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4 years ago

Simplify (2^3)^–2 ment to put this

Mathematics

2 answers:

bekas [8.4K]4 years ago

6 0

Answer:

1/64

Step-by-step explanation:

$\left(2^3\right)^{-2}\\\\\mathrm{Apply\:exponent\:rule}:\quad \left(a^b\right)^c=a^{bc},\\\left(2^3\right)^{-2}=2^{3\left(-2\right)}\\\\3\left(-2\right)=-6\\\\=2^{-6}\\\\=\frac{1}{2^6}\\\\2^6=64\\\\=\frac{1}{64}$

aleksley [76]4 years ago

4 0

Answer:

$\frac{1}{64}$

Step-by-step explanation:

$\left(2^3\right)^{-2}\\\\\mathrm{Apply\:exponent\:rule}:\quad \left(a^b\right)^c=a^{bc},\:\quad \:a\ge 0\\\left(2^3\right)^{-2}=2^{3\left(-2\right)}\\\\3\left(-2\right)=-6\\=2^{-6}\\\\\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}\\2^{-6}=\frac{1}{2^6}\\\\2^6=64\\\\=\frac{1}{64}$

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Sacramento County high school seniors have an average SAT score of 1,020. From a random sample of 144 Sacramento High School stu

kari74 [83]

Answer:

We need to conduct a hypothesis in order to check if the high school students are representative of the overall population (or if the true mean is 1020 or not), the system of hypothesis would be:

Null hypothesis: $\mu = 1020$

Alternative hypothesis: $\mu \neq 1020$

Step-by-step explanation:

Previous concepts and data given

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".

$\bar X=1100$ represent the sample mean

$s=144$ represent the sample standard deviation

n=144 represent the sample selected

$\alpha$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the high school students are representative of the overall population (or if the true mean is 1020 or not), the system of hypothesis would be:

Null hypothesis: $\mu = 1020$

Alternative hypothesis: $\mu \neq 1020$

If we analyze the size for the sample is > 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{1100-1020}{\frac{144}{\sqrt{144}}}=6.67$

P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=144-1 =143$

Then since is a two sided test the p value would be:

$p_v =2*P(t_{144}>6.67)=2.59x10^{-10}$

Conclusion

If we compare the p value and a significance level for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the population mean is significant different from 1020 at 5% of significance.