<span>(a)
</span>P (all
B+) = (0.12)(0.12)(0.12)=0.001728=0.1728%
<span>(b)
</span>P (no
B+) = (0.88)(0.88)(0.88) = 0.681472=68.15%
<span>(c)
</span>P (not
one B+) + P (no B+) = 1
P (not one B+) = 1 – P (no B+)
<span> = 1 – 0.681472 </span>
P (at least 1 B+) = 0.318528=31.85%
<span>(d)
</span>B. The event in part (a) is unusual because its
probability is less than or equal to 0.05.
Answer:
<em>0m</em>
Step-by-step explanation:
Before the ball is thrown, it is at the ground level. At the ground level, the height is zero. Substitute h = 0 into the equation;
h(x)=-x^2+10x-16
0=-x^2+10x-16
x^2-10x+16 = 0
Factorize
x^2-8x-2x+16 = 0
x(x-8) - 2 (x-8) = 0
x-2 = 0 and x - 8 = 0
x = 2 and 8
At x = 2
h(2) = -2^2 + 10(2) - 16
h(2) = -4 + 4
h(2) = 0
<em>Hence the height of the ball at the time it is thrown is 0m</em>
Absolutely not because
if you see as fraction
8/7 and 15/16 is not equal at all
if it had to be equal if should've been 16/14
