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umka2103 [35]
3 years ago
11

which are perfect square trinomials? select two options A. x^2 -9 B. x^2 -100 C. x^2 - 4x + 4 D. x^2 +10x +25 E. x^2 +15x + 36​

Mathematics
2 answers:
salantis [7]3 years ago
5 0
C and D
x^2 - 4x + 4 = (x-2)^2
x^2 + 10x +25 = (x+5)^2
Dmitry [639]3 years ago
4 0

Answer:

C. x^2 - 4x + 4

D. x^2 +10x +25

Step-by-step explanation:

A trinomial ax^2+bx+c is perfect square trinomial if,

D=b^2 - 4ac = 0,

A. x^2 -9

By comparing,

a = 1, b = 0, c = -9,

0^2 - 4\times 1\times -9\neq 0

⇒ it is not a perfect square trinomial.

B. x^2 -100

By comparing,

a = 1, b = 0, c = -100,

0^2 - 4\times 1\times -100\neq 0

⇒ it is not a perfect square trinomial.

C. x^2 - 4x + 4

By comparing,

a = 1, b = -4, c = 4,

(-4)^2 - 4\times 1\times 4=16-16=0

⇒ it is a perfect square trinomial.

D. x^2 +10x +25

By comparing,

a = 1, b = 10, c = 25,

10^2 - 4\times 1\times 25=100-100=0

⇒ it is a perfect square trinomial.

E.  x^2 +15x + 36

By comparing,

a = 1, b = 15, c = 36,

15^2 - 4\times 1\times 36=225-144=81\neq 0

⇒ it is not a perfect square trinomial.

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Answer:

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Step-by-step explanation:

6×3-1

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A group of 40 children attend a baseball game. Each child received either a hotdog or a bag of popcorn. Hotdogs were $2.25 and p
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Answer:

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Step-by-step explanation:

Had all received popcorn, the bill would have been 40×$1.75 = $70. The bill was $13.50 more than that. Each hot dog purchased in place of popcorn adds $0.50 to the bill, so the number of hot dogs must be ...

  $13.50/$0.50 = 27

Of course, the remainder of the 40 items were popcorn, so 13 bags of popcorn.

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Suppose that the population of the scores of all high school seniors who took the SAT Math (SAT-M) test this year follows a Norm
Vlad1618 [11]

Answer:

Confidence =1-\alpha=1-0.01=0.99

And then the confidence level would be given by 99%

Step-by-step explanation:

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Assuming the X follows a normal distribution

X \sim N(\mu, \sigma)

The distribution for the sample mean is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

\bar x =512 represent the sample mean

\sigma=100 represent the population standard deviation

n= 100 sample size selected.

The confidence interval is given by this formula:

\bar X \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}   (1)

The marginof error for this case is given by Me=25.76. And we know that the formula for the margin of error is given by:

Me=z_{\alpha/2} \frac{\sigma}{\sqrt{n}}

25.76=z_{\alpha/2} \frac{100}{\sqrt{100}}

And we can find the critical value z_{\alpha/2} like this:

z_{\alpha/2}=\frac{25.76(\sqrt{100})}{100}=2.576

And we know that on the right tail of the z score =2.576 we have \alpha/2 of the total area. We can find the area on the right of the z score using this excel code:

"=1-NORM.DIST(2.576,0,1,TRUE)" or using a table of the normal standard distribution, and we got 0.004998=\alpha/2, so then \alpha=0.00498*2=0.009995 \approx 0.01, and then we can find the confidence like this:

Confidence =1-\alpha=1-0.01=0.99

And then the confidence level would be given by 99%

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