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podryga [215]
3 years ago
5

What do i do i need help

Mathematics
2 answers:
Reika [66]3 years ago
8 0
Help with what exactly?
pav-90 [236]3 years ago
5 0
Help with what exactly?
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The values of a, b and c for the equation 8x2 - 2x = 1 are:
meriva

Answer:

The values are B

a= 8,b = -2 c= -1

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3 years ago
Move the decimals to the lines to order them from least to greatest.
wel
Least to greatest: 0.7 0.9 0.73 0.81
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What is 322 divided by 15 ?
Murrr4er [49]
   Your answer is 21 7/15
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2 years ago
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A student is getting ready to take an important oral examination and is concerned about the possibility of having an "on" day or
Tamiku [17]

Answer:

The students should request an examination with 5 examiners.

Step-by-step explanation:

Let <em>X</em> denote the event that the student has an “on” day, and let <em>Y</em> denote the

denote the event that he passes the examination. Then,

P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})

The events (Y|X) follows a Binomial distribution with probability of success 0.80 and the events (Y|X^{c}) follows a Binomial distribution with probability of success 0.40.

It is provided that the student believes that he is twice as likely to have an off day as he is to have an on day. Then,

P(X)=2\cdot P(X^{c})

Then,

P(X)+P(X^{c})=1

⇒

2P(X^{c})+P(X^{c})=1\\\\3P(X^{c})=1\\\\P(X^{c})=\frac{1}{3}

Then,

P(X)=1-P(X^{c})\\=1-\frac{1}{3}\\=\frac{2}{3}

Compute the probability that the students passes if request an examination with 3 examiners as follows:

P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})

        =[\sum\limits^{3}_{x=2}{{3\choose x}(0.80)^{x}(1-0.80)^{3-x}}]\times\frac{2}{3}+[\sum\limits^{3}_{x=2}{{3\choose x}(0.40)^{3}(1-0.40)^{3-x}}]\times\frac{1}{3}

       =0.715

The probability that the students passes if request an examination with 3 examiners is 0.715.

Compute the probability that the students passes if request an examination with 5 examiners as follows:

P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})

        =[\sum\limits^{5}_{x=3}{{5\choose x}(0.80)^{x}(1-0.80)^{5-x}}]\times\frac{2}{3}+[\sum\limits^{5}_{x=3}{{5\choose x}(0.40)^{x}(1-0.40)^{5-x}}]\times\frac{1}{3}

       =0.734

The probability that the students passes if request an examination with 5 examiners is 0.734.

As the probability of passing is more in case of 5 examiners, the students should request an examination with 5 examiners.

8 0
2 years ago
Find the area of circle with diameter of 52 cm. Use ≈3.14. <br> The area of circle is approximately?
Likurg_2 [28]

Answer:

169.56

Step-by-step explanation:

4 0
3 years ago
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