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2 years ago

8

Alison has 3 containers with 25 crayons in each.She also has 4 boxes of markers with 12 markers in each box. She gives 10 crayon

s to a friend.How many crayons and markers does Alison have now?

1 answer:

svet-max [94.6K]2 years ago

6 0

3 x 25 = 75
4 x 12 = 48
75 - 10 = 65
65 + 48 = 113
Answer: 113 markers and crayons or 65 crayons and 48 markers

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Elsa pumped 12 gallons of water into her pool. This was done over a period of 2 minutes at a constant rate. What was the change

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Answer:

it was 6 because do 12 / 2 will equal 6.

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2 years ago

A car starts with a dull tank of gas. After driving around 5he city, 1/7 of the gas has been used. With the rest of the gas in t

Sati [7]

Answer:

$\frac{2}{7}$

Step-by-step explanation:

Given:

A car starts with a dull tank of gas

1/7 of the gas has been used around the city.

With the rest of the gas in the car, the car can travel to and from Ottawa three times.

Question asked:

What fractions of a tank of gas does each complete trip to Ottawa use?

Solution:

Fuel used around the city = $\frac{1}{7}$

Remaining fuel after driving around the city = 1 - $\frac{1}{7}$

= $\frac{7 - 1}{7} = \frac{6}{7}$

According to question:

As from the rest of the gas in the car that is $\frac{6}{7}$ , the car can complete 3 trip to Ottawa which means,

By unitary method:

The car can complete 3 trip by using = $\frac{6}{7}$ tank of gas.

The car can complete 1 trip by using = $\frac{6}{7} \div 3$

= $\frac{6}{7} \times\frac{1}{3}$

= $\frac{6}{21}$

= $\frac{2}{7}$ tank of gas

Thus, $\frac{2}{7}$ tank of gas used for each complete trip to Ottawa.

5 0

3 years ago

Use each model to predict the life expectancy of residents of a country for which the average annual income is $80,000

Mandarinka [93]

The life expectancies of residents of a country for which the average annual income is $80,000 for the three models are 12309.9352, 172.2436 and 4828.1393

The life expectancies of the models are given as:

$y =69.9352 + 0.1530\times (income)$ --- model 1

$y =4.2436 + 0.0021\times (income)$ --- model 2

$y =4.1393 + 0.0603\times (income)$ --- model 3

Given that the average annual income is $80,000;

We simply substitute 80000 for income in the equations of the three models.

So, we have:

<u>Model 1</u>

$y =69.9352 + 0.1530\times (income)$

$y =69.9352 + 0.1530\times 80000$

$y =12309.9352$

<u>Model 2</u>

$y =4.2436 + 0.0021\times (income)$

$y =4.2436 + 0.0021\times 80000$

$y =172.2436$

<u>Model 3</u>

$y =4.1393 + 0.0603\times (income)$

$y =4.1393 + 0.0603\times 80000$

$y =4828.1393$

Hence, the life expectancies are 12309.9352, 172.2436 and 4828.1393

Read more about linear models at:

brainly.com/question/8609070

4 0

2 years ago

Dwayne's family has many pets. at the house they have both cats and birds. between all the cats and birds, they have 16 eyes and

densk [106]

Let C be the number of cats and B be the number of birds.
Each cat has 4 legs while each bird has 2 legs. Thus,
4C + 2B = 22
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Each cat and each bird has 2 eyes
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Substituting 2B
2C + 22 - 4C = 16
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6 0

2 years ago

Read 2 more answers

Write (1/3i)-(-6+2/3i) as a complex number in standard form

Eddi Din [679]

Answer:

$6+\frac{i}{3}$

Step-by-step explanation:

$\frac{1}{3\imath}-(-6+\frac{2}{3\imath})$

$\frac{1}{3\imath}+6-\frac{2}{3\imath}$

taking like terms together

$\frac{1}{3\imath}-\frac{2}{3\imath}+6$

taking LCM

$\frac{1-2}{3\imath}+6$

$\frac{-1}{3\imath}+6$

taking LCM

$\frac{-1+18\imath}{3\imath}$

splitting the term

$\frac{-1+18\imath}{3\imath}$

splitting the term

$-\frac{1}{3\imath}+\frac{18\imath}{3\imath}$

$-\frac{1\times3\imath}{3\imath \times \imath}+6$

$-\frac{i}{3\imath^2}+6$

we know that

$\imath^2=-1$

putting this value in above equation

$\frac{\imath}{3}+6$

7 0

3 years ago