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stiv31 [10]
3 years ago
10

A phone company charges a base fee of $12 per month plus an additional charge per minute.the monthly phone cost c can be represe

nted by this equation: c=12+a •m, where a is the additional charge per minute, and m is the number of minutes used.
What equation can be used to find the number of minutes a customer used I we know a and c
Mathematics
2 answers:
shtirl [24]3 years ago
8 0

Answer:

Hi there!

Your answer is

(C-12)/A = M

Step-by-step explanation

c=12+a * m

SOLVE FOR M!

C=12+AM

-12

C-12+AM

Divide out A to isolate M

(C-12)/A = M

kupik [55]3 years ago
4 0

Answer:

I believe it's this:

A. M= (C-12)/a  (Not sure.)

Step-by-step explanation:

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A construction company plans to invest in a building project. There is a 30% chance that the company will lose $30,000, a 40% ch
Gnom [1K]

Answer: Option 'A' is correct.

Step-by-step explanation:

Since we have given that

30% chance that the company will lose $30000.

40% chance of a break even that there is no loss and no profit.

30% chance that the company will profit $ 60000.

As we know the formula for "Expectation":

So, Expected value will be

\frac{30}{100}\times (-30000)+\frac{40}{100}\times 0+\frac{30}{100}\times 60000\\\\=03\times (-30000)+0.4\times 0+0.3\times 60000\\\\=-9000+18000\\\\=\$9000

Expected value is $9000. So, the company should proceed with the project.

Hence, Option 'A' is correct.

4 0
3 years ago
Find the constant of proportionality k. Then write an equation for the relationship between x and y.
egoroff_w [7]

Answer:

k= 5

Equation is: y=5x

3 0
3 years ago
Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis
lesya [120]

Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = <u><em>the interval of time between the eruption</em></u>

So, X ~ Normal(\mu=75, \sigma^{2} =20)

The z-score probability distribution for the normal distribution is given by;

                            Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

 

    P(X > 84 min) = P( \frac{X-\mu}{\sigma} > \frac{84-75}{20} ) = P(Z > 0.45) = 1 - P(Z \leq 0.45)

                                                        = 1 - 0.6736 = <u>0.3264</u>

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{13} } } ) = P(Z > 1.62) = 1 - P(Z \leq 1.62)

                                                        = 1 - 0.9474 = <u>0.0526</u>

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{20} } } ) = P(Z > 2.01) = 1 - P(Z \leq 2.01)

                                                        = 1 - 0.9778 = <u>0.0222</u>

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

8 0
3 years ago
In Kailah's class the ratio of girls to boys is 13 to 8. If there were 16 boys, how many girls will there be?
VladimirAG [237]

Answer:

I believe that the answer is 26

Step-by-step explanation:

The ratio of girls to boys is 13 to 8 so there are 13 girls for every 8 boys so if there are 16 boys (8X2) then there are 26 girls (13X2)

5 0
3 years ago
Read 2 more answers
Brett painted three walls. Each wall was 9 ft tall and 12 ft long. How much wall area did he paint
julsineya [31]

Answer:

First we need to calculate the are of each wall, since we alredy knew the length (l) and the width (w) which is the height of the wall in this case:

A = wl = 9 . 12 = 108 (ft²)

We also know that he painted 3 walls, we need to multiply our first result by 3, in other words, the area of wall that Brett painted is the sum of the area of three walls: 108 . 3 = 324 (ft²)

8 0
4 years ago
Read 2 more answers
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