Hey there,
15 games = 7 hits
1 game = 7 / 15
= <span>0.46666666666
45 games = </span><span>0.46666666666 x 45
= 21
Hope this helps :))
<em>~Top♥</em>
</span>
5x^2-4x+1 + -3x^2+x-3= 2x^2-3x-2
Answer:
there is 1/2 a chance you will roll an odd number.
Answer:
1.
T mBAC = mB'A'C'
F 2mABC = mA'B'C'
F BC = 2B'C'
T 2XA = XA'
2
D'(-2/3; -1)
E'(-1;1)
F'(1;1)
G'(1;-1)
3
the centre is L(0;-2)
the scale factor is 4
length J'K' = 4JK
the measure of L is equal the measure of L'
<u>the</u><u> </u><u>table</u><u>:</u>
K(4;2) 4 4 16 16 0+16 -2+16 K'(16;14)
Answer:
P(X ≥ 1) = 0.50
Step-by-step explanation:
Given that:
The word "supercalifragilisticexpialidocious" has 34 letters in which 'i' appears 7 times in the word.
Then; the probability of success = 7/34 = 0.20588
Using Binomial distribution to determine the probability; we have:

where;
x = 0,1,2,...n and 0 < β < 1
and x represents the number of successes.
However; since the letter is drawn thrice; the probability that the letter "i" is drawn at least once can be computed as:
P(X ≥ 1) = 1 - P(X< 1)
P(X ≥ 1) = 1 - P(X =0)
![P(X \ge 1) = 1 - \bigg [ {^3C__0} (0.21)^0 (1-0.21)^{3-0} \bigg]](https://tex.z-dn.net/?f=P%28X%20%5Cge%201%29%20%3D%20%201%20-%20%5Cbigg%20%5B%20%7B%5E3C__0%7D%20%280.21%29%5E0%20%281-0.21%29%5E%7B3-0%7D%20%5Cbigg%5D)
![P(X \ge 1) = 1 - \bigg [ 1 \times 1 (0.79)^{3} \bigg]](https://tex.z-dn.net/?f=P%28X%20%5Cge%201%29%20%3D%20%201%20-%20%5Cbigg%20%5B%201%20%5Ctimes%201%20%280.79%29%5E%7B3%7D%20%5Cbigg%5D)
P(X ≥ 1) = 1 - 0.50
P(X ≥ 1) = 0.50