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Lunna [17]
2 years ago
3

PLZZ HELP What are the values of a, b, and c in the quadratic equation 0 = one-halfx2 – 3x – 2? a = one-half, b = 3, c = 2 a = o

ne-half, b = –3, c = –2 a = one-half, b = 3, c = –2 a = one-half, b = –3, c = 2

Mathematics
2 answers:
earnstyle [38]2 years ago
6 0

Answer:

a = 1/2, b = -3, and c = -2

(second option shown)

Step-by-step explanation:

Recall that the standard form of a quadratic equation is normally given as:

ax^2+bx+c=0

Therefore the coefficient "a" that accompanies the x^2 in your given expression is: "1/2",

The coefficient "b" that accompanies the term in "x" in your case is: "-3"

and the coefficient "c" that is the constant term in your case is "-2"

So the correct answer is the second option shown

Vilka [71]2 years ago
6 0

Answer:

B

Step-by-step explanation:

did it on edge

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Answer: Parents.

Step-by-step explanation:

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Here, Sharon wants to estimate the percentage of parents that use cloth diapers.

i.e.Researcher = Sharon

Since Sharon will reach parents about her objective of whether they use cloth diapers.

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Harry predicts that if he can cycle 5 miles in 1 hour, he can travel 20 miles in 4 hours. What do you think might be a problem w
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1 year ago
The monthly profit for a small company that makes long-sleeve T-shirts depends on the price per shirt. If the price is too high,
Andrej [43]

Answer:

A. $18

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C. p\in (16,20)

Step-by-step explanation:

Function:

f(p) = -50p^2 + 1,800p - 16,000

<u>Parts A and B:</u>

The price that generates the maximum profit is ate vertex of parabola. Find the coordinates of the vertex:

p_v=\dfrac{-b}{2a}=\dfrac{-1,800}{2\cdot (-50)}=\dfrac{1,800}{100}=18\\ \\f(p_v)=-50\cdot 18^2+1,800\cdot 18-16,000=200

The price that generates the maximum profit is $18

The maximum profit is $200

<u>Part C:</u>

The company breaks even when the profit is positive. From the graph of the function you can see that the graph of the function is over p-axis for all p\in (16,20), so the positive profit is for p\in (16,20)

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Bingel [31]

Answer:

We need a sample size of at least 719

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\alpha = \frac{1-0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

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How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?

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(\sqrt{n})^{2} = (\frac{1.96*4.103}{0.3})^{2}

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Rouding up

We need a sample size of at least 719

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Step-by-step explanation:

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4 0
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