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3 years ago

Urgent math help needed!! Multiple choice question. Will mark brainliest!!!

Mathematics

1 answer:

muminat3 years ago

8 0

Total number of 10 grader students = 260 students.

Total number of students who rides the bus and they are of 10th grader = 150 students.

Probability (a student rides the bus given that they are a 10th grader) = $\frac{Total \ number \ of \ students \ rides \ the \ bus \ and \ 10th \ grader}{Total \ number \ of \ 10 \ grader \ students}$

Therefore,

P(E) = $\frac{150}{260}$

Dividing top and bottom by 10, we get

P(E) = 15/26.

In decimals 0.577.

Therrefore, correct option is C) 150/260 = 15/26 = .577.

You might be interested in

Local extreme Value for each of the Following A) F(x)=x^² - 4x² +5

kobusy [5.1K]

Answer:

max{x²-4x²+5} = 5 at x = 0

Step-by-step explanation:

1. Find the critical numbers by finding the first derivative of f(x), set it to 0 and solve for x.

$f'(x)=0$

We get:

$f(x) = -3x^2+5\\f'(x) = -6x\\-6x = 0\\x = 0$

So the critical number is x = 0.

2. Evaluate the first derivative by plugging in the critical number and see if the derivative is positive or negative on both sides:

$f'(x)$ is positive when the x < 0 (for example: -6*(-1)=+)

$f'(x)$ is negative when the x > 0 (for example: -6*(1)=-)

Therefore, you have a local maximum.

Now just get the Y value by plugging in the critical number in the original function. $f(0)=5$

local maximum is (0,5)

4 0

1 year ago

The monthly fees for the local pool are $8 per month

Liono4ka [1.6K]

Given:

Monthly fees for the local pool are $8 per month and $2 per visit.

Hector pays $34 in pool fees total for the month.

To find:

The number of times he visit the pool.

Solution:

We have,

Monthly fee of pool = $8

Additional fee = $2 per visit

Let Hector visit x times.

Additional fee for x times = $2x

Total fee = Monthly fee + Additional fee

$34=8+2x$

$34-8=2x$

$26=2x$

Divide both sides by 2.

$\dfrac{26}{2}=x$

$13=x$

Therefore, Hector visit the pool 13 times.

3 0

2 years ago

Can someone check whether its correct or no? this is supposed to be the steps in integration by parts

Gwar [14]

Answer:

$\displaystyle - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}$

Step-by-step explanation:

$\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}$

Given integral:

$\displaystyle -\int \dfrac{\sin(2x)}{e^{2x}}\:\text{d}x$

$\textsf{Rewrite }\dfrac{1}{e^{2x}} \textsf{ as }e^{-2x} \textsf{ and bring the negative inside the integral}:$

$\implies \displaystyle \int -e^{-2x}\sin(2x)\:\text{d}x$

Using <u>integration by parts</u>:

$\textsf{Let }\:u=\sin (2x) \implies \dfrac{\text{d}u}{\text{d}x}=2 \cos (2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

Therefore:

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\sin (2x)- \int \dfrac{1}{2}e^{-2x} \cdot 2 \cos (2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\sin (2x)- \int e^{-2x} \cos (2x)\:\text{d}x\end{aligned}$

$\displaystyle \textsf{For }\:-\int e^{-2x} \cos (2x)\:\text{d}x \quad \textsf{integrate by parts}:$

$\textsf{Let }\:u=\cos(2x) \implies \dfrac{\text{d}u}{\text{d}x}=-2 \sin(2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\cos(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\cos(2x)- \int \dfrac{1}{2}e^{-2x} \cdot -2 \sin(2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x\end{aligned}$

Therefore:

$\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x$