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3 years ago

A number is chosen at random from 2 to 16 inclusive. What is the probability that the number is prime?

Mathematics

1 answer:

Mkey [24]3 years ago

3 0

<h3>Answer: 2/5</h3>

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Explanation:

A = list of primes between 2 and 16, inclusive for both endpoints

A = {2,3,5,7,11,13}

B = list of integers between 2 and 16, inclusive for both endpoints

B = {2,3,4,...,14,15,16}

The dots indicate the pattern continues, which means we don't have to write out every value.

There are n(A) = 6 items in set A and n(B) = 16-2+1 = 15 items in set B. I'm using the formula that a set like {p,p+1,p+2,...,q-2,q-1,q} has q-p+1 elements in it. This only works for integers and they must be consecutive. Also q > p.

Once we know how many items are in each set, we divide the two counts

P(finding a prime between 2 and 16) = n(A)/n(B) = 6/15 = 2/5

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Two possible solutions of √11-2x=√x^2+4x+4 are –7 and 1. Which statement is true? A) Only x = –7 is an extraneous solution.

11111nata11111 [884]

Answer:

Option D)Neither solution is extraneous.

Step-by-step explanation:

we have

$\sqrt{11-2x}=\sqrt{x^{2}+4x+4}$

we know that

two possible solutions are x=-7 and x=1

<u><em>Verify each solution</em></u>

Substitute each value of x in the expression above and interpret the results

1) For x=-7

$\sqrt{11-2(-7)}=\sqrt{-7^{2}+4(-7)+4}$

$\sqrt{25}=\sqrt{25}$

$5=5$ ----> is true

therefore

x=-7 is not a an extraneous solution

2) For x=1

$\sqrt{11-2(1)}=\sqrt{1^{2}+4(1)+4}$

$\sqrt{9}=\sqrt{9}$

$3=3$ ----> is true

therefore

x=1 is not a an extraneous solution

therefore

Neither solution is extraneous

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

You are given 1000 one dollar bills and 10 envelopes. Put the bills into the envelopes in such a way that someone can ask you fo

taurus [48]

Answer:

We can do it with envelopes with amounts $1,$2,$4,$8,$16,$32,$64,$128,$256 and $489

Step-by-step explanation:

Observe that, in binary system, 1023=1111111111. That is, with 10 digits we can express up to number 1023.

This give us the idea to put in each envelope an amount of money equal to the positional value of each digit in the representation of 1023. That is, we will put the bills in envelopes with amounts of money equal to $1,$2,$4,$8,$16,$32,$64,$128,$256 and $512.

However, a little modification must be done, since we do not have $1023, only $1,000. To solve this, the last envelope should have $489 instead of 512.

Observe that:

1+2+4+8+16+32+64+128+256+489=1000
Since each one of the first 9 envelopes represents a position in a binary system, we can represent every natural number from zero up to 511.
If we want to give an amount "x" which is greater than $511, we can use our $489 envelope. Then we would just need to combine the other 9 to obtain x-489 dollars. Since $x-489\leq511$ , by 2) we know that this would be possible.

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3 years ago

Which inequality can be used to represent a number x that is less than equal to 8 and greater than -3

KIM [24]

Answer: x = -2,-1, 0, 1, 2, 3, 4, 5, 6, 7, 8

Solution:

A number line is used to represent the possible numbers that x can be.

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So it will be represented as,

-3 < x ≤ 8

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valina [46]

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3 years ago

A number is chosen at random from 2 to 16 inclusive. What is the probability that the number is prime?​

A number is chosen at random from 2 to 16 inclusive. What is the probability that the number is prime?