step 1
<span>compute the average: add the values and divide by 6
Average =(44+ 46+40+34+29+41)/6=39
step 2
</span><span>Compute the deviations from the average
dev: (44-39)=5,
</span>dev: (46-39)=7
dev: (40-39)=1
dev: (34-39)=-5
dev: (29-39)=-10
dev: (41-39)=2
step 3
<span>Square the deviations and add
sum (dev^2): 5^2+7^2+1</span>^2+-5^2+-10^2+2^2
sum (dev^2): 25+49+1+25+100+4-----> 204
step 4
<span>Divide step #3 by the sample size=6
(typically you divide by sample size-1 to get the sample standard deviation,
but you are assuming the 6 values are the population,
so
no need to subtract 1, from the sample size.
This result is the variance
Variance =204/6=34
step 5
</span><span>Standard deviation = sqrt(variance)
standard deviation= </span>√<span>(34)------> 5.83
the answer is
5.83</span>
Answer:
89.4 in²
Step-by-step explanation:
I'm too lazy to explain.
Answer:
Hello!
Step-by-step explanation:
How do we solve for j when there is no equation with j in it to solve for j. <em><u>Hope we can help you with this when we get a equation! </u></em>
Answer:
5.1 days
Step-by-step explanation:
Given in the question,
initial amount of substance = 34 grams
k-value = 0.137
To find the half life of this substance we will use following formula

here N(0) is initial amount of substance
t is time in days
Plug values in the formula

1/2 = e^{-0.137t}
Take logarithm on both sides
ln(1/2) = ln( e^{-0.137t})
ln(1/2) = -0.137t
t = ln(1/2) / -0.137
t = 5.059
t ≈ 5.1 days (nearest to tenths)
Answer:
a) 
With:


b) 

c) 

d) 


Step-by-step explanation:
For this case we know the following propoertis for the random variable X

We select a sample size of n = 81
Part a
Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

With:


Part b
We want this probability:

We can use the z score formula given by:

And if we find the z score for 89 we got:


Part c

We can use the z score formula given by:

And if we find the z score for 75.65 we got:


Part d
We want this probability:

We find the z scores:


