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3 years ago

Find the quotient 3/7 and 9/14

Mathematics

1 answer:

wolverine [178]3 years ago

7 0

Dividing in fractions means that you multiply, and flip the second number's numerator and denominator.

So, it turns out to be 3/7 x 14/9, which is 2/3
Therefore, the answer is 2/3

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Find the probability that x <= -3

Slav-nsk [51]

Answer:

30%

Step-by-step explanation:

there are two numbers that fit in the x<=3 category, those being -3 and -5, so you add the two decimals for them (.17 and .13) to get .30 and move the decimal two times to the right to make it in percentage form.

6 0

3 years ago

Inverse of y=(x/2)^2

Lady bird [3.3K]

Tell me if the answer is true or wrong

6 0

3 years ago

What is the value of x? Enter your answer in the box. mm

xenn [34]

136mm?? Isn't it 68 x 2 because Y-K and V-Y are the same, so 68 x 2=136? I don't know

7 0

3 years ago

Read 2 more answers

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 X denote the event that the student has an “on” day, and let Y 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

3 years ago

THE FIRST AND BEST ANSWER GETS BRAINLIEST

wolverine [178]

If it’s 5 points no one is gonna answer

4 0

3 years ago