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SpyIntel [72]
3 years ago
8

Hanna charges an hourly babysitting fee of 6$ for two children,m 7$ for 3 children, and $8 for four or five children.

Mathematics
1 answer:
masya89 [10]3 years ago
6 0
The answer is the third one where:




The relationship is linear for c but not f

WORK:
C GOES UP BY ONE EACH
F STAYS THE SAME FOR TWO, THEN MOVES UP ONE, THEN ONE AGAIN AND STAYS ON 8 TWO TIMES
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100 PLZ HELP NO LINK AND IF YOU GUESS YOU GET REPORTED
Vika [28.1K]
The answer should be (-1/3,3/4) if I did the math correct
6 0
2 years ago
Suppose you have a mean standardized score of 1200 points with a standard deviation of 200 points. This data is normally distrib
Keith_Richards [23]
The score is given by:
z=(x-mean)/s.d
In the question above:
x=1300
mean=1200
s.d=200
Thus:
z=(1300-1200)/200
=100/200
=0.5
The answer is 0.5
5 0
3 years ago
I need help whats the closer 3 or 7 whats in the mittle
maks197457 [2]
34567

5 is in the middle
8 0
3 years ago
Look at the graph shown below:
lina2011 [118]

Answer:

Well in order to find the equation you are looking for you first must find the slope of the line.  Since it gives you two points at (-4, 1) and at (0, 4), use these points to find the slope.  

y = (3/4)x + 4 is the answer

Step-by-step explanation:

Tell me if it is wrong

<em> I would appreciate a brainliest</em>

Have a great day! :3

5 0
3 years ago
Read 2 more answers
1)How many ways can the letters in the word BOOKKEEPER be arranged? (Must show set-up )
sergeinik [125]
Question 1

There are 5 letters (B, O, K, E, R) and there is a total of 10 letters to make up the word.
There are \frac{10!}{(10-6)!6!} ways of arranging the letters, which equal to 210 ways

Question 2

There are seven swimmers in total.
There are \frac{7!}{(7-1)!1!} ways of choosing the first winner, which is 7 ways
There are \frac{6!}{(6-1)!1!} ways of choosing the second winner, which is 6 ways
There are \frac{5!}{(5-1)!1!} ways of choosing the third winner, which is 5 ways
There are 7×6×5=210 ways of choosing first, second, and third winner

Question 3

The probability of eating an orange and a red candy is \frac{15}{31}×\frac{9}{30}, which equals to \frac{9}{62}

The probability of eating two green candies is \frac{7}{31}×\frac{6}{30} which equals to \frac{7}{155}
3 0
3 years ago
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