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2 years ago

Hey can u guys plz help me? its one more question

Mathematics

2 answers:

jeyben [28]2 years ago

6 0

Answer:

its 7.5 do you need help

Zolol [24]2 years ago

5 0

Answer:

7.5

Step-by-step explanation:

thats were the two lines intersect meaning the cost was the same there

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For his son's birthday party, Mr. Mori bought four equally-priced pizzas and a $3 bag of potato chips. If he spent $39, find the

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the pizza cost $49 dollars

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2 years ago

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Need answer asap, correct answer will get brainliest

Cloud [144]

Answer:

x+5

Step-by-step explanation:

When you move to the right or left , that's on the x axis. And since it's to the right, the x- coordinate becomes larger. Hope this helps!

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The sum of 22 and a number is 41

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Step-by-step explanation:

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2 years ago

Pineapple Corporation (PC) maintains that their cans have always contained an average of 12 ounces of fruit. The production grou

krok68 [10]

Answer:

We accept the null hypothesis, that is, that the mean weight of the cans is still of 12 ounces of fruit.

Step-by-step explanation:

Pineapple Corporation (PC) maintains that their cans have always contained an average of 12 ounces of fruit.

This means that the null hypothesis is: $H_0: \mu = 12$

The production group believes that the mean weight has changed.

This means that the alternate hypothesis is:

$H_a: \mu \neq 12$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

12 is tested at the null hypothesis:

This means that $\mu = 12$

They take a sample of 15 cans and find a sample mean of 12.05 ounces and a sample standard deviation of .08 ounces.

This means, respectibely, that $n = 15, X = 12.05, \sigma = 0.08$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{12.05 - 12}{\frac{0.08}{\sqrt{15}}}$

$z = 2.42$

Pvalue of the test:

We are testing if the mean is different from a value, which means that the pvalue is 2 multiplied by 1 subtracted by the pvalue of z = 2.42.

Looking at the z-table, z = 2.42 has a pvalue of 0.9922

1 - 0.9922 = 0.0078

2*0.0078 = 0.0156

What conclusion can we make from the appropriate hypothesis test at the .01 level of significance?

0.0156 > 0.01. This means that at the 0.01 level, we accept the null hypothesis, that is, that the mean weight of the cans is still of 12 ounces of fruit.

6 0

2 years ago

Help with q25 please. Thanks.

Westkost [7]

First, I'll make f(x) = sin(px) + cos(px) because this expression shows up quite a lot, and such a substitution makes life a bit easier for us.

Let's apply the first derivative of this f(x) function.

$f(x) = \sin(px)+\cos(px)\\\\f'(x) = \frac{d}{dx}[f(x)]\\\\f'(x) = \frac{d}{dx}[\sin(px)+\cos(px)]\\\\f'(x) = \frac{d}{dx}[\sin(px)]+\frac{d}{dx}[\cos(px)]\\\\f'(x) = p\cos(px)-p\sin(px)\\\\ f'(x) = p(\cos(px)-\sin(px))\\\\$

Now apply the derivative to that to get the second derivative

$f''(x) = \frac{d}{dx}[f'(x)]\\\\f''(x) = \frac{d}{dx}[p(\cos(px)-\sin(px))]\\\\ f''(x) = p*\left(\frac{d}{dx}[\cos(px)]-\frac{d}{dx}[\sin(px)]\right)\\\\ f''(x) = p*\left(-p\sin(px)-p\cos(px)\right)\\\\ f''(x) = -p^2*\left(\sin(px)+\cos(px)\right)\\\\ f''(x) = -p^2*f(x)\\\\$

We can see that f '' (x) is just a scalar multiple of f(x). That multiple of course being -p^2.

Keep in mind that we haven't actually found dy/dx yet, or its second derivative counterpart either.

-----------------------------------

Let's compute dy/dx. We'll use f(x) as defined earlier.

$y = \ln\left(\sin(px)+\cos(px)\right)\\\\y = \ln\left(f(x)\right)\\\\\frac{dy}{dx} = \frac{d}{dx}\left[y\right]\\\\\frac{dy}{dx} = \frac{d}{dx}\left[\ln\left(f(x)\right)\right]\\\\\frac{dy}{dx} = \frac{1}{f(x)}*\frac{d}{dx}\left[f(x)\right]\\\\\frac{dy}{dx} = \frac{f'(x)}{f(x)}\\\\$

Use the chain rule here.

There's no need to plug in the expressions f(x) or f ' (x) as you'll see in the last section below.

Now use the quotient rule to find the second derivative of y

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{dy}{dx}\right]\\\\\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{f'(x)}{f(x)}\right]\\\\\frac{d^2y}{dx^2} = \frac{f''(x)*f(x)-f'(x)*f'(x)}{(f(x))^2}\\\\\frac{d^2y}{dx^2} = \frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2}\\\\$

If you need a refresher on the quotient rule, then

$\frac{d}{dx}\left[\frac{P}{Q}\right] = \frac{P'*Q - P*Q'}{Q^2}\\\\$

where P and Q are functions of x.

-----------------------------------

This then means

$\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2} + \left(\frac{f'(x)}{f(x)}\right)^2 + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2} +\frac{(f'(x))^2}{(f(x))^2} + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2+(f'(x))^2}{(f(x))^2} + p^2\\\\\frac{f''(x)*f(x)}{(f(x))^2} + p^2\\\\$

Note the cancellation of -(f ' (x))^2 with (f ' (x))^2

------------------------------------

Let's then replace f '' (x) with -p^2*f(x)

This allows us to form ( f(x) )^2 in the numerator to cancel out with the denominator.

$\frac{f''(x)*f(x)}{(f(x))^2} + p^2\\\\\frac{-p^2*f(x)*f(x)}{(f(x))^2} + p^2\\\\\frac{-p^2*(f(x))^2}{(f(x))^2} + p^2\\\\-p^2 + p^2\\\\0\\\\$

So this concludes the proof that $\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + p^2 = 0\\\\$ when $y = \ln\left(\sin(px)+\cos(px)\right)\\\\$

Side note: This is an example of showing that the given y function is a solution to the given second order linear differential equation.

7 0

2 years ago