Which of the following equations have no solutions?

24 less than 4 times a number is 60. what is the number?

kondaur [170]

Hey there! :D

Make an algebraic expression.

Let the 'number' equal x.

4x-24= 60

Add 24 on both sides.

4x= 84

Divide both sides by 4.

x=21

The number is 21.

I hope this helps!
~kaikers

8 0

3 years ago

Charles smith makes $9.75 an hour he works four hours on Monday six hours on Tuesday five hours on Wednesday five hours on Thurs

ladessa [460]

Simple. Multiple

Monday: 6x9.75
Tuesday &Wednesday: 5x9.75
Thursday & Friday: 7x9.75

4 0

3 years ago

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Two cars leave Phoenix and travel along roads 90 degrees apart. If Car 1 leaves 30 minutes earlier than Car 2 and averages 42 mp

romanna [79]

Answer:

C. 210 miles

Step-by-step explanation:

We have been given that two cars leave Phoenix and travel along roads 90 degrees apart. Car 1 leaves 30 minutes earlier than Car 2 and averages 42 mph.

We will use distance formula and Pythagoras theorem to solve our given problem.

$\text{Distance}=\text{Speed}\times \text{Time}$

$\text{Distance covered by Car 1 in 3.5 hours}=\frac{42\text{ Miles}}{\text{Hour}}\times \text{3.5 hour}$

$\text{Distance covered by Car 1 in 3.5 hours}=147\text{ Miles}$ .

Since car 1 leaves 30 minutes before car 2, so car 2 will travel for only 3 hours when car 1 will travel for 3.5 hours.

$\text{Distance covered by Car 2 in 3 hours}=\frac{50\text{ Miles}}{\text{Hour}}\times \text{3 hour}$

$\text{Distance covered by Car 2 in 3 hours}=150\text{ Miles}$

Since both car travel along roads 90 degree apart, therefore, the distance between both cars after Car 1 has traveled 3.5 hours would be hypotenuse with legs 147 and 150.

$\text{Distance between both cars}=\sqrt{147^2+150^2}$

$\text{Distance between both cars}=\sqrt{21609+22500}$

$\text{Distance between both cars}=\sqrt{44109}$

$\text{Distance between both cars}=210.021427\approx 210$

Therefore, the both cars will be 210 miles apart and option C is the correct choice.

3 0

3 years ago

A large operator of timeshare complexes requires anyone interested in making a purchase to first visit the site of interest. His

morpeh [17]

Answer:

There is a 21.053% probability that this person made a day visit.

There is a 39.474% probability that this person made a one night visit.

There is a 39.474% probability that this person made a two night visit.

Step-by-step explanation:

We have these following percentages

20% select a day visit

50% select a one-night visit

30% select a two-night visit

40% of the day visitors make a purchase

30% of one night visitors make a purchase

50% of two night visitors make a purchase

The first step to solve this problem is finding the probability that a randomly selected visitor makes a purchase. So:

$P = 0.2(0.4) + 0.5(0.3) + 0.3(0.5) = 0.38$

There is a 38% probability that a randomly selected visitor makes a purchase.

Now, as for the questions, we can formulate them as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

Suppose a visitor is randomly selected and is found to have made a purchase.

How likely is it that this person made a day visit?

What is the probability that this person made a day visit, given that she made a purchase?

P(B) is the probability that the person made a day visit. So $P(B) = 0.20$

P(A/B) is the probability that the person who made a day visit made a purchase. So $P(A/B) = 0.4$

P(A) is the probability that the person made a purchase. So $P(A) = 0.38$

So

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.4*0.2}{0.38} = 0.21053$

There is a 21.053% probability that this person made a day visit.

How likely is it that this person made a one-night visit?

What is the probability that this person made a one night visit, given that she made a purchase?

P(B) is the probability that the person made a one night visit. So $P(B) = 0.50$

P(A/B) is the probability that the person who made a one night visit made a purchase. So $P(A/B) = 0.3$

P(A) is the probability that the person made a purchase. So $P(A) = 0.38$

So

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.5*0.3}{0.38} = 0.39474$

There is a 39.474% probability that this person made a one night visit.

How likely is it that this person made a two-night visit?

What is the probability that this person made a two night visit, given that she made a purchase?

P(B) is the probability that the person made a two night visit. So $P(B) = 0.30$

P(A/B) is the probability that the person who made a two night visit made a purchase. So $P(A/B) = 0.5$

P(A) is the probability that the person made a purchase. So $P(A) = 0.38$

So

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.3*0.5}{0.38} = 0.39474$

There is a 39.474% probability that this person made a two night visit.

3 0

3 years ago

What is the answer to this?

Elanso [62]

The answer is 16b^2c square root of 5c

3 0

2 years ago

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