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zhuklara [117]
3 years ago
10

Autum school holding a voleturing challeng. Students are encouraged to volunteer for at least 1 1/2 hours per week. Volenteers a

bou the same amount each week. During a 3- week period she volunteers for 6 3/4 hours. Autumn wants to compare her volunteers rate with the challenge rate
Mathematics
2 answers:
Pani-rosa [81]3 years ago
8 0

Given :

Challenge given is to volunteer for about 1 1/2 hours per week .

Autum did at a rate of 6 3/4 hours per week .

To Find :

The comparison ( ratio )  her volunteers rate with the challenge rate .

Solution :

Simplifying challenge rate in in normal fraction , 1\dfrac{1}{2}=\dfrac{2+1}{2}=\dfrac{3}{2} .

Also , simplifying her rate , 6\dfrac{3}{4}=\dfrac{(6\times 4)+3}{4}=\dfrac{27}{4} .

Now , ratio of his work by challenge is :

R=\dfrac{\dfrac{27}{4}}{\dfrac{3}{2}}\\\\\\R=\dfrac{27\times 2}{4\times 3}\\\\R=\dfrac{9}{2}

Therefore , the ratio is 9/2 or 4 1/2 .

Hence , this is the required solution .

Anit [1.1K]3 years ago
6 0

Answer:

its 2 1/4

Step-by-step explanation:

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The difference between 3 times a number x and 2 is 19. What is the value of x?
Ber [7]

Answer:

the value of x is 7.

Step-by-step explanation:

2+19=21

21÷3=7

Check:

3×7=21

21-2=19

Therefore, 7 is the answer.

6 0
3 years ago
What is the substitution for 3x+4y^2 if x= -2 and y=3?
solniwko [45]

Answer:

<h2>30</h2>

Step-by-step explanation:

3x+4y^2\\\\\text{Substitute the value of x = -2 and y = 3 to the expression:}\\\\3(-2)+4(3^2)\qquad\text{use PEMDAS}\\\\=-6+4(9)=-6+36=30

5 0
4 years ago
Combine like terms<br> 30x + 6y - 24x - y =
Irina18 [472]

Answer:

6x+5y

Step-by-step explanation:

Combining like terms (x and y)

we get

30x-24x=6x

6y-y=5y

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8 0
3 years ago
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I need help <br>perform the indicated operation <br>​
RoseWind [281]
I’m pretty sure it’s x-3y over 2
6 0
3 years ago
If 8 identical blackboards are to be divided among 4 schools,how many divisions are possible? How many, if each school mustrecei
MAXImum [283]

Answer:

There are 165 ways to distribute the blackboards between the schools. If at least 1 blackboard goes to each school, then we only have 35 ways.

Step-by-step explanation:

Essentially, this is a problem of balls and sticks. The 8 identical blackboards can be represented as 8 balls, and you assign them to each school by using 3 sticks. Basically each school receives an amount of blackboards equivalent to the amount of balls between 2 sticks: The first school gets all the balls before the first stick, the second school gets all the balls between stick 1 and stick 2, the third school gets the balls between sticks 2 and 3 and the last school gets all remaining balls.

 The problem reduces to take 11 consecutive spots which we will use to localize the balls and the sticks and select 3 places to put the sticks. The amount of ways to do this is {11 \choose 3} = 165 . As a result, we have 165 ways to distribute the blackboards.

If each school needs at least 1 blackboard you can give 1 blackbooard to each of them first and distribute the remaining 4 the same way we did before. This time there will be 4 balls and 3 sticks, so we have to put 3 sticks in 7 spaces (if a school takes what it is between 2 sticks that doesnt have balls between, then that school only gets the first blackboard we assigned to it previously). The amount of ways to localize the sticks is {7 \choose 3} = 35. Thus, there are only 35 ways to distribute the blackboards in this case.

4 0
3 years ago
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