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2 years ago

Twenty people bought raffle tickets, including

Mathematics

1 answer:

pantera1 [17]2 years ago

3 0

Answer:

Step-by-step explanation:

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4x - 2 + y = 6 - 2x for x

Vlada [557]

Answer:

Step-by-step explanation:

8 0

3 years ago

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$

= $(\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5$

= $(\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]$

= $(\frac{3}{4^4})[1+\frac{1}{16}]$

= $(\frac{3}{256})[\frac{17}{16}]$

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0

3 years ago

Helppppppp pleaseeee it’s mathhhhhh

Anuta_ua [19.1K]

Answer:

Step-by-step explanation:

6x+25=10x-11

-4x=-36

x=8

6(8)+25

48+25=73

I am not sure on this one but hope I helped

3 0

2 years ago

HELP ME PLS IF IT LINEAR OR NONLINEAR

klasskru [66]

I think it’s linear. no numbers repeat. i could be wrong but LOL

4 0

2 years ago

Read 2 more answers

Rolling a die what is the probability of rolling two 4s in a row

muminat

Probability =

(number of ways to succeed) / (total possible outcomes) .

There are 6 total possible outcomes of rolling one honest die.
Only one of them is a ' 4 '.

-- So the probability of rolling a '4' on the first roll is 1/6 .

-- The probability of rolling a '4' on the second roll is 1/6 .

-- The probability of BOTH events is

(1/6) x (1/6) = 1/36 = (2 and 7/9) percent .

6 0

3 years ago