A figure can be covered by 28 unit squares,without any gaps or overlaps.what is the area of the figure?
1 answer:
Remember that the area of a square is just one of its sides squared. 
where
is the area of the square
is one of the sides of the square
Unit square is a square whose sides have length 1, so lets use our formula to find the volume of one unit square:
We now know that each one of the squares has area 1 unit squared. Since there are 28 unit square in our figure, we are going to multiply the area of a unit square by 28 to find the area of our figure:
Area of the figure=
units squared
We can conclude that the area of the figure is 28 units squared.
where
Unit square is a square whose sides have length 1, so lets use our formula to find the volume of one unit square:
We now know that each one of the squares has area 1 unit squared. Since there are 28 unit square in our figure, we are going to multiply the area of a unit square by 28 to find the area of our figure:
Area of the figure=
We can conclude that the area of the figure is 28 units squared.
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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!