1.If a number is chosen at random from the integers 5 to 25 inclusive , find the probability that the number is a multiple of 5 or 3. 2.Good Limes =10
Good Apples = 8
Bad Limes = 6
Bad Apples 6
The information above shows the number of limes and apples of the same size in a bag . If two of the fruits are picked at random , one at a time without replacement .Find the probability that :
I. Both are good limes
II. Both are good fruits
III. One is a good apple and the other a bad lime

Solution given:

total outcomes between 5 to25 inclusive

n[T]=25-5+1=21

multiple of 5n[5]=5,10,15,20,25=5

multiple of 3n[3]=6,9,12,15,18,21,24=7

now

probability of getting multiple of 5p[5]=5/21

and

probability of getting multiple of 3 p[3]=7/21=1/3

again

the probability that the number is a multiple of 5 or 3 P[5or 3]=p[5]+p[3]=5/21+1/3=4/7

2:

.Good Limes n[GL] =10

Good Apples n[GA]= 8

Bad Limes n[BL] = 6

Bad Apples n[BA]= 6

total fruits n[T]=10+8+6+6=30

no of good apple n[G]=10+8=18

no of bad apple n[B]=6+6=12

again

I. Both are good limes

=

=10/30*9/29=3/29

II.Both are good fruits

=

=6/30*5/29=1/29

III. One is a good apple and the other a bad lime

=n[G]/n[T] *n[B]/(n[T]-1)

=18/30*12/29=36/145

d. 8

step-by-step explanation:

the given radical expression is:

we can rewrite this as:

we express as square numbers to get;

the exponents will now cancel out to give.

the correct answer is d