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4 years ago

What are 2 fractions that equal 5/6??

Mathematics

2 answers:

elena-14-01-66 [18.8K]4 years ago

7 0

Answer:

5/6 10/12 15/18 20/24

25/30 30/36 35/42 40/48

45/54 50/60 55/66 60/72

65/78 70/84 75/90 80/96

85/102 90/108 95/114 100/120

105/126 110/132 115/138 120/144

125/150 130/156 135/162 140/168

145/174 150/180 155/186 160/192

165/198 170/204 175/210 180/216

185/222 190/228 195/234 200/240

205/246 210/252 215/258 220/264

225/270 230/276 235/282 240/288

245/294 250/300 255/306 260/312

265/318 270/324 275/330 280/336

285/342 290/348 295/354 300/360

305/366 310/372 315/378 320/384

325/390 330/396 335/402 340/408

345/414 350/420 355/426 360/432

365/438 370/444 375/450 380/456

385/462 390/468 395/474 400/480

405/486 410/492 415/498 420/504

425/510 430/516 435/522 440/528

445/534 450/540 455/546 460/552

465/558 470/564 475/570 480/576

485/582 490/588 495/594 500/600

Step-by-step explanation:

all of these are equivalent to 5/6

Helen [10]4 years ago

5 0

Answer:

10/12 and 30/36

Step-by-step explanation:

Because those two fractions are equal to 5/6

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Step-by-step explanation:

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3 years ago

An individual who has automobile insurance from a certain company is randomly selected. Let y be the number of moving violations

Hoochie [10]

Answer:

a) $E(Y)= \sum_{i=1}^n Y_i P(Y_i)$

And replacing we got:

$E(Y) = 0*0.45 +1*0.2 +2*0.3 +3*0.05= 0.95$

b) $E(80Y^2) =80[ 0^2*0.45 +1^2*0.2 +2^2*0.3 +3^2*0.05]= 148$

Step-by-step explanation:

Previous concepts

In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".

The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).

And the standard deviation of a random variable X is just the square root of the variance.

Solution to the problem

Part a

We have the following distribution function:

Y 0 1 2 3

P(Y) 0.45 0.2 0.3 0.05

And we can calculate the expected value with the following formula:

$E(Y)= \sum_{i=1}^n Y_i P(Y_i)$

And replacing we got:

$E(Y) = 0*0.45 +1*0.2 +2*0.3 +3*0.05= 0.95$

Part b

For this case the new expected value would be given by:

$E(80Y^2)= \sum_{i=1}^n 80Y^2_i P(Y_i)$

And replacing we got

$E(80Y^2) =80[ 0^2*0.45 +1^2*0.2 +2^2*0.3 +3^2*0.05]= 148$

5 0

3 years ago