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3 years ago

I need helpp please

Mathematics

2 answers:

Bumek [7]3 years ago

4 0

Answer:

138.75 or D

Step-by-step explanation:

15(3.5+5.75)

IrinaK [193]3 years ago

3 0

Answer:

Total is $138.75 which is answer D

Step-by-step explanation:

15*3.5= $52.50

15*5.75= $86.25

Total is $138.75

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Which number line represents the solution set for the inequality –4(x + 3) ≤ –2 – 2x?

Yuliya22 [10]

–4(x + 3) ≤ –2 – 2x

>>.....-4x -12 ≤ -2 -2x

>> -12 +2 ≤ +2x

>> - 10 ≤ 2x

>> -5 ≤ x............>> x >= -5

This answer is not represented in the pictures you attached

The line starts in x = -5 and goes up to infinity

3 0

3 years ago

Read 2 more answers

Which equation has infinite solutions?

AnnyKZ [126]

The equation that has an infinite number of solutions is $2x + 3 = \frac{1}{2}(4x + 2) + 2$

<h3>How to determine the equation?</h3>

An equation that has an infinite number of solutions would be in the form

a = a

This means that both sides of the equation would be the same

Start by simplifying the options

3(x – 1) = x + 2(x + 1) + 1

3x - 3 = x + 3x + 2 + 1

3x - 3 = 4x + 3

Evaluate

x = 6 ----- one solution

x – 4(x + 1) = –3(x + 1) + 1

x - 4x - 4 = -3x - 3 + 1

-3x - 4 = -3x - 2

-4 = -2 ---- no solution

$2x + 3 = \frac{1}{2}(4x + 2) + 2$

2x + 3 = 2x + 1 + 2

2x + 3 = 2x + 3

Subtract 2x

3 = 3 ---- infinite solution

Hence, the equation that has an infinite number of solutions is $2x + 3 = \frac{1}{2}(4x + 2) + 2$

Read more about equations at:

brainly.com/question/15349799

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Complete question

Which equation has infinite solutions?

3(x – 1) = x + 2(x + 1) + 1

x – 4(x + 1) = –3(x + 1) + 1

$2x + 3 = \frac{1}{2}(4x + 2) + 2$

$\frac 13(6x - 3) = 3(x + 1) - x - 2$

5 0

2 years ago

The lifetime for a certain type of tre has been shown to be five years the mean) with a standard deviation of 6 months. In a nor

olga_2 [115]

Answer:

2.28%

Step-by-step explanation:

mean = 5 years = 60 months

standard deviation = 6 months

X = number of years the tires will last

P = probability of ...

z = z-score

6 years = 72 months

z = (# of months - mean) / standard deviation

z = (72 - 60)/6

z = 12/6

z = 2

P(X > 72) = 1 - P(X < 72)

= 1 - P(z < 2)

(using a Standard Normal Probabilities Table we can see that P(z < 2) = .9772)

So:

= 1 - .9772

= 0.0228 OR 2.28%

6 0

3 years ago

Ming spent half of her weekly allowance buying pizza. To earn more money her parents let her clean the oven for $6. What is her

Gelneren [198K]

Answer:

$16

Step-by-step explanation:

she has $14 now

her parents let her earn 6

so $14 - $6 = $8

so thats what was left over from her allowance. she spent half of her allowance to get to $8 so you double the money

$8 * 2 = $16

$16 is her weekly allowance

7 0

3 years ago

Suppose that five ones and four zeros are arranged around a circle. Between any two equal bits you insert a 0 and between any tw

PolarNik [594]

Answer:

Using backward reasoning we want to show that "We can never get nine 0's".

Step-by-step explanation:

Basically in order to create nine 0's, the previous step had to have all 0's or all 1's. There is no other way possible, because between any two equal bits you insert a 0.

If we consider two cases for the second-to-last step:

There were 9 0's:

We obtain nine 0's if all bits in the previous step were the same, thus all bit were 0's or all bits were 1's. If the previous step contained all 0's, then we have the same case as the current iteration step. Since initially the circle did not contain only 0's, the circle had to contain something else than only 0's at some point and thus there exists a point where the circle contained only 1's.

There were 9 1's:

A circle contains only 1's, if every pair of the consecutive nine digits is different. However this is impossible, because there are five 1's and four 0's (we have an odd number of bits!), thus if the 1's and 0's alternate, then we obtain that 1's that will be next to each other (which would result in a 1 in the next step). Thus, we obtained a contradiction and thus assumption that the circle contains nine 0's after iteratins the procedure is false. This then means that you can never get nine 0's.

To summarize, in order to create nine 0's, the previous step had to have all 0's or al 1's. As we didn't start the arrange with all 0's, the only way is having all 1's, but having all 1's will not be possible in our case since we have an odd number of bits.

5 0

3 years ago