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3 years ago

Yeah anyone else bored while ur in school so ur browsing brainly??

Mathematics

1 answer:

Margarita [4]3 years ago

8 0

Answer:

I'm Not bored im currently Dying while in school :)

Step-by-step explanation:

You might be interested in

(Please help me with this one it is due in ten minutes :'(

Flura [38]

Answer:

I think the answer is C

Step-by-step explanation:

First you need to find 2 coordinates on the graph. (2,3) (4,6)

Use the formula M= y2-y1 ÷ x2-x1

Then m= 6-3 ÷ 4-2

Then m= 3÷2 = 1.5

So the slope should be 1.5 or 1 1/2

6 0

3 years ago

Jenna borrowed $5,000 for 3 years and had to pay $1,350 simple interest at the end of that time. What RATE of interest did she p

mote1985 [20]

Answer:

Simple Interest =PRT/100

Rate= I ×100/PT

=1350×100/5000×3

135000/15000

=9%

7 0

2 years ago

HElp me find the IQR please

patriot [66]

The interquartile range is 4

8 0

2 years ago

In a random sample of students who took the SAT test, 427 had paid for coaching courses and the remaining 2733 had not. Calculat

Dvinal [7]

Answer:

The 95% confidence interval for the proportion of students who get coaching on the SAT is (0.1232, 0.147).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

427 had paid for coaching courses and the remaining 2733 had not.

This means that $n = 427 + 2733 = 3160, \pi = \frac{427}{3160} = 0.1351$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1351 - 1.96\sqrt{\frac{0.1351*0.8649}{3160}} = 0.1232$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1351 + 1.96\sqrt{\frac{0.1351*0.8649}{3160}} = 0.147$

The 95% confidence interval for the proportion of students who get coaching on the SAT is (0.1232, 0.147).

8 0

3 years ago

Find the indefinite integral. (Note: Solve by the simplest method—not all require integration by parts. Use C for the constant o

umka2103 [35]

Answer:

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-1/2 ln⁡(1+x^2 )+C

Step-by-step explanation:

∫▒〖1st .2nd dx=1st∫▒〖2nd dx〗-∫▒〖(derivative of 1st) dx∫▒〖2nd dx〗〗〗

Let 1st=arctan⁡(x)

And 2nd=1

∫▒〖arctan⁡(x).1 dx=arctan⁡(x) ∫▒〖1 dx〗-∫▒〖(derivative of arctan(x))dx∫▒〖1 dx〗〗〗

As we know that

derivative of arctan(x)=1/(1+x^2 )

∫▒〖1 dx〗=x

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-∫▒〖(1/(1+x^2 ))dx.x〗…………Eq1

Let’s solve ∫▒(1/(1+x^2 ))dx by substitution now

Let 1+x^2=u

du=2xdx

Multiply and divide ∫▒〖(1/(1+x^2 ))dx.x〗 by 2 we get

1/2 ∫▒〖(2/(1+x^2 ))dx.x〗=1/2 ∫▒(2xdx/u)

1/2 ∫▒(2xdx/u) =1/2 ∫▒(du/u)

1/2 ∫▒(2xdx/u) =1/2 ln⁡(u)+C

1/2 ∫▒(2xdx/u) =1/2 ln⁡(1+x^2 )+C

Putting values in Eq1 we get

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-1/2 ln⁡(1+x^2 )+C (required soultion)

3 0

3 years ago

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