The value of c is 81/64 for the expression a perfect square trinomial.
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What is an Expression?</h3>
An expression can be defined as a mathematical statement that consists of variables, constants and mathematical operators simultaneously.
It is given that an expression
x² +9/4x +c is a perfect square polynomial
c =?
(a+b)² = a² +2ab+b²
2ab = 9/4
b = 9/8
c = b² = 81/64
Therefore the value of c is 81/64 for the expression a perfect square trinomial.
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Answer:
2x^2-3x-5=y
(2x-5)(x+1)=y
(2x-5)(x+1)=0
2x-5=0
2x=5
x=2.5
x+1=0
x=-1
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Step-by-step explanation:
The answer is A because if you multiply you’ll get 70°
Answer:
C = 6.75n + 50
Step-by-step explanation:
The equation for the cost of renting the party room can be written in the form;
C= mn+k ......i
Where C = cost
n = number of people
Substituting the 2 cases into the equation we have,
Case 1
117.5 = 10m + k ......1
Case 2
151.25 = 15m + k .....2
Subtracting eqn 1 from 2 we have
151.25 - 117.50 = 15m - 10m
5m = 33.75
m = 33.75/5 = 6.75
Substituting m = 6.75 into eqn 1, we have
117.50 = 10(6.75) + k
k = 117.5 - 67.50
k = 50
Therefore, rewriting the eqn i, we have
C = 6.75n + 50
Which data set has an outlier? 25, 36, 44, 51, 62, 77 3, 3, 3, 7, 9, 9, 10, 14 8, 17, 18, 20, 20, 21, 23, 26, 31, 39 63, 65, 66,
umka21 [38]
It's hard to tell where one set ends and the next starts. I think it's
A. 25, 36, 44, 51, 62, 77
B. 3, 3, 3, 7, 9, 9, 10, 14
C. 8, 17, 18, 20, 20, 21, 23, 26, 31, 39
D. 63, 65, 66, 69, 71, 78, 80, 81, 82, 82
Let's go through them.
A. 25, 36, 44, 51, 62, 77
That looks OK, standard deviation around 20, mean around 50, points with 2 standard deviations of the mean.
B. 3, 3, 3, 7, 9, 9, 10, 14
Average around 7, sigma around 4, within 2 sigma, seems ok.
C. 8, 17, 18, 20, 20, 21, 23, 26, 31, 39
Average around 20, sigma around 8, that 39 is hanging out there past two sigma. Let's reserve judgement and compare to the next one.
D. 63, 65, 66, 69, 71, 78, 80, 81, 82, 82
Average around 74, sigma 8, seems very tight.
I guess we conclude C has the outlier 39. That one doesn't seem like much of an outlier to me; I was looking for a lone point hanging out at five or six sigma.