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2 years ago

Plz help i dont know the answer

Mathematics

1 answer:

Yuliya22 [10]2 years ago

4 0

Answer:

Im not 100 percent sure but I think you have to do this

Step-by-step explanation:

Add all numbers together to find total students

Then add the students in Chemistry and Physics and both together

Then find what that number is in percent to the total students

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Consider the probability distribution of x, where x is the number of job applications completed by a college senior through the

Oksi-84 [34.3K]

The expected number out of 1000 selected college seniors that completed 1 job application through the career centers is given by 0.011 x 1000 = 11 which is close to 14.

The expected number out of 1000 selected college seniors that completed 2 job application through the career centers is given by 0.115 x 1000 = 115 which is far away from 15.

The expected number out of 1000 selected college seniors that completed 3 job application through the career centers is given by 0.123 x 1000 = 123 which is close to 130.

Therefore, the result that would be suprising is "15 seniors completed 2 job applications through the career center."

3 0

3 years ago

Hya think of a number "m" . She square it, add "y" to the answer and then subtract 7 from it. The final answer is "A". 1) Derive

Bond [772]

<h2><u>Solution (1)</u> :</h2>

Given, to find A we have to :

square m
Add y to m²
Subtract 7 from m² + y

From the question, the following equation can be formed :

$=\tt A = {m}^{2} + y - 7$

Therefore, the formula for finding A = m² + y - 7

<h2><u>Solution (2)</u> :</h2>

The value of A we can derive from the formula is :

$=\tt A = {m}^{2} + y - 7$

Value of m = 3 (given)

Which means :

$=\tt A = {3}^{2} + y - 7$

$=\tt \: A = 9 + y - 7$

$=\tt A = 9 - 7 + y$

$\color{plum} =\tt A = 2 + y$

Thus, the value of A = 2+y

Therefore, the value of A = <u>2+y</u>

7 0

2 years ago

Find a particular solution to the nonhomogeneous differential equation y??+4y?+5y=?10x+e^(?x).

Firdavs [7]

Answer:

A) Particular solution:

$2x+\frac{1}{2}e^{-x}-\frac{8}{5}$

B) Homogeneous solution:

$y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))$

C) The most general solution is

$y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}$

Step-by-step explanation:

Given non homogeneous ODE is

$y''+4y'+5y=10x+e^{-x}---(1)$

To find homogeneous solution:

$D^{2}+4D+5=0\\D^{2}+4D+4-4+5=0\\\\(D+2)^{2}=-1\\D+2=\pm iD=-2 \pm i\\y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))---(2)$

To find particular solution:

$y_{p}=Ax+B+Ce^{-x}\\\\y'_{p}=A-Ce^{-x}\\y''_{p}=Ce^{-x}\\$

Substituting $y_{p},y'_{p},y''_{p}$ in (1)

$y''_{p}+4y'_{p}+5y_{p}=10x+e^{-x}\\Ce^{-x}+4(A-Ce^{-x})+5(Ax+B+Ce^{-x})=10x+e^{-x}\\$

Equating the coefficients

$5Ax+2Ce^{-x}+4A+5B=10x+e^{-x}\\5A=10\\A=2\\4A+5B=0\\B=-\frac{4A}{5}B=-\frac{8}{5}2C=1\\C=\frac{1}{2}\\So,\\y_{p}=2x+\frac{1}{2}e^{-x}-\frac{8}{5}---(3)\\$