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uysha [10]
2 years ago
14

The telephone company offers two billing plans for local calls. Plan 1 charges ​$27 per month for unlimited calls and Plan 2 cha

rges ​$11 per month plus ​$0.05 per call. A. Use an inequality to find the number of monthly calls for which Plan 1 is more economical than Plan 2.
Mathematics
1 answer:
mylen [45]2 years ago
3 0

Answer:

<h2>0.05x+11<27 for x<=320</h2>

Step-by-step explanation:

Step one:

Plan 1

charges = $27 monthly

the total charges is expressed as

y=27------1

Plan 2

charges= $11 monthly

extral= $0.05 per call

let x be the number of calls

the total is expressed linearly as

y=0.05x+11--------2

Step two:

equating 12 and 2 we have

27=0.05x+11

collect like terms

27-11=0.05x

16=0.05x

divide both sides by 0.05

x=16/0.05

x=320

the inequality for which plan A is more economical is

0.05x+11<27 for x<=320

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3 years ago
In a recent year, Washington State public school students taking a mathematics assessment test had a mean score of 276.1 and a s
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Answer:

a) \mu_{\bar x} =\mu = 276.1

\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3

b) From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)

c) P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)

P(Z\geq2.070)=1-P(Z

Step-by-step explanation:

Let X the random variable the represent the scores for the test analyzed. We know that:

\mu=E(X) = 276.1 , \sigma=Sd(X) = 34.4

And we select a sample size of 64.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Part a

For this case the mean and standard error for the sample mean would be given by:

\mu_{\bar x} =\mu = 276.1

\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3

Part b

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)

Part c

For this case we want this probability:

P(\bar X \geq 285)

And we can use the z score defined as:

z=\frac{\bar x -\mu}{\sigma_{\bar x}}

And using this we got:

P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)

And using a calculator, excel or the normal standard table we have that:

P(Z\geq2.070)=1-P(Z

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Answer:

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Step-by-step explanation:

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We have 18 boxes that weigh 7.2 lbs

18x = 7.2

Divide each side by 18

18x/18 = 7.2/18

x =.4

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Step-by-step explanation:

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