Solve the equation 88,405 x 54 = ___. (4 points) a 4,420,250 b 4,597,060 c 4,685,465 d 4,773,870

Jan's office contains 12 file cabinets.

expeople1 [14]

57 im pretty sure maybe

6 0

1 year ago

Which of the following theorems cannot be used to prove two lines are parallel?

mariarad [96]

Answer:

Option H

Step-by-step explanation:

Option F

If two lines are cut by a transversal so that a pair of corresponding angles are congruent, then the lines are parallel.

True.

Option G

If two lines are cut by a transversal so that a pair of alternate exterior angles are congruent, then the lines are parallel.

True

Option H

If two line are cut by a transversal so that a pair of vertical angles are congruent, then the lines are parallel.

Since, vertical angles don't prove the lines cut by a transversal are parallel.

So the statement is False.

Option J

If two lines are are cut by a transversal so that a pair of same side interior angles are supplementary, then the lines are parallel.

True.

6 0

3 years ago

Subtract 12h+1 from 34h+4 . what is the answer? a.−1/4h+5 b.1/4h+5 c.−1/4h d.1/4h+3

Lemur [1.5K]

Answer:

D

Step-by-step explanation:

We would like to solve:

(3/4h + 4) - (1/2h + 1)

This is equivalent to:

3/4h + 4 - 1/2h - 1

First, combine like terms; that is, combine the terms with h in them and combine the terms without h in them:

3/4h - 1/2h + 4 - 1

The common denominator for 3/4 and 1/2 is 4, so:

3/4h - 1/2h = 3/4h - 2/4h = 1/4h

Put this back in:

1/4h + 4 - 1

1/4h + 3

The answer is thus D.

~ an aesthetics love

8 0

3 years ago

One year, the population of a city was 64,000. Several years later it was 53,120. Find the percent decrease.

ohaa [14]

Answer:

17% i think

Step-by-step explanation:

4 0

2 years ago

A representative from the National Football League's Marketing Division randomly selects people on a random street in Kansas Cit

Orlov [11]

Using the binomial distribution, we have that:

a) 0.1024 = 10.24% probability that the marketing representative must select 4 people to find one who attended the last home football game.

b) 0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.

c) The expected number of people is 4, with a variance of 20.

For each person, there are only two possible outcomes. Either they attended a game, or they did not. The probability of a person attending a game is independent of any other person, which means that the binomial distribution is used.

Binomial probability distribution

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

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

The parameters are:

p is the probability of a success on a single trial.

n is the number of trials.

p is the probability of a success on a single trial.

The expected number of trials before q successes is given by:

$E = \frac{q(1-p)}{p}$

The variance is:

$V = \frac{q(1-p)}{p^2}$

In this problem, 0.2 probability of a finding a person who attended the last football game, thus $p = 0.2$ .

Item a:

None of the first three attended, which is P(X = 0) when n = 3.
Fourth attended, with 0.2 probability.

Thus:

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

$P(X = 0) = C_{3,0}.(0.2)^{0}.(0.8)^{3} = 0.512$

$0.2(0.512) = 0.1024$

0.1024 = 10.24% probability that the marketing representative must select 4 people to find one who attended the last home football game.

Item b:

This is the probability that none of the first six went, which is P(X = 0) when n = 6.

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

$P(X = 0) = C_{6,0}.(0.2)^{0}.(0.8)^{6} = 0.2621$

0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.

Item c:

One person, thus $q = 1$ .

The expected value is:

$E = \frac{q(1-p)}{p} = \frac{0.8}{0.2} = 4$

The variance is:

$V = \frac{0.8}{0.04} = 20$

The expected number of people is 4, with a variance of 20.

A similar problem is given at brainly.com/question/24756209

3 0

1 year ago