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3 years ago

C = 0.20m+2.00 what is the charge for 2.7 mile ride?

1 answer:

3 0

M = number of miles
m = 2.7 is plugged into the equation to get
C = 0.20*m + 2.00
C = 0.20*2.7 + 2.00
C = 0.54 + 2.00
C = 2.54
The cost is $2.54

You might be interested in

Mr and mrs. Dorsey and their three children are fling to Springfield. The cost of each ticket is $179. Estimate how much the tic

Soloha48 [4]

First figure out how many people there are all together. So mrs and mr is 2 there three children so 2+3=5 so next round 179 to 200 and multiply 200x5. To do that you can use the mental math way witch would be to take the whole number out of 200 and make it 2. Then take the 5 in 200x5 and multiply 2x5. Then add the 2 zeros there were. And your estamated answer would be 1,000.
2+3=5
179=200
200x5
2x5=10
1,000
To figure out your real answer you will have to take 179 and multiply it by 5 so it will look like this:
34
179
x
5
895
Soto conculde the answer would be:

Estamated: $1,000

Exact: $895

Hope this helps!

7 0

3 years ago

Two students are inviting friends to a mutual party,each student is inviting 5 of their 15 mutual friends, what is the probabili

Nesterboy [21]

Answer:

Step-by-step explanation:

7 0

4 years ago

What is the slope of the line? A. -1 B. 0 C. 1 D. 20

sp2606 [1]

The answer is B) 100%

3 0

3 years ago

Read 2 more answers

What is the solution to –2(5y – 5) – 3y ≤ –7y + 22?

Gre4nikov [31]

-2(5y - 5) - 3y < = -7y + 22
-10y + 10 - 3y < = -7y + 22
-13y + 10 < = -7y + 22
-13y + 7y < = 22 - 10
-6y < = 12
y > = -12/6
y > = -2....answer B

6 0

3 years ago

Read 2 more answers

The physical plant at the main campus of a large state university recieves daily requests to replace florecent lightbulbs. The d

jek_recluse [69]

Answer:

50%

Step-by-step explanation:

68-95-99.7 rule

68% of all values lie within the 1 standard deviation from mean $(\mu-\sigma,\mu+\sigma)$

95% of all values lie within the 1 standard deviation from mean $(\mu-1\sigma,\mu+1\sigma)$

99.7% of all values lie within the 1 standard deviation from mean $(\mu-3\sigma,\mu+3\sigma)$

The distribution of the number of daily requests is bell-shaped and has a mean of 55 and a standard deviation of 4.

$\mu = 55\\\sigma = 4$

68% of all values lie within the 1 standard deviation from mean $(\mu-\sigma,\mu+\sigma)$ = $(55-4,55+4)$ = $(51,59)$

95% of all values lie within the 2 standard deviation from mean $(\mu-1\sigma,\mu+1\sigma)$ = $(55-2(4),55+2(4))$ = $(47,63)$

99.7% of all values lie within the 3 standard deviation from mean $(\mu-3\sigma,\mu+3\sigma)$ = $(55-3(4),55+3(4))$ = $(43,67)$

Refer the attached figure

P(43<x<55)=2.5%+13.5%+34%=50%

Hence The approximate percentage of light bulb replacement requests numbering between 43 and 55 is 50%

4 0

3 years ago