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3 years ago

6

Mr. Shively’s TV cabinet is 30 inches tall and 40 inches wide. Best Buy is selling 50 inch, 55 inch, 60 inch, and 65 inch TVs. O

f those options, which one should he buy? Explain your answer using mathematics , please explain using Pythagorean theorem.

1 answer:

Simora [160]3 years ago

4 0

Answer:

He should buy a 50 inch TV

Step-by-step explanation:

Using Pythagorean theorem the TV cabinet has a diagonal of :

$\sqrt{30^{2} + 40^{2} } = 50$

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What is 9/10 + 7/15

Yakvenalex [24]

$\huge\text{Hey there!}$

$\mathsf{\dfrac{9}{10}+\dfrac{7}{15}}$

$\large\textsf{FIRST: FIND the LCD (Lowest Common Denominator) then solve}\\\large\textsf{from there!}$

$\large\textsf{If you have calculated it correctly, you should have came up with \underline{\bf 30}}\\\large\textsf{as your LCD (Lowest Common Denominator).}$

$\mathsf{= \dfrac{9\times3}{10\times3}+ \dfrac{7\times2}{15\times2}}$

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$\mathsf{= \dfrac{27+14}{30}}$

$\mathsf{27+ 14=\bf 41}\\\\\mathsf{30+0=\bf 30}$

$\mathsf{= \dfrac{41}{30}}\large\textsf{ which you could convert to }\mathsf{1 \dfrac{11}{30}}$

$\boxed{\boxed{\large\textsf{ANSWER: }\bf \dfrac{41}{30} \large\textsf{ or }\mathsf{\bf 1 \dfrac{11}{30}\large\textsf{ because they both equal the same thing}}}}}\huge\checkmark$

$\large\text{Good luck on your assignment and enjoy your day!}$

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2 years ago

Read 2 more answers

A marketing research firm wishes to compare the prices charged by two supermarket chains—Miller’s and Albert’s. The research fir

AVprozaik [17]

Answer:

a) $F=\frac{s^2_2}{s^2_1}=\frac{1.84^2}{1.4^2}=1.727 \approx 1.73$

b) For this case since we have a two tailed test we have two critical values, and we can find the two values since we need that on each tail the area would be $\alpha/2 =0.025$ , the distribution is the F with 9 df for the numerator and denominator, and we can use the following excel codes to find the critical values:

"=F.INV(0.025,9,9)"

"=F.INV(1-0.025,9,9)"

The two critical values are 0.248 and 4.026, so the rejection zone would be: $F>4.026 \cup F$ since our calculated value is not on the rejection zone we fail to rejec the null hypothesis

Step-by-step explanation:

Data given and notation

$n_1 = 10$ represent the sampe size for the Miller's Stores

$n_2 =10$ represent the sample size for the Albert's stores

$\bar X_1 =121.92$ represent the sample mean for Miller's store

$\bar X_2 =114.81$ represent the sample mean for Albert's store

$s_1 = 1.4$ represent the sample deviation for the Miller's store

$s_2 = 1.84$ represent the sample deviation for the Albert's stores

$s^2_2 = 12.25$ represent the sample variance for the utility stocks

$\alpha=0.05$ represent the significance level provided

Confidence =0.95 or 95%

F test is a statistical test that uses a F Statistic to compare two population variances, with the sample deviations s1 and s2. The F statistic is always positive number since the variance it's always higher than 0. The statistic is given by:

$F=\frac{s^2_2}{s^2_1}$

Solution to the problem

System of hypothesis

We want to test if the variation for th two groups is the same, so the system of hypothesis are:

H0: $\sigma^2_1 = \sigma^2_2$

H1: $\sigma^2_1 \new \sigma^2_2$

a) Calculate the statistic

Now we can calculate the statistic like this:

$F=\frac{s^2_2}{s^2_1}=\frac{1.84^2}{1.4^2}=1.727 \approx 1.73$

Now we can calculate the p value but first we need to calculate the degrees of freedom for the statistic. For the numerator we have $n_1 -1 =10-1=9$ and for the denominator we have $n_2 -1 =10-1=9$ and the F statistic have 9 degrees of freedom for the numerator and 9 for the denominator. And the P value is given by:

P value

$p_v =2*P(F_{9,9}>1.727)=0.428$

And we can use the following excel code to find the p value:"=2*(1-F.DIST(1.727,9,9,TRUE))"

b) Critical value

For this case since we have a two tailed test we have two critical values, and we can find the two values since we need that on each tail the area would be $\alpha/2 =0.025$ , the distribution is the F with 9 df for the numerator and denominator, and we can use the following excel codes to find the critical values:

"=F.INV(0.025,9,9)"

"=F.INV(1-0.025,9,9)"

The two critical values are 0.248 and 4.026, so the rejection zone would be: $F>4.026 \cup F$ since our calculated value is not on the rejection zone we fail to rejec the null hypothesis

Since our calcu

Conclusion

Since the $p_v > \alpha$ we have enough evidence to FAIL to reject the null hypothesis. And we can say that we don't have enough evidence to conclude that the two deviations are different at 5% of significance.

8 0

3 years ago