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Tpy6a [65]
2 years ago
14

Jane’s cell phone plan is $58 each month plus $0.20 per minute for each minute over 200 minutes of call time. Jane’s cell phone

bill last month was $67.00. Write an equation that can be used to determine the number of minutes over 200 that Jane was billed. How many minutes over 200 was Jane billed last month?
m = number of minutes over 200

what is the equation

what is the number of minutes
Mathematics
1 answer:
Lyrx [107]2 years ago
6 0

Answer:

m = (67$ - 58$)/0.20$

_____________________

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30 POINTS !!!
Soloha48 [4]

Each ticket costs $1.25

Jermaine bought 25 of them, which costs 25 x 1.25

25 x 1.25 = 31.25

He spent a total of $43.75 from the tickets and price of admission

we know the tickets cost 31.25, so the admission cost was 43.75 - 31.25

43.75 - 31.25 = $12.5

b) y = 1.25x + 12.5

Where y is the total cost, 1.25 is the price of each ticket, x is how many tickets a person bought, and 12.5 is the admission fee.

Hope it helps :)

7 0
3 years ago
I need help please help me :(
mario62 [17]

Answer:

A=9

Step-by-step explanation:

15÷(3+2)×3

=15÷5×3

=3×3

=9

Hope this helps.

5 0
2 years ago
Find the length of the midsegment RT
olga55 [171]
To find the mid segment, it is half of the sum of the bases
for number 13 it's 7+13=20 20/2=10
number 14 it's 12+9=21 21/2=10.5
number 15 it's 22+18=40 40/2=20
4 0
2 years ago
I'm a sophomore taking trigonometry any advice
krok68 [10]
I took it 9th grade last year, what’s up?
4 0
3 years ago
Find the indicated probability or percentage for the sampling error. The distribution of weekly salaries at a large company is r
Flauer [41]

Answer:

The probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 80 employees will be at most $75 is 0.9297.

Step-by-step explanation:

According to the Central Limit Theorem if we have a non-normal population with mean <em>μ</em> and standard deviation <em>σ</em> and appropriately huge random samples (<em>n</em> > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Then, the mean of the distribution of sample means is given by,

\mu_{\bar x}=\mu

And the standard deviation of the distribution of sample means is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

The information provided is:

\mu=\$1000\\\sigma=\$370\\n=80

As <em>n</em> = 80 > 30, the central limit theorem can be used to approximate the sampling distribution of sample mean weekly salaries.

Let \bar X represent the sample mean weekly salaries.

The distribution of \bar X is: \bar X\sim N(\$1000,\ \$41.37)

Now we need to compute the probability of the sampling error made in estimating the mean weekly salary to be at most $75.

The sampling error is the the difference between the estimated value of the parameter and the actual value of the parameter, i.e. in this case the sampling error is, |\bar X-\mu|= 75.

Compute the probability as follows:

P(-75

                                     =P(-1.81

Thus, the probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 80 employees will be at most $75 is 0.9297.

3 0
3 years ago
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