Sin(5x) cos(3x)-1/2sin(8x) = sin(x) cos(x)

Show work please~

love history [14]

For every 1 cup of blue paint, $\frac{10}{4}$ cups of red paint are needed

For every 1 cup of red paint, $\frac{4}{10}$ cup of blue paint is needed

For every 4 cups of red paint, $\frac{16}{10}$ cups of blue paint are needed

<em><u>Solution:</u></em>

Given that, there are 3 1/3 red cups of paint for every 1 1/3 cups of blue paint

Therefore, ratio is

$Red : blue = 3\frac{1}{3} : 1\frac{1}{3}\\\\Red : blue = \frac{10}{3} : \frac{4}{3}$

<h3><u>For every 1 cup of blue paint, ___ cups of red paint are needed</u></h3>

Let "x" be the cups of red paint needed

Then we get,

$\frac{10}{3} : \frac{4}{3}\\\\x : 1$

This forms a proportion

$1 \times \frac{10}{3} = \frac{4}{3} \times x\\\\x = \frac{10}{4}$

Therefore, 10/4 cups of red are needed for 1 cup of blue

<h3><u>For every 1 cup of red paint, ___ cup of blue paint is needed</u></h3>

Let "x" be the cups of blue paint needed

Then, we get

$\frac{10}{3} : \frac{4}{3}\\\\1 : x$

This forms a proportion

$\frac{10}{3} \times x = \frac{4}{3} \times 1\\\\x = \frac{4}{10}$

Thus, 4/10 cups of blue are needed for 1 cup of red paint

<h3><u>For every 4 cups of red paint,___ cups of blue paint are needed</u></h3>

Let "x" be the cups of blue paint needed

Then, we get

$\frac{10}{3} : \frac{4}{3}\\\\4 : x$

This forms a proportion

$\frac{10}{3} \times x = \frac{4}{3} \times 4\\\\x = \frac{16}{10}$

Thus 16/10 cups of blue paint are needed for every 4 cups of red paint

5 0

3 years ago

The scientist performs additional analyses and observes that the number of major earthquakes does appear to be decreasing but wo

kifflom [539]

Answer:

Step-by-step explanation:

Hello!

A regression model was determined in order to predict the number of earthquakes above magnitude 7.0 regarding the year.

^Y= 164.67 - 0.07Xi

Y: earthquake above magnitude 7.0

X: year

The researcher wants to test the claim that the regression is statistically significant, i.e. if the year is a good predictor of the number of earthquakes with magnitude above 7.0 If he is correct, you'd expect the slope to be different from zero: β ≠ 0, if the claim is not correct, then the slope will be equal to zero: β = 0

The hypotheses are:

H₀: β = 0

H₁: β ≠ 0

α: 0.05

The statistic for this test is a student's t: $t= \frac{b - \beta }{Sb} ~~t_{n-2}$

The calculated value is in the regression output $t_{H_0}= -3.82$

This test is two-tailed, meaning that the rejection region is divided in two and you'll reject the null hypothesis to small values of t or to high values of t, the p-value for this test will also be divided in two.

The p-value is the probability of obtaining a value as extreme as the one calculated under the null hypothesis:

p-value: $P(t_{n-2}\leq -3.82) + P(t_{n-2}\geq 3.82)$

As you can see to calculate it you need the information of the sample size to determine the degrees of freedom of the distribution.

If you want to use the rejection region approach, the sample size is also needed to determine the critical values.

But since this test is two tailed at α: 0.05 and there was a confidence interval with confidence level 0.95 (which is complementary to the level of significance) you can use it to decide whether to reject the null hypothesis.

Using the CI, the decision rule is as follows:

If the CI includes the "zero", do not reject the null hypothesis.

If the CI doesn't include the "zero", reject the null hypothesis.

The calculated interval for the slope is: [-0.11; -0.04]

As you can see, both limits of the interval are negative and do not include the zero, so the decision is to reject the null hypothesis.

At a 5% significance level, you can conclude that the relationship between the year and the number of earthquakes above magnitude 7.0 is statistically significant.

I hope this helps!

(full output in attachment)