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SpyIntel [72]
3 years ago
5

A phone company offers two monthly plans. Plan A costs $10 plus an additional $0.15 for each minute of calls. Plan B costs $19 p

lus and additional $0.10 for each minute of calls.
For what amount of calling do the two plants cost the same?

What is the cost when the tow plans cost the same?
Mathematics
1 answer:
Len [333]3 years ago
7 0

Answer:

Step-by-step explanation:

A phone company offers two monthly plans. Plan A costs $10 plus an additional $0.15 for each minute of calls. Plan B costs $19 plus and additional $0.10 for each minute of calls.

For what amount of calling do the two plants cost the same?

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Write an equation of the form y=mx for the pairs (-1,4) and (1,-4)
hram777 [196]
Y = mx
We just need to find the slope:

y2 - y1 / x2 -x1 = -4 - 4 / 1 - - 1 = -8/2 = -4/1 = -4

so:  y = -4x

Hope that helps! :)
5 0
3 years ago
71+50+6x-1=180<br> excuse the 121
UNO [17]

Answer:

9 1/3

Step-by-step explanation:

That is wrong i tried to answer it but messed up so don't use it

6 0
2 years ago
Read 2 more answers
Fiona made a score of 76 on her midterm exam for her to get a b in the course the average of her midterm exam and final exam mus
qaws [65]

Answer:

84 <= x <= 102

Step-by-step explanation:

Fiona scored 76 on her 1st midterm. Lets call x the score she will get on her 2nd exam.

We know she needs between 80 and 89 inclusive so get a b. So the question is, how much does she need to get on the 2nd test? Or, what values x can take for her to get a b?

The average she gets between both test is the arithmetic mean, this is, the sum of both tests divided by 2:

(76+x)/2

So, she need this value to be between 80 and 89. Lets use inequalities:

80 <= (76+x)/2 <= 89

We can solve it in a similar way we do for equalities. Lets start by multiplying terms in every side by 2, to eliminate the 2 dividing the central term:

2*80 <= 76+x <= 89*2

160 <= 76+x <= 178

Now, lets subtract 76 in every side so we get x alone in the center:

160 - 76 <= x <= 178 - 76

84 <= x <= 102

This is, for getting a b she need her second score to be equal or greater than 84 and less or equal to 102.

Lets verify for some values. For example, if she gets 88:

(76+84)/2 = 160/2 =80---> verifies!

And any value greater than 84 will give her a better score. Lets see what happens with 102"

(76+102)/2 = 178/2 = 89 ---> verifies

Any score greater than 102 will giver here a total sore greater than 89 so it won't be a b.

So, if she scores 102 she will get 89 (the statement does not say the range of scores, it is not usually common that the pass 100, but as the statement does not restrict it we consider here. We the maximum score is 100 we will use 84 <= x <= 100).

3 0
3 years ago
In 2018, Mike Krzyewski and John Calipari topped the list of highest paid college basketball coaches (Sports Illustrated website
expeople1 [14]

From the data given, we estimate the population mean and population standard deviation. Then, we use this estimate to find a 95% confidence interval for the population variance and the population standard deviation.

Sample:

Salaries in millions of dollars: 2.2, 1.5, 0.5, 1.3, 2.4, 1.5, 2.7, 0.3, 2.0, 0.3

Question a:

The mean is the sum of all values divided by the number of values. So

\overline{x} = \frac{2.2 + 1.5 + 0.5 + 1.3 + 2.4 + 1.5 + 2.7 + 0.3 + 2.0 + 0.3}{10} = 1.42

The sample mean salary is of 1.42 million.

Question b:

The standard deviation is the square root of the difference squared between each value and the mean, divided by one less than the number of values.

So

s = \sqrt{\frac{(2.2-1.42)^2 + (1.5-1.42)^2 + (0.5-1.42)^2 + (1.3-1.42)^2 + (2.4-1.42)^2 + (1.5-1.42)^2 + (2.7-1.42)^2 + ...}{9}} = 0.8772

Thus, the estimate for the population standard deviation is of 0.8772 million.

Question c:

The sample size is n = 10

The significance level is \alpha = 1 - 0.05 = 0.95

The estimate, which is the sample standard deviation, is of s = 0.8772.

Now, we have to find the critical values for the Pearson distribution. They are:

\chi^2_{\frac{\alpha}{2},n-1} = \chi^2_{0.025,9} = 19.0228

\chi^2_{1-\frac{\alpha}{2},n-1} = \chi^2_{0.975,9} = 2.7004

The confidence interval for the population variance is:

\frac{(n-1)s^2}{\chi^2_{\frac{\alpha}{2},n-1}} < \sigma^2 < \frac{(n-1)s^2}{\chi^2_{1-\frac{\alpha}{2},n-1}}

\frac{9*0.8772^2}{19.0228} < \sigma^2 < \frac{9*0.8772^2}{2.7004}

0.3641 < \sigma^2 < 2.5646

Thus, the 95% confidence interval for the population variance is (0.3641, 2.5646)

Question d:

Standard deviation is the square root of variance, so:

\sqrt{0.3641} = 0.6034

\sqrt{2.5646} = 1.6014

The 95% confidence interval for the population standard deviation is (0.6034, 1.6014).

For more on confidence intervals for population mean/standard deviation, you can check brainly.com/question/13807706

4 0
2 years ago
What are expressions like 4(3+2) and 4(3)+4(2) called
Mariulka [41]
The distribution process
4 0
3 years ago
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