Answer:
what the heck
was that also i'm country
Step-by-step explanation:
<span>Exactly 33/532, or about 6.2%
This is a conditional probability, So what we're looking for is the probability of 2 gumballs being selected both being red. So let's pick the first gumball.
There is a total of 50+150+100+100 = 400 gumballs in the machine. Of them, 100 of the gumballs are red. So there's a 100/400 = 1/4 probability of the 1st gumball selected being red.
Now there's only 399 gumballs in the machine and the probability of selecting another red one is 99/399 = 33/133.
So the combined probability of both of the 1st 2 gumballs being red is
1/4 * 33/133 = 33/532, or about 0.062030075 = 6.2%</span>
Given that a room is shaped like a golden rectangle, and the length is 29 ft with the ratio of golden rectangle being (1+√5):2, thus the width of the room will be:
ratio of golden triangle=(length if the room)/(width of the room)
let the width be x
thus plugging the values in the expression we get:
29/x=(1+√5)/2
solving for x we get:
x/29=2/(1+√5)
thus
x=(29×2)/(1+√5)
answer is:
x=58/(1+√5)
or
byrationalizing the denominator by multiplying both the numerator and the denominator by (1-√5)
58/(1+√5)×(1-√5)/(1-√5)
=[58(1-√5)]/1-5
=(58√5-58)/4
There are 10 balls in the Urn Total.
Red: 6
Green: 4
Question One: The probability that five red and two green is selected is likely. (as that is over half for both)
Question Two: Impossible. There is only 6 red balls, and 7 are taken from the urn. Thus it would at most be possible for 6 red and 1 green.
Question Three: At least four is likely, as there is more red then green in the Urn.
Hope I helped!
(Mark Brainliest if you can please!)
648.566 Km is the answer. :)
