50+8
50 represents tens place value, while 8 represents the ones place value.
Situations:
1) he makes 1st shot + he makes 2nd shot
P(A) = 0.4 · 0.4 = 0.16 ( 16 % )
2) he makes 1 st shot + he misses 2nd shot
P(B) = 0.4 · 0.6 = 0.24 ( 24 % )
3) he misses 1st shot and he has no more attempts
P (C) = 0.6 ( 60 %)
0.16 · 2 + 0.24 · 1 + 0.6 · 0 = 0.32 + 0.34 = 0.56
The expected value is 0.56 points.
Answer:
99.85%
Step-by-step explanation:
The lifespans of meerkats in a particular zoo are normally distributed. The average meerkat lives 10.4 years; the standard deviation is 1.9 years.
Use the empirical rule (68-95-99.7%) to estimate the probability of a meerkat living less than 16.1 years.
Solution:
The empirical rule states that for a normal distribution most of the data fall within three standard deviations (σ) of the mean (µ). That is 68% of the data falls within the first standard deviation (µ ± σ), 95% falls within the first two standard deviations (µ ± 2σ), and 99.7% falls within the first three standard deviations (µ ± 3σ).
Therefore:
68% falls within (10.4 ± 1.9). 68% falls within 8.5 years to 12.3 years
95% falls within (10.4 ± 2*1.9). 95% falls within 6.6 years to 14.2 years
99.7% falls within (10.4 ± 3*1.9). 68% falls within 4.7 years to 16.1 years
Probability of a meerkat living less than 16.1 years = 100% - (100% - 99.7%)/2 = 100% - 0.15% = 99.85%
Translation of 7 units to the right ⇒ subtract 7 units to the argument ⇒ f(x-7)
Translation of 5 units up ⇒ add 5 units to the function ⇒ f(x) + 5
g(x) = f (x-7) + 5 = (x-7)^2 + 5
The subtraction of 7 units to the argument of the function (this is x) translates the function 7 units to the right.
Adding 5 units to the function f, translates the graph 5 units up.
