That would be true because the adjacent side to B is 4 and the hypotenuse is 5 and the formula for cos b is adjacent/hypotenuse
Hope this helps!
Answer:
X = 28+3(6)
X = 28 +18
<u>X = 46</u>
Thus, the <u>value of X </u>will be <u>46</u>.
Hope it helps.
Answer:
Full proof below
Step-by-step explanation:

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 