95% of red lights last between 2.5 and 3.5 minutes.
<u>Step-by-step explanation:</u>
In this case,
- The mean M is 3 and
- The standard deviation SD is given as 0.25
Assume the bell shaped graph of normal distribution,
The center of the graph is mean which is 3 minutes.
We move one space to the right side of mean ⇒ M + SD
⇒ 3+0.25 = 3.25 minutes.
Again we move one more space to the right of mean ⇒ M + 2SD
⇒ 3 + (0.25×2) = 3.5 minutes.
Similarly,
Move one space to the left side of mean ⇒ M - SD
⇒ 3-0.25 = 2.75 minutes.
Again we move one more space to the left of mean ⇒ M - 2SD
⇒ 3 - (0.25×2) =2.5 minutes.
The questions asks to approximately what percent of red lights last between 2.5 and 3.5 minutes.
Notice 2.5 and 3.5 fall within 2 standard deviations, and that 95% of the data is within 2 standard deviations. (Refer to bell-shaped graph)
Therefore, the percent of red lights that last between 2.5 and 3.5 minutes is 95%
Answer:
A. 502.4 cm^3
Step-by-step explanation:
r=8/2=4 cm
V=pi*r^2*h=3.14*16*10=502.4 cm^3 (A)
Answer:
C. 60 ft
Step-by-step explanation:
If triangles ABC and EDC are in a 1:1 relation, they are congruent, and
AC = EC
5x - 5 = 3x + 9
5x = 3x + 14
2x = 14
x = 7
AC = 5x + 5 = 5(7) - 5 = 35 - 5 = 30 ft
EC = 3x + 9 = 3(7) + 9 = 21 + 9 = 30 ft
Distance between top and bottom of bridge = AC + EC = 30 + 30 = 60 ft
Apply slip and slide
a^2-3a-4
(a-4)(a+1)
(a-2)(2a+1)
Find zeros
a-2=0
a=2
2a+1=0
2a=-1
a=-1/2
Final answer: {-1/2, 2}
Answer:
a) 151lb.
b) 6.25 lb
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean 
and standard deviation 
, a large sample size can be approximated to a normal distribution with mean and standard deviation 
.
In this problem, we have that:
So
a) The expected value of the sample mean of the weights is 151 lb.
(b) What is the standard deviation of the sampling distribution of the sample mean weight?
This is 
