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3241004551 [841]
3 years ago
5

A particular chemistry book costs $6 less than a particular physics book, while two such chemistry books and three such physics

books cost a total of $123. Construct two simultaneous equations and solve them using the substitution method.
Mathematics
1 answer:
sladkih [1.3K]3 years ago
3 0

Answer:

Step-by-step explanation:

Let's say the Physics book costs $x

The Chemistry book will cost $x - $6

Two such Chemistry books will cost;

2($x - $6) = $(2x - 12)

Three such Physics books cost $3x

$(2x - 12) + $3x = $123

$5x - $12 = $123

$5x = $111

x = $22\frac{1}{5}

So the Physics book costs;  $22\frac{1}{5}

and the Chemistry book costs  $22\frac{1}{5} - 6 =  $16\frac{1}{5}

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A researcher collected data from a sample of correctional officers and found that on average they made 47,173 dollars a year wit
ollegr [7]

Answer:

a) z = \frac{40000- 47173}{6364} = -1.127

b) z = \frac{50000- 47173}{6364} = 0.444

c) P(X>40000)=P(\frac{X-\mu}{\sigma}>\frac{40000-\mu}{\sigma})=P(Z>\frac{40000-47173}{6364})=P(z>-1.127)

And we can find this probability using the complement rule and the normal standard distribution or excel and we got:

P(z>-1.127)=1-P(z

Step-by-step explanation:

Previous concepts

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the amount of money of a population of interet, and we know the following info:  

Where \bar X=47173 and s=6364

We can assume that the sample size is large and the estimators can be used as a good description for the parameters  \mu \sigma

The z score for 40000 would be:

z = \frac{40000- 47173}{6364} = -1.127

Part b

The z score for 50000 would be:

z = \frac{50000- 47173}{6364} = 0.444

Part c

We are interested on this probability

P(X>40000)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X>40000)=P(\frac{X-\mu}{\sigma}>\frac{40000-\mu}{\sigma})=P(Z>\frac{40000-47173}{6364})=P(z>-1.127)

And we can find this probability using the complement rule and the normal standard distribution or excel and we got:

P(z>-1.127)=1-P(z

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