$\Rightarrow\overline{A B}=\overline{D E}$

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3 years ago

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Mice21 [21]

Differnce of 2 pefect squares
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p^2-9=(p-3)(p+3)

factored is
(p-3)(p+3)(p^2+9)

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A state end-of-grade exam in American History is a multiple-choice test that has 50 questions with 4 answer choices for each que

Assoli18 [71]

Answer:

Q1) The student has a 0.01% probability of passing the test.

Q2) She has a 99.91% probability of passing in the test.

Step-by-step explanation:

For each question, there are only two possible outcomes. Either he gets it correct, or he gets it wrong. So we solve this problem using the binomial probability distribution.

Binomial probability distribution

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

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

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

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

And $\pi$ is the probability of X happening.

For this problem, we have that:

Question 1.

There are 50 questions, so $n = 50$ .

The student is going to guess each question, so he has a $\pi = \frac{1}{4} = 0.25$ probability of getting it right.

He needs to get at least 25 question right.

So we need to find $P(X \geq 25)$ .

Using a binomial probability calculator, with $n = 50$ and $\pi = 0.25$ we get that $P(X \geq 25) = 0.0001$ .

This means that the student has a 0.01% probability of passing the test.

Question 2.

Now, we need to find $P(X \geq 25)$ with $\pi = 0.70$ . So $P(X \geq 25) = 0.9991$

She has a 99.91% probability of passing in the test.

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