Answer:
C) There is not sufficient evidence to support the claim that the mean attendance is greater than 523.
Step-by-step explanation:
Let μ be the the average attendance at games of the football team
The claim: the average attendance at games is over 523
Null and alternative hypotheses are:
: μ=523
: μ>523
The conclusion is failure to reject the null hypothesis.
This means that <em>test statistic</em> is lower than <em>critical value</em>. Therefore it is not significant, there is no significant evidence to accept the <em>alternative</em> hypothesis.
That is no significant evidence that the average attendance at games of the football team is greater than 523.
Answer:
A.8 B.(-1,4) C.(1,4)
Step-by-step explanation:
First, know that (3,2)=(x1,y1) and (-5,6)=(x2,y2)
A. The diameter is 8, given the diference between x1 and x2: 3-(-5)=8
B. The center point is given by (x1+x2)/2 and (y1+y2)/2
(x1+x2)/2 and (y1+y2)/2 = (3-5)/2 and (2+6)/2 = (-1,4)
C. The symmetric point of C about the x-axis is (1,4)
2. 8 books total 2 books fiction= 2/8 fiction
4. 6 ferns total 2 have been watered= 4/6 havent been
6. 3/4=6/8 clamshelss
9514 1404 393
Answer:
a) w(4w-15)
b) w²
c) w(4w -15) = w²
d) w = 5
e) 5 by 5
Step-by-step explanation:
a) If w is the width, and the length is 15 less than 4 times the width, then the length is 4w-15. The area is the product of length and width.
A = w(4w -15)
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b) If w is the side length, the area of the square is (also) the product of length and width:
A = w²
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c) Equating the expressions for area, we have ...
w(4w -15) = w²
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d) we can subtract the right side to get ...
4w² -15w -w² = 0
3w(w -5) = 0
This has solutions w=0 and w=5. Only the positive solution is sensible in this problem.
The side length of the square is 5 units.
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e) The rectangle is 5 units wide, and 4(5)-15 = 5 units long.
The rectangle and square have the same width and the same area, so the rectangle must be a square.
