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4 years ago

Percent of times you roll odd on a standered die

Mathematics

2 answers:

zzz [600]4 years ago

6 0

Answer:

50%

Step-by-step explanation:

On a standard die the numbers are 1, 2, 3, 4, 5, and 6.

The odd numbers are 1, 3, and 5

that means there are 3 out of 6 numbers that are odd on a standard die.

so the probability is 3/6 or 1/2 which translates to 50%

Gelneren [198K]4 years ago

3 0

1/3 = 33.333...%
There are 6 numbers, and 3 odd numbers.

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HELP PLEASE!!!!!!!!!!!!!!!!!!!

pochemuha

Answer:

h=20

i=75

g=55

Step-by-step explanation:

20 is equal to h

75 is equal to i

g looks like its in the middle of them so you would subtract the two and be left with 55

g=55

im not the best at math so im sorry if its wrong.

6 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 <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

3 years ago

PLEASE HELP!!!!!

ivanzaharov [21]

Answer:

0.468 or 65/139, 46.8%

Step-by-step explanation:

P(green) = 45 / 139

P(yellow) = 20 / 139

Add them together for either one:

65 / 139 = 0.468 = 46.8%

6 0

2 years ago

K, m,n each stand for a whole number.

mrs_skeptik [129]

You need to solve a system comprised of 3 equations.

1) k+m+n=1500
2) m=3n
3) k=2n
Replace equations 2) and 3) in 1) => 2n+3n+n=1500 => 6n=1500 => n=250

By reaplacing n in 2) => m=3*250=750

In 3) => k=2*250=500

7 0

3 years ago

Write 11/12 demoniator 84 to a equivalent fraction

Pepsi [2]

Answer:

77/84

Step-by-step explanation:

84 ÷ 12 = 7

Multiply the numerator and denominator by 7

$\frac{11}{12}=\frac{11*7}{12*7}=\frac{77}{84}$

6 0

3 years ago