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3 years ago

-(1+7x)=6(-7-x)+36 Alg 2

Mathematics

1 answer:

olganol [36]3 years ago

7 0

Let's solve your equation step-by-step.

−(1+7x)=6(−7−x)+36

Step 1: Simplify both sides of the equation.

−(1+7x)=6(−7−x)+36

−7x+−1=(6)(−7)+(6)(−x)+36(Distribute)

−7x+−1=−42+−6x+36

−7x−1=(−6x)+(−42+36)(Combine Like Terms)

−7x−1=−6x+−6

−7x−1=−6x−6

Step 2: Add 6x to both sides.

−7x−1+6x=−6x−6+6x

−x−1=−6

Step 3: Add 1 to both sides.

−x−1+1=−6+1

−x=−5

Step 4: Divide both sides by -1.

−x

−1

−5

−1

x=5

Answer:

x=5

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Probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

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We are given that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15.

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Let $\bar X$ = sample average time spent studying

The z-score probability distribution for sample mean is given by;

Z = $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ ~ N(0,1)

where, $\mu$ = population mean hours spent studying = 25 hours

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The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the average time spent studying for the sample was between 29 and 30 hours studying is given by = P(29 hours < $\bar X$ < 30 hours)

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P( $\bar X$ < 30 hours) = P( $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ < $\frac{ 30-25}{\frac{15}{\sqrt{36} } }} }$ ) = P(Z < 2) = 0.97725

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Therefore, P(29 hours < $\bar X$ < 30 hours) = 0.97725 - 0.94520 = 0.0321

Hence, the probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

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