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DIA [1.3K]
4 years ago
7

Kirk has 42 pieces of candy to divide evenly between his three children if he puts the pieces into three boxes how many pieces o

f candy are there per box
Mathematics
1 answer:
noname [10]4 years ago
3 0

So what you would do is, 42 divided by 3, which would equal 14, so Kirk would put 14 pieces of candy in a box. I hope this could help.

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which inequality represents the following situation 3/5 times 5 less than a number is no more than 27
andrey2020 [161]

The answer is \frac{3}{5}(x-5) \leq 27


Let's build the expression piece by piece, translating sentences into formulas:


"3/5 times..." \to \frac{3}{5} \times \ldots


"5 less than a number..." \to x-5 \ldots


So, the first block becomes


"3/5 times 5 less than a number..." \to \frac{3}{5}(x-5) \ldots


"...is no more than 27" \ldots \leq 27


So, putting the pieces together, you have the answer



6 0
3 years ago
Marty rides his bicycle from home to work at an average rate of 12 miles per hour. If it takes him 20 minutes to get to work, ho
Crank
It would be 4 miles! :)

4 0
3 years ago
What’s 40 41 42 43 44 45 46 47 48 49 50 IQR pls fast!!!!!
aliina [53]

Answer:

IQR = 6

Step-by-step explanation:

Given:

The given set of data is:

40 41 42 43 44 45 46 47 48 49 50

Number of terms, n=11

Therefore, the median is given as (\frac{n+1}{2})^{th} term which is the sixth term. Median = 45

(40 41 42 43 44) <u>45</u> (46 47 48 49 50 )

Now, we find the median of the lower half (Q_1) and upper half (Q_3)of the data that are before and after the median.

The lower half of the data has the values 40 41 <u>42</u> 43 44.

Number of terms are 5. So, median is the third term which is, Q_1=42

The upper half of the data has the values 46 47 <u>48</u> 49 50.

Number of terms are 5. So, median is the third term which is, Q_3=48

Now, IQR is given as the difference of upper half median and lower half median.

IQR=Q_3-Q_1\\IQR=48-42=6

4 0
4 years ago
Read 2 more answers
Angelo brought apples and bananas at the fruit stand. He bought 20 pieces of fruit and spent $11.50 apples cost $.50 and bananas
Jlenok [28]

Angelo bought 14 apples and 6 bananas

<em><u>Solution:</u></em>

Let "a" be the number of apples bought

Let "b" be the number of bananas bought

Cost of 1 apple = $ 0.50

Cost of 1 banana = $ 0.75

<em><u>He bought 20 pieces of fruit. Therefore,</u></em>

number of apples bought + number of bananas bought = 20

a + b = 20 -------- eqn 1

<em><u>He spent $ 11.50. Therefore, we frame a equation as:</u></em>

number of apples x Cost of 1 apple + number of bananas x Cost of 1 banana = 11.50

a \times 0.50 + b \times 0.75 = 11.50

0.50a + 0.75b = 11.50 ------- eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

From eqn 1,

a = 20 - b ------ eqn 3

<em><u>Substitute eqn 3 in eqn 2</u></em>

0.50(20 - b) + 0.75b = 11.50

10 - 0.5b + 0.75b = 11.50

0.25b = 11.50 - 10

0.25b = 1.5

Divide both sides of equation by 0.25

<h3>b = 6</h3>

<em><u>Substitute b = 6 in eqn 3</u></em>

a = 20 - 6

<h3>a = 14</h3>

Thus he bought 14 apples and 6 bananas

3 0
4 years ago
Help me with this question
Law Incorporation [45]

greatest to least: 6.32, 6, 4.27, 4, 3.5 .

I got 4.27 by multiplying 4 x 11, adding three which makes it 47 and then divided it by 11

6 0
3 years ago
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