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1 year ago

10

Maureen spent sh 207 to buy 7 exercise books and 4 pens while Sharon spent sh 165 to buy 5 exercise books and 5 pens of same typ

e.Find the cost of each item

1 answer:

Umnica [9.8K]1 year ago

3 0

Answer:

book 25; pen 8

Step-by-step explanation:

If we let b and p represent the cost of a book and a pen, respectively, then the two purchases can be written as ...

7b +4p = 207

5b +5p = 165

Multiply the second equation by 4/5 and subtract the result from the first equation:

(7b +4p) -4/5(5b +5p) = (207) -4/5(165)

3b = 75 . . . . simplify

b = 25 . . . . . divide by 3

Dividing the second equation by 5 gives ...

b + p = 33

Solving for p, we have ...

p = 33 -b = 33 -25 = 8

The cost of an exercise book is 25; the cost of a pen is 8.

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Step-by-step explanation:

Data given and notation

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$\alpha$ represent the significance level for the hypothesis test.

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Alternative hypothesis: $\mu \neq 37100$

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2 years ago

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Answer:

a) $\bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})$

b) $Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5$

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d) $P(\bar X >28.25)=0.9997=99.97\%$

e) $P(28.25$

Step-by-step explanation:

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".

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".

Let X the random variabl length of advertisements produced by Majesty Video Production Inc. We know from the problem that the distribution for the random variable X is given by:

$X\sim N(\mu =30,\sigma =2)$

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

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b. What is the standard error of the mean time?

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In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

$z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}$

And we want to find this probability:

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In order to answer this question we can use the z score in order to find the probabilities, the formula is given by:

$z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}$

And we want to find this probability:

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2 years ago