How

P³ = 1/8 please help me if anybody can .

ivanzaharov [21]

Answer:

$p= \dfrac{1}{2}$

Step-by-step explanation:

Given equation:

$p^3=\dfrac{1}{8}$

Cube root both sides:

$\implies \sqrt[3]{p^3}= \sqrt[3]{\dfrac{1}{8}}$

$\implies p= \sqrt[3]{\dfrac{1}{8}}$

$\textsf{Apply exponent rule} \quad \sqrt[n]{a}=a^{\frac{1}{n}}:$

$\implies p= \left(\dfrac{1}{8}\right)^{\frac{1}{3}}$

$\textsf{Apply exponent rule} \quad \left(\dfrac{a}{b}\right)^c=\dfrac{a^c}{b^c}:$

$\implies p= \dfrac{1^{\frac{1}{3}}}{8^{\frac{1}{3}}}$

$\textsf{Apply exponent rule} \quad 1^a=1:$

$\implies p= \dfrac{1}{8^{\frac{1}{3}}}$

Rewrite 8 as 2³:

$\implies p= \dfrac{1}{(2^3)^{\frac{1}{3}}}$

$\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}:$

$\implies p= \dfrac{1}{2^{(3 \cdot \frac{1}{3})}}$

Simplify:

$\implies p= \dfrac{1}{2^{\frac{3}{3}}}$

$\implies p= \dfrac{1}{2^{1}}$

$\implies p= \dfrac{1}{2}$

3 0

1 year ago

Read 2 more answers

Luda [366]

Answer:

a) H0: There is no association between level of education and TV station preference (Independence)

H1: There is association between level of education and TV station preference (No independence)

b) $\chi^2 = \frac{(15-10)^2}{10}+\frac{(15-20)^2}{20}+\frac{(10-10)^2}{10}+\frac{(5-10)^2}{10}+\frac{(25-10)^2}{10}+\frac{(10-20)^2}{20} =33.75$

c) $\chi^2_{crit}=5.991$

d) Since the p value is lower than the significance level we enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that we have dependence between the two variables analyzed.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

High school Some College Bachelor or higher Total

Public Broadcasting 15 15 10 40

Commercial stations 5 25 10 40

Total 20 40 20 80

We need to conduct a chi square test in order to check the following hypothesis:

Part a

H0: There is no association between level of education and TV station preference (Independence)

H1: There is association between level of education and TV station preference (No independence)

The level os significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

Part b

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{20*40}{80}=10$

$E_{2} =\frac{40*40}{80}=20$

$E_{3} =\frac{20*40}{80}=10$

$E_{4} =\frac{20*40}{80}=10$

$E_{5} =\frac{40*40}{80}=20$

$E_{6} =\frac{20*40}{80}=10$

And the expected values are given by:

High school Some College Bachelor or higher Total

Public Broadcasting 10 20 10 40

Commercial stations 10 10 20 40

Total 20 30 30 80

Part b

And now we can calculate the statistic:

$\chi^2 = \frac{(15-10)^2}{10}+\frac{(15-20)^2}{20}+\frac{(10-10)^2}{10}+\frac{(5-10)^2}{10}+\frac{(25-10)^2}{10}+\frac{(10-20)^2}{20} =33.75$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(2-1)(3-1)=2$

Part c

In order to find the critical value we need to look on the right tail of the chi square distribution with 2 degrees of freedom a value that accumulates 0.05 of the area. And this value is $\chi^2_{crit}=5.991$

Part d

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >33.75)=2.23x10^{-7}$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(33.75,2,TRUE)"

Since the p value is lower than the significance level we enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that we have dependence between the two variables analyzed.

7 0

4 years ago

What is the value of the expression 9+n/3-6 when n is 12?

Nina [5.8K]

-7 (9+12/3-6)=21/-3=-7

4 0

3 years ago

Read 2 more answers

Mr. Evans is paid $9.20 per hour for the first 40 hours he works in a week.He is paid 1.5 that rate for each hour after that . l

stiks02 [169]

No, Mr.Evans will make under $400. His base pay, 9.20 times the hours he worked at regular pay(40) would equal $368. Now, take those extra 2.25 hours he worked at 1.5 times his base pay of 9.20, which would give you $13.80. Now, multiply 13.80 and 2.25, and you get 31.05. These two amount added, would equal $399.05, leaving Mr. Evans at just under $400 earned that week.

3 0

3 years ago

The number of books read by a mathematician 2 years after he turned 25 is given by the polynomial

stich3 [128]

Evaluating a function means to plug the required value, substituting every x occurrence with that value.

So, you have

$f(x)=3x^2+80x \implies f(2) = 3\cdot 2^2+80\cdot 2 = 3\cdot 4 + 160 = 12+160=172$

5 0

3 years ago