(2x^2)^-4 how to solve this problem

What happens to the graph of a periodic function if the amplitude, a, is between 0 and 1

svetoff [14.1K]

When 0 < a < 1, the amplitude of the graph decreases, causing the slopes of the graph to appear more "flat".

5 0

3 years ago

Write the pair of fractions as a pair of fractions with a common denominator 1/4 and 5/12

zavuch27 [327]

See picture for answer.

4 0

4 years ago

Question Help

hram777 [196]

Answer:

.3%, 21/7000=.003

Step-by-step explanation:

to make it by percent you multipy by 100

5 0

3 years ago

Read 2 more answers

Você é um comandante de uma espaçonave. Sua missão é chegar até Alfa Centauro em cinco anos. A distância do Sol até Alfa Centaur

Deffense [45]

Answer:

The spaceship will get to Alpha Centaur in time

Step-by-step explanation:

<u><em>The complete question in English is</em></u>

You're the commander of a spaceship. Your mission is to reach Alpha Centaur in five years. The distance from the sun to Alpha Centaur is 2.5 x 10^13 miles. The distance from Earth to Sun is approximately 9.3 x 10^7

miles. Your spaceship can travel at the speed of light. You know that light can travel a distance of 5.88 x 10^12 miles in a year. Can you get to Alpha Centaur in time?

step 1

Find the time it takes for the spaceship to travel from Earth to the Sun

The distance from the Earth to Sun is equal to

$9.3*10^7\ miles$

The spaceship can travel at the speed of light

The light can travel a distance of $5.88*10^{12}\ miles$ in a year

so

using a proportion

$\frac{5.88*10^{12}\ miles}{1\ year}=\frac{9.3*10^7\ miles}{x}\\\\x=(9.3*10^7)/5.88*10^{12}\\\\x=1.58*10^{-5} \ years$

This time is very small in years

Convert to minutes

$1.58*10^{-5} \ years=1.58*10^{-5} (365)(24)(60)=8.3\ min$

step 2

Find the time it takes for the spaceship to travel from the Sun to Alpha Centaur

The distance from the the Sun to Alpha Centaur is equal to

$2.5*10^{13}\ miles$

The spaceship can travel at the speed of light

The light can travel a distance of $5.88*10^{12}\ miles$ in a year

so

using a proportion

$\frac{5.88*10^{12}\ miles}{1\ year}=\frac{2.5*10^{13}\ miles}{x}\\\\x=(2.5*10^{13})/5.88*10^{12}\\\\x=4.25\ years$

$4.25\ years< 5\ years$

therefore

The spaceship will get to Alpha Centaur in time

3 0

3 years ago

An automotive part must be machined to close tolerances to be acceptable to customers. Production specifications call for a maxi

Vilka [71]

Answer:

H0: $\sigma \leq 0.0004$

H1: $\sigma >0.0004$

$t=(30-1) [\frac{0.0224}{0.02}]^2 =36.25$

For this case since we have a right tailed test the p value is given by:

$p_v = P(\Chi^2_{29}>36.25)=0.166$

If we compare the p value and the significance level given we see that $p_v >\alpha$ so then we have enough evidence to FAIL to reject the null hypothesis that the true population variance it's lower or equal than 0.0004 at 5% of significance.

Step-by-step explanation:

Previous concepts and notation

The chi-square test is used to check if the standard deviation of a population is equal to a specified value. We can conduct the test "two-sided test or a one-sided test".

n = 30 sample size

$s^2 =0.0005$ represent the sample variance

s= 0.0224 represent the sample deviation

$\sigma_o =0.095$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic

$\alpha=0.05$ significance level

State the null and alternative hypothesis

On this case we want to check if the population variance is higher than 0.0004 (the specification), so the system of hypothesis are:

H0: $\sigma \leq 0.0004$

H1: $\sigma >0.0004$

In order to check the hypothesis we need to calculate th statistic given by the following formula:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

$t=(30-1) [\frac{0.0224}{0.02}]^2 =36.25$

What is the critical value for the test statistic at an α = 0.05 significance level?

Since is a right tailed test the critical zone it's on the right tail of the distribution. On this case we need a quantile on the chi square distribution with df=30-1=19 degrees of freedom that accumulates 0.05 of the area on the right tail and 0.95 on the left tail.

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.95,29)". And our critical value would be $\Chi^2 =42.56$

Since our calculated value is lower than the critical value we to reject the null hypothesis.

What is the approximate p-value of the test?

For this case since we have a right tailed test the p value is given by:

$p_v = P(\Chi^2_{29}>36.25)=0.166$

If we compare the p value and the significance level given we see that $p_v >\alpha$ so then we have enough evidence to FAIL to reject the null hypothesis that the true population variance it's lower or equal than 0.0004 at 5% of significance.

6 0

3 years ago