Arandom two -digit number (10-99)is drawn find each probability
p (32)
p (odd number )
p (a multiple of 5 )

Step-by-step explanation:

Given that two digit numbers are used and one digit is drawn randomly.

Sample space = {10,11,...99}

n(S) = 90

32 is only one number favourable

Hence P(32) = 1/90

Odd numbers are exactly 45.

Hence prob (odd number) = 45/90 = 1/2

Multiples of 5 are 10,15,95

There are exactly 18 numbers.

Hence P(a multiple of 5) = 18/90 =1/5

0.16 or 16%

step-by-step explanation:

lets take a coin for example.

if you toss it twice, it has a 1/4 chance of landing on heads both times because 1/2 * 1/2 is 1/4.

with the cards, the chance of picking dog is 0.4. having two people pick one is like picking a card twice in a row. so 0.4 * 0.4 is 0.16, or 16%

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?