Two steamers leave a port on a river at the same time, traveling in opposite directions. Each is traveling 22 miles per hour. Ho
w long will it take for them to be 110 miles apart
2 answers:
Answer:
2.5 hours.
Step-by-step explanation:
Let x represent time taken by each steamer.
We have been given that two steamers leave a port on a river at the same time, traveling in opposite directions.
Since both steamers has same speed, so distance traveled by each steamer in x hours would be .
We have been given that both steamers traveled a distance of 110 miles, so we can represent this information in an equation as:
Let us solve for x:
Therefore, it will take both steamers 2.5 hours to be 110 miles apart.
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Answer:
Step-by-step explanation:
Hello!
You have the variable
X: Area that can be painted with a can of spray paint (feet²)
The variable has a normal distribution with mean μ= 25 feet² and standard deviation δ= 3 feet²
since the variable has a normal distribution, you have to convert it to standard normal distribution to be able to use the tabulated accumulated probabilities.
a.
P(X>27)
First step is to standardize the value of X using Z= (X-μ)/ δ ~N(0;1)
P(Z>(27-25)/3)
P(Z>0.67)
Now that you have the corresponding Z value you can look for it in the table, but since tha table has probabilities of , you have to do the following conervertion:
P(Z>0.67)= 1 - P(Z≤0.67)= 1 - 0.74857= 0.25143
b.
There was a sample of 20 cans taken and you need to calculate the probability of painting on average an area of 540 feet².
The sample mean has the same distribution as the variable it is ariginated from, but it's variability is affected by the sample size, so it has a normal distribution with parameners:
X[bar]~N(μ;δ²/n)
So the Z you have to use to standardize the value of the sample mean is Z=(X[bar]-μ)/(δ/√n)~N(0;1)
To paint 540 feet² using 20 cans you have to paint around 540/20= 27 feet² per can.
c.
P(X≤27) = P(Z≤(27-25)/(3/√20))= P(Z≤2.98)= 0.999
d.
No. If the distribution is skewed and not normal, you cannot use the normal distribution to calculate the probabilities. You could use the central limit theorem to approximate the sampling distribution to normal if the sample size was 30 or grater but this is not the case.
I hope it helps!
Answer:
<em>Mean of the sample = 27.83</em>
<em> The variance of the the sample = 106.96</em>
<em> </em><em>Standard deviation of the sample = 10.34</em>
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given random sample of six employees
x 26 32 29 16 45 19
mean of the sample
Mean of the given data = 27.83
<u>Step(ii):-</u>
<u>Given data</u>
x : 26 32 29 16 45 19
x - x⁻ : -1.83 4.17 1.17 -11.83 17.17 -8.83
(x - x⁻)² : 3.3489 17.3889 1.3689 139.9489 294.80 77.9689
∑ (x-x⁻)² = 534.8245
Given sample size 'n' =6
The variance of given data
S² = ∑(x-x⁻)² / n-1
The variance of the given sample = 106.9649
<u> Step(iii):-</u>
Standard deviation of the given data
Standard deviation of the sample = 10.3423