The area of a rectangle is A=LW, the area of a square is A=S^2.
W=S-2 and L=2S-3
And we are told that the areas of each figure are the same.
S^2=LW, using L and W found above we have:
S^2=(2S-3)(S-2) perform indicated multiplication on right side
S^2=2S^2-4S-3S+6 combine like terms on right side
S^2=2S^2-7S+6 subtract S^2 from both sides
S^2-7S+6=0 factor:
S^2-S-6S+6=0
S(S-1)-6(S-1)=0
(S-6)(S-1)=0, since W=S-2, and W>0, S>2 so:
S=6 is the only valid value for S. Now we can find the dimensions of the rectangle...
W=S-2 and L=2S-3 given that S=6 in
W=4 in and L=9 in
So the width of the rectangle is 4 inches and the length of the rectangle is 9 inches.
Answer: D. 72 m
Step-by-step explanation:
The formula for the volume of a cone is Volume = pi times radius squared and then times the height, and then divided by 3.
Since the diameter is 12, the radius is 6. Applying the formula, 6 squared is 36, times pi is 36 pi. The height is 6 so 36 * 6 = 216.
But be careful, we're not done yet... We still have to divide by 3
216 / 3 = 72
we can be sure this answer is correct because the other answers are clearly too big
here's your answer, hope this helps :)
Answer:
We see that the p value is lower than the significance level given of 0.01 so then we can conclude that the true mean for the nicotine content is significantly higher than 40 mg
Step-by-step explanation:
Information provided
represent the average nicotine content
represent the sample standard deviation
sample size
represent the value to check
represent the significance level for the hypothesis test.
t would represent the statistic
represent the p value for the test
System of hypothesis
We want to verify if the nicotine content of the cigarettes exceeds 40 mg , the system of hypothesis are:
Null hypothesis:
Alternative hypothesis:
The statistic for this case would be:
(1)
And replcing we got:
The degrees of freedom are:
The p value would be:
We see that the p value is lower than the significance level given of 0.01 so then we can conclude that the true mean for the nicotine content is significantly higher than 40 mg