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4 years ago

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For her birthday, Sarah’s parents got her a new phone. The phone’s plan costs $59.99 a month and allows for unlimited calls, unl

imited web access, and 100 text messages a month. In addition, each text message over the 100 included with the plan is $0.10 . If Sarah’s parents budget $78.00 a month for her cell phone bill, what is the maximum number of text messages that she can send in one month?

1 answer:

S_A_V [24]4 years ago

5 0

78.00 = 59.99+0.10x

10.01=0.10x

x= 180.1 so she can send 180 texts plus the 100 free ones for a total of 280 texts

double check: 180 x 0.10 = 18.00 + 59.99 = 77.99

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Answer:

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Step-by-step explanation:

<u>Arithmetic Sequences</u>

The arithmetic sequences are identified because any term n is obtained by adding or subtracting a fixed number to the previous term. That number is called the common difference.

The equation to calculate the nth term of an arithmetic sequence is:

$a_n=a_1+(n-1)r$

Where

an = nth term

a1 = first term

r = common difference

n = number of the term

We are given the first terms of a sequence:

-12, -28, -44,...

Find the common difference by subtracting consecutive terms:

r = -28 - (-12) = -16

r = -44 - (-28) = -16

The first term is a1 = -12. Now we calculate the term n=61:

$a_{61}=-12+(61-1)(-16)$

$a_{61}=-12-60*16=-12-960$

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Determine if the graph is symmetric about the x-axis, the y-axis, or the origin. r = 2 cos 5θ

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Looking at the graph, we'll notice that there is a local maximum at x=0 and it looks similar on both sides of the y axis, therefore making it symmetric around the y axis given the options

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Answer:

a) $72-1.52\frac{10}{\sqrt{150}}=70.76$

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So on this case the 87% confidence interval would be given by (70.76;73.24)

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And the margin of error is given by:

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And replacing we can solve for the critical value $t_{\alpha/2}$

And we got:

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And we can find the significance level with this formula:

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And the confidence level would be 1-0.000332= 0.999668

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Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=72$ represent the sample mean

$\mu$ population mean (variable of interest)

$s=10$ represent the sample standard deviation

n=150 represent the sample size

87% confidence interval

Part a

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The degrees of freedom are given by:

$df = 150-1=149$

Since the Confidence is 0.87 or 87%, the value of $\alpha=0.13$ and $\alpha/2 =0.065$ , and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.065,149)".And we see that $z_{\alpha/2}=1.52$

Now we have everything in order to replace into formula (1):

$72-1.52\frac{10}{\sqrt{150}}=70.76$

$72+1.52\frac{10}{\sqrt{150}}=73.24$

So on this case the 87% confidence interval would be given by (70.76;73.24)

Part b

For this case the confidence interval is given by: (69,75)

We can estimate the margin of error like this:

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And the margin of error is given by:

$ME=t_{\alpha/2}\frac{s}{\sqrt{n}}$

And replacing we can solve for the critical value $t_{\alpha/2}$

And we got:

$t_{\alpha/2}= \frac{ME*\sqrt{n}}{s}= \frac{3* \sqrt{150}}{10}= 3.674$

And we can find the significance level with this formula:

"=2*T.DIST(-3.674;149;TRUE)"

And the confidence level would be 1-0.000332= 0.999668

And then that correspond to 99.96%

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3 years ago

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Answer:

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Step-by-step explanation:

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