<span><span>9<span>(6j+2+j)</span></span><span>9<span>(6j+2+j)</span></span></span>Add <span><span>6j</span><span>6j</span></span> and <span>jj</span> to get <span><span>7j</span><span>7j</span></span>.<span><span>9<span>(7j+2)</span></span><span>9<span>(7j+2)</span></span></span>Apply the distributive property.<span><span>9<span>(7j)</span>+9⋅2</span><span>9<span>(7j)</span>+9⋅2</span></span>Multiply <span>77</span> by <span>99</span> to get <span>6363</span>.<span><span>63j+9⋅2</span><span>63j+9⋅2</span></span>Multiply <span>99</span> by <span>22</span> to get <span>1818</span>.<span>63j+<span>18</span></span>
B, it’s the point that the lines cross
Answer:
The total number of apples is 17
Step-by-step explanation:
(a)
Number of green apples = 4
P ( that all three apples are green)
= (limit 4 and 3) / (limit 17 and 3)
= 4! / 3!(4−3)!
= 17! / 3!(17−3)!
= 1 / 170
= 0.005
(b)
Number of red apples = 8
P(that all three apple are green)
=(limit 8 and 3) / (limit 17 and 3)
= 8! / 3!(8−3)!
= 17! / 3!(17−3)!
= 7 / 85
= 0.082
Now,
P(that no three apple is red)= 1−0.082 = 0.918
c)
P(of selecting 4 apple that contain at least 2 red apples)
= (limit 8 and 2) /( limit 17 and 2) X (2/15) + (limit 8 and 3) /(limit 17 / 3) X (1/14)
= (7/34) × (2/15) + (7/85) × (1/14)
= 0.027 + 0.058
= 0.085
d)
P(that the second apple selected is yellow given that the first apple is red)
= (limit 5 and 1) / ( limit 16 and 1)
= 5! / 1!(5−1)!
= 16! / 1!(16−1)!
= 5 /16
= 0.3125
204204 - 16 = 204188
204188 / 4 = 51,047
width = 51,047
length = 51,055
