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SOVA2 [1]
3 years ago
14

Paul had some candy to give to his three children. He first took four pieces for himself and then evenly divided the rest among

his children. Each child received two pieces. With how many pieces did he
start?
Mathematics
2 answers:
Vesna [10]3 years ago
6 0
He started out with 10 pieces
jekas [21]3 years ago
3 0

Answer:

10 pieces

Step-by-step explanation:

Work backwards

Multiply 3 by 2 to get 6

Add 4 to 6 to get 10

Check:

10 - 4 = 6

6/3 = 2

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Use function notation to write the equation of the line
blondinia [14]

f(x)=mx+b\to y=mx+b

We have two points:

(-3,\ 0);\ (-4,\ 7)

The point slope form of the line:

y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}

Substitute

m=\dfrac{7-0}{-4-(-3)}=\dfrac{7}{-1}=-7

y-0=-7(x-(-3))\\\\y=-7(x+3)\\\\y=-7x-21

Answer:

f(x)=-7x-21

8 0
3 years ago
Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the
Alexxandr [17]

Answer:

Step-by-step explanation:

The table can be computed as:

Advertising Expenses ($ million)     Number of companies

25 up to 35                                                    4

35 up to 45                                                    19

45 up to 55                                                    27

55 up to 65                                                    16

65 up to 75                                                     9

TOTAL                                                            75

Let's find the probabilities first:

P(25 - 35) = P \Big(\dfrac{25-50.93}{10.80}

For 35 up to 45

P(35 - 45) = P \Big(\dfrac{35-50.93}{10.80}

For 45 up to 55

P(45 - 55) = P \Big(\dfrac{45-50.93}{10.80}

For 55 up to 65

P(55 - 65) = P \Big(\dfrac{55-50.93}{10.80}

For 65 up to 75

P(65 - 75) = P \Big(\dfrac{65-50.93}{10.80}

Chi-Square Table can be computed as follows:

Expense   No of   Probabilities(P)  Expe                (O-E)^2   \dfrac{(O-E)^2}{E}

             compa                                 cted E (n*p)

             nies (O)  

25-35           4      0.0612    75*0.0612 = 4.59        0.3481       0.0758

35-45           19     0.2218   75*0.2218 = 16.635     5.5932       0.3362

45-55           27     0.3568   75*0.3568 = 26.76     0.0576      0.021

55-65           16      0.2552   75*0.2552 = 19.14      9.8596      0.5151

65-75           9      0.0839     75*0.0839 = 6.2925   7.331         1.1650

                                                                                           \sum \dfrac{(O-E)^2}{E}= 2.0492                                                                                                      

Using the Chi-square formula:

X^2 = \dfrac{(O-E)^2}{E} \\ \\ Chi-square  \ X^2 = 2.0942

Null hypothesis:

H_o: \text{The population of advertising expenses follows a normal distribution}

Alternative hypothesis:  

H_a: \text{The population of advertising expenses does not follows a normal distribution}

Assume that:

\alpha = 0.02

degree of freedom:

= n-1

= 5 -1

= 4

Critical value from X^2 = 11.667

Decision rule: To reject H_o  \  if \  X^2  test statistics is greater than X^2 tabulated.

Conclusion: Since X^2 = 2.0942 is less than critical value 11.667. Then we fail to reject H_o

6 0
3 years ago
What is LCM of 2 4 8​
slavikrds [6]

Answer:

the LCM would be 8 based on the following set of multiples: Multiples of 2: 2, 4, 6, 8, 10, 12, 14, ... Multiples of 8: 8, 16, 24, 32, 40, 48, 56, ...

Step-by-step explanation:

3 0
3 years ago
Previous
PilotLPTM [1.2K]

just realized this is an old question

7 0
3 years ago
What is the measure of the missing angle?
AURORKA [14]

Answer:

Es 60

Step-by-step explanation:

7 0
3 years ago
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