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enot [183]
2 years ago
10

A cab company charges according to the following rate chart. Which graph correctly represents the charge, y, in dollars for a ca

b ride of x miles? Cab Rates by Mile Miles Charge Flag drop fee (0 miles) $5.00 g miles $1.00 per mile miles $2.00 per mile for each mile over 10
Mathematics
1 answer:
77julia77 [94]2 years ago
4 0

The graph is the one that correctly represents the linear piece-wise function is the first graph.

<h3>How to solve for the graph</h3>

First we have a linear function as

Mx + b

mc= slope, this is the rate of change, how y changes as x changes by 1.

b = y-intercept, y value as x = 0

y = 5 + x

x ≤ 10

at 10 we have 5 + 10 = 15

at 2 dollars

y = 15 + 2(x - 10)

at 14

y = 15 + 2(14 - 10)

15 + 2*4

= 23

Hence we can conclude that the first graph here is the correct one.

Read more on graphs here:

brainly.com/question/22295833

#SPJ1

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The ratio of girls to boys in the junior high band is 5 to 7. At the beginning of year, there were 72 students in the band. By t
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Ratio is 5girls:7boys with total 72 students
5+7= 12
72➗12= 6
6(5:7)= 30 girls : 42 boys

New ratio is 3:4 with 48 boys

48➗4= 12

So for girls 3x12= 36

So started with 30 girls ended with 36 girls... 6 girls joined band during the school year.
7 0
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10–{22–[(−9)+(−11)]} help i'M HAVING A bad day :( fix it
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Answer:

-32

Step-by-step explanation:

10–{22–[(−9)+(−11)]}

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10–{22–[(-20)]}

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7 0
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The ball game begins at 2:00 pm.It takes Ying 30 minutes to get to the ballpark.At what time should Ying leave home to get to th
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What fraction is equivalent 2/6
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4 0
3 years ago
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If you can explain your answer that’d be great!! Thank you!
Rasek [7]

Answer:

A(t)=A_{0}e^{\frac{ln(\frac{1}{2})}{22}t}

Step-by-step explanation:

This half life exponential decay equation goes by the formula:

A(t)=A_{0}e^{kt}

Where

k=\frac{ln(\frac{1}{2})}{Half-Life}

Since half life is given as 22, we plug that into "Half-Life" in the formula for k and then plug in the formula for k into the exponential decay formula:

So,

k=\frac{ln(\frac{1}{2})}{Half-Life}\\k=\frac{ln(\frac{1}{2})}{22}

Now

A(t)=A_{0}e^{\frac{ln(\frac{1}{2})}{22}t}

third choice is correct.

8 0
3 years ago
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