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3 years ago

Which table represents y as. Function of x?

Mathematics

1 answer:

RoseWind [281]3 years ago

6 0

I would say j but i can’t get my mind on which one

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There are 1,023 M&M's in a bag. Out of 1,023 M&M's, 132 M&M's are blue, 69 are green, 68 are brown, 63 are olive, 62

lesantik [10]

The probability that the person pulls 1 blue M&M given that one is only allowed to pull 5 random M&M's per day will be 0.6450.

<h3>How to calculate probability?</h3>

From the information given, there are 1,023 M&M's in a bag and out of 1,023 M&M's, 132 M&M's are blue.

Therefore, the probability of getting a blue M&M will be:

= 132/1023

= 0.1290

Therefore, since the person is given the chance to pull 5 random M&M's per day, the probability will be:

= 0.1290 × 5

= 0.6450

Learn more about probability on:

brainly.com/question/24756209

#SPJ1

4 0

2 years ago

(06.04 HC) The graph shows Wilson's science scores versus the number of hours spent doing science homework. A graph titled Wilso

alexandr1967 [171]

Answer:

The answer 55

Step-by-step explanation:

Because it goes up by five for they gave us the amount of point they get for 5 hours and the amount of point he had for 5 hours was 50 so we just needed to add 5 more and 50+5= 55

I don't know if I made myself clear but good luck! :)

5 0

3 years ago

MATH HOMEWORK HELP!! >> Solving systems by GRAPHING:

grin007 [14]

I dont know the coordinates because i dont feel like putting them in right now buttttt........

main street bus equation is y=2x+40

County bus line equation is y=3x+20

thats all i've got.

7 0

3 years ago

How to do this question

RoseWind [281]

Idek hun I don’t see anything

3 0

3 years ago

Suppose you just received a shipment of nine televisions. Three of the televisions are defective. If two televisions are randoml

Temka [501]

<h2>Answer:</h2>

The probability that both televisions work is: 0.42

The probability at least one of the two televisions does not work is:

0.5833

<h2>Step-by-step explanation:</h2>

There are a total of 9 televisions.

It is given that:

Three of the televisions are defective.

This means that the number of televisions which are non-defective are:

9-3=6

The probability that both televisions work is calculated by:

$=\dfrac{6_C_2}{9_C_2}$

( Since 6 televisions are in working conditions and out of these 6 2 are to be selected.

and the total outcome is the selection of 2 televisions from a total of 9 televisions)

Hence, we get:

$=\dfrac{\dfrac{6!}{2!\times (6-2)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{\dfrac{6!}{2!\times 4!}}{\dfrac{9!}{2!\times 7!}}\\\\\\=\dfrac{5}{12}\\\\\\=0.42$

The probability at least one of the two televisions does not work:

Is equal to the probability that one does not work+probability both do not work.

Probability one does not work is calculated by:

$=\dfrac{3_C_1\times 6_C_1}{9_C_2}\\\\\\=\dfrac{\dfrac{3!}{1!\times (3-1)!}\times \dfrac{6!}{1!\times (6-1)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{3\times 6}{36}\\\\\\=\dfrac{1}{2}\\\\\\=0.5$

and the probability both do not work is:

$=\dfrac{3_C_2}{9_C_2}\\\\\\=\dfrac{1}{12}\\\\\\=0.0833$

Hence, Probability that atleast does not work is:

0.5+0.0833=0.5833

4 0

4 years ago