Need help with this math

A phone company charges a base fee of $12 per month plus an additional charge per minute. The monthly phone cost C can be repres

Rina8888 [55]

Answer:

It’s actually m = (C - 12) /3 But I don’t see it in the 4 option you have.

Step-by-step explanation:

C = 12+ am

C - 12 = am

(C - 12) /a = m

3 0

2 years ago

Read 2 more answers

Suppose twenty-two communities have an average of = 123.6 reported cases of larceny per year. assume that σ is known to be 36.8

Delvig [45]

We are given the following data:

Average = m = 123.6
Population standard deviation = σ= psd = 36.8
Sample Size = n = 22

We are to find the confidence intervals for 90%, 95% and 98% confidence level.

Since the population standard deviation is known, and sample size is not too small, we can use standard normal distribution to find the confidence intervals.

Part 1) 90% Confidence Interval
z value for 90% confidence interval = 1.645

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-1.645* \frac{36.8}{ \sqrt{22}}=110.69$

Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+1.645* \frac{36.8}{ \sqrt{22}}=136.51$

Thus the 90% confidence interval will be (110.69, 136.51)

Part 2) 95% Confidence Interval
z value for 95% confidence interval = 1.96

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-1.96* \frac{36.8}{ \sqrt{22}}=108.22$

Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+1.96* \frac{36.8}{ \sqrt{22}}=138.98$

Thus the 95% confidence interval will be (108.22, 138.98)

Part 3) 98% Confidence Interval
z value for 98% confidence interval = 2.327

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-2.327* \frac{36.8}{ \sqrt{22}}=105.34$
Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+2.327* \frac{36.8}{ \sqrt{22}}=141.86$

Thus the 98% confidence interval will be (105.34, 141.86)

Part 4) Comparison of Confidence Intervals
The 90% confidence interval is: (110.69, 136.51)
The 95% confidence interval is: (108.22, 138.98)
The 98% confidence interval is: (105.34, 141.86)

As the level of confidence is increasing, the width of confidence interval is also increasing. So we can conclude that increasing the confidence level increases the width of confidence intervals.

3 0

3 years ago

What is a polynomial function in standard form with zeroes 1, 2, -2. and -3

Juli2301 [7.4K]

<span>A polynomial with the given zeros can be represented as
f(x) = (x-1)(x-2)(x+2)(x+3).
Note that if you set f(x) = 0, then 1,2,-2, and -3 certainly are the solutions. From here, we simply multiply/expand out the polynomial. We can do this in a variety of ways, one of which is taking the left two and right two products separately. We have
(x-1)(x-2) = x^2 - 3x + 2
and
(x+2)(x+3) = x^2 + 5x + 6.
This gives that
f(x) = (x^2 - 3x + 2) (x^2 + 5x + 6).
Expanding this expression out then gives us our answer as
f(x) = x^4 + 2x^3 - 7x^2 - 8x + 12
or answer choice B.</span>

3 0

3 years ago

The population of a small town can be described by the equation PN equals 4000+70n. Where n is the number of years after 2005. E

ruslelena [56]

Answer:

The current population (in 2005) is 4000 because it is the constant. The slope tells us how many people are added to the population, so 70 per year. For ex. 2008 (2008- 2005 = 3=> 3 x 70 = 210 => 210 + 4000 = 4210).

8 0

3 years ago

Intellectual development (Perry) scores were determined for 21 students in a first-year, project-based design course. (Recall th

Anit [1.1K]

Answer:

The 99% confidence interval is (3.0493, 3.4907).

We are 99% sure that the true mean of the students Perry score is in the above interval.

Step-by-step explanation:

Our sample size is 21.

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

$df = 21-1 = 20$ .

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.99}{2} = \frac{0.01}{2} = 0.005$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 20 and 0.005 in the two-sided t-distribution table, we have $T = 2.528$