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butalik [34]
3 years ago
15

H(z) = -(x + 1)(2 – 7) Simplify

Mathematics
1 answer:
mestny [16]3 years ago
8 0

Answer:

h=5x+6/z

Step-by-step explanation:

hz=(-(x+1))(2-7)

hz/z=5x+5/z

h=5x+5/z

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A company is interviewing potential employees. Suppose that each candidate is either qualified, or unqualified with given probab
stira [4]

Answer:

P(C=1|T=1)=q(\sum_{i=15}^{20}\binom{20}{i} p^i(1-p)^{20-i})( \sum_{i=15}^{20}\binom{20}{i}[qp^i(1-p)^{20-i} + (1-q)p^{20-i}(1-p)^i])^{-1}

Step-by-step explanation:

Hi!

Lets define:

C = 1  if candidate is qualified

C = 0 if candidate is not qualified

A = 1 correct answer

A = 0 wrong answer

T = 1 test passed

T = 0 test failed

We know that:

P(C=1)=q\\P(A=1 | C=1) = p\\P(A=0 | C=0) = p

The test consist of 20 questions. The answers are indpendent, then the number of correct answers X has a binomial distribution (conditional on the candidate qualification):

P(X=x | C=1)=f_1(x)=\binom{20}{x}p^x(1-p)^{20-x}\\P(X=x | C=0)=f_0(x)=\binom{20}{x}(1-p)^xp^{20-x}

The probability of at least 15 (P(T=1))correct answers is:

P(X\geq 15|C=1)=\sum_{i=15}^{20}f_1(i)\\P(X\geq 15|C=0)=\sum_{i=15}^{20}f_0(i)\\

We need to calculate the conditional probabiliy P(C=1 |T=1). We use Bayes theorem:

P(C=1|T=1)=\frac{P(T=1|C=1)P(C=1)}{P(T=1)}\\P(T=1) = qP(T=1|C=1) + (1-q)P(T=1|C=0)

P(T=1)=q\sum_{i=15}^{20}f_1(i) + (1-q)\sum_{i=15}^{20}f_0(i)\\P(T=1)=\sum_{i=15}^{20}\binom{20}{i}[qp^i(1-p)^{20-i} + (1-q)p^{20-i}(1-p)^i)]

P(C=1|T=1)=q(\sum_{i=15}^{20}\binom{20}{i} p^i(1-p)^{20-i})( \sum_{i=15}^{20}\binom{20}{i}[qp^i(1-p)^{20-i} + (1-q)p^{20-i}(1-p)^i])^{-1}

5 0
3 years ago
Find the area of this shaded region.
4vir4ik [10]

Answer:9.13

Step-by-step explanation:There are three lines from the center to the sides. So they are all the same. you can see that the radius is 8 from one of the sides. You can find the area of the big triangle by knowing that 2 of the sides are 8 and 360-270=90 and that is a right triangle so we can use the pythagorean theorem to find the hypotenuse. 8^2+8^2=√128. And we can find the area by doing (8^2+8^2)/2=32 so each triangle is 16 units^2. To find out the lengths of the left triangle we have (.5*√128)^2=√32. So 8^2-32=x² x=√32. if the radius is 8 and the one side of the radius is √32 it would be 8-√32. and the other line touching the shaded area it √128 so we can say that the area of the non-little circle is the area of the whole circle-the almost circle. The area of the whole circle it 8^2*pi≈201.06. The area of the almost circle is 201.06*.75=150.79 because 270/360=.75 plus the area of the triangle (32) so the area is 182.79. So the area of the little thing is 201.06-182.79=18.27 divided by 2 =9.13

4 0
3 years ago
PLS HELP PLS HELp PLEASE
Orlov [11]
Download Photomath, it actually helps with problems like these!
6 0
3 years ago
HELP ME ASAP!!!!!!!! :)
insens350 [35]

Answer:

Step-by-step explanation:

(B) 0.752 and 0.232323....

4 0
3 years ago
To the nearest tenth, what is the area of the shaded segment when BN = 8 ft?
laila [671]

Answer:


Step-by-step explanation:

Answer is


  39.3 ft²


======


We need to find the area of that sector of the circle.

Then we subtract that by the area of the ΔBNW.


Area of a sector of a circle where \theta is the cent. ang of the sector (in degrees) and r is the radius:


  A_{\text{sector}} = \pi r^2 \frac{\theta}{360}


We know that the radius is 8 ft; it is given to us. The central angle is also given as 120°


Number crunching:

 

  \begin{aligned}
A_{\text{sector}} &= \pi(8)^2 \frac{120}{360} \\
&= 64 \pi \cdot \frac{1}{3} \\
&= \frac{64}{3}\pi
\end{aligned}


Area of a triangle is given by


  A_{\text{triangle}} = \frac{1}{2}ab\sin C


where a and b are sides that are not opposite to angle C.


Well we know two sides of the triangle; they're identical sides because they're both the same length of the radius. We also know an angle in the triangle that is not opposite to either of our known sides.

C = 120° and a = b = 8 ft, therefore


  \begin{aligned}
A_{\text{triangle}} &= \tfrac{1}{2}(8)(8) \sin120 \\
&= 32\sin 120
\end{aligned}


Area of the shaded segment (use a calculator, degree mode)


  \begin{aligned}
A_\text{shaded} &= A_{\text{sector}} - A_{\text{triangle}} \\
&= \tfrac{64}{3}\pi - 32\sin 120 \\
&= 39.3\text{ ft}^2
\end{aligned}


Read more on Brainly.com - brainly.com/question/4113822#readmore


7 0
3 years ago
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