In a survey conducted with 600 participants across the United States, 450 were found to have studied science in college. If we w
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Answer:
Options (3) and (6)
Step-by-step explanation:
ΔABC is a dilated using a scale factor of to produce image triangle ΔA'B'C'.
Since, dilation is a rigid transformation,
Angles of both the triangles will be unchanged or congruent.
m∠A = m∠A' and m∠B = m∠B'
Since, sides of ΔA'B'C' = of the sides of ΔABC
Area of ΔA'B'C' =
Area of ΔABC > Area of ΔA'B'C'
Since, angles of ΔABC and ΔA'B'C' are congruent, both the triangles will be similar.
ΔABC ~ ΔA'B'C'
Therefore, Option (3) and Option (6) are the correct options.
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!