Answer:
i don't know nothing about that
Answer:
10
Step-by-step explanation:
Answer:
0.30
Step-by-step explanation:
Probability of stopping at first signal = 0.36 ;
P(stop 1) = P(x) = 0.36
Probability of stopping at second signal = 0.54;
P(stop 2) = P(y) = 0.54
Probability of stopping at atleast one of the two signals:
P(x U y) = 0.6
Stopping at both signals :
P(xny) = p(x) + p(y) - p(xUy)
P(xny) = 0.36 + 0.54 - 0.6
P(xny) = 0.3
Stopping at x but not y
P(x n y') = P(x) - P(xny) = 0.36 - 0.3 = 0.06
Stopping at y but not x
P(y n x') = P(y) - P(xny) = 0.54 - 0.3 = 0.24
Probability of stopping at exactly 1 signal :
P(x n y') or P(y n x') = 0.06 + 0.24 = 0.30
100, because in 3,492, it is in the hundreds place. In 704, it is in the one's place. 100/1 is 100. So 100 times greater.
Answer:
<u>A. Mean = 402.5</u>
<u>B. Variance = 77,556.25</u>
<u>C. Standard Deviation = 278.49</u>
Step-by-step explanation:
Let's calculate the mean, variance and standard deviation of the set of numbers given:
A. Mean = (45 + 340 + 400 + 825)/4 = 1,610/4 =<u> 402.5</u>
B. Variance [(45 - 402.5)² + (340 - 402.5)² + (400 - 402.5)² + (825 - 402.5)²]/4 = [(127,806.25 + 3,906.25 +6.25 + 178,506.25/4 =<u> </u><u>77,556.25</u>
C. Standard Deviation = √Variance = √77,556.25 =<u> 278.49</u>