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Vilka [71]
2 years ago
15

Please help. Is algebra. PLEASE HELP NO LINKS OR FILES

Mathematics
2 answers:
Temka [501]2 years ago
4 0

Question 9

(7y - 3z)^2

(7y - 3z)(7y - 3z)

We now use the FOIL method.

49y^2 -42yz + 9z^2

Question 10

(3b + 2)(3b - 2)

Use FOIL.

9b^2 - 4

Question 11

(6x + 5)(3x + 7)

Use FOIL.

18x^2 + 42x + 15x + 35

18x^2 + 57x + 35

I just did your homework. What's going to happen when I am not sitting next to you in class? Think about it.

BabaBlast [244]2 years ago
3 0
Question 9. d

Question 10. b

Question 11.b
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Please answer correctly !!!!!! Will mark brainliest answer !!!!!!!!!
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Answer:

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Step-by-step explanation:

(x + 1.5)² - 2.25 - 28 = 0

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3 years ago
Simplify the expression<br> 26 · 268
Gre4nikov [31]

Answer:

26 * 26^8 = 5.4295037e+12.

Step-by-step explanation:

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6 0
2 years ago
Which Two Numbers add Up To 11 and Multiply To 10
alekssr [168]


The two numbers are 10 and 1. When you add 10 and 1, it adds up to 11. When you multiply 10 by 1, the answer is 10.

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8 0
2 years ago
Community college students conduct a survey at their college. They ask "Do you plan to transfer to a university to pursue a bacc
Tamiku [17]

Answer:

0.8 - 2.58\sqrt{\frac{0.8(1-0.8)}{100}}=0.697

0.8 + 2.58\sqrt{\frac{0.8(1-0.8)}{100}}=0.903

The 99% confidence interval would be given by (0.697;0.903)

And the best option is : 0.697 to 0.903

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

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".  

Solution to the problem

For this case the conditions we have the conditions to assume a normal distribution for the population proportion since:

np=100*0.8=80 >10 , n(1-p) = 10*(1-0.8) =20 >10

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.005. And the critical value would be given by:

z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58

The confidence interval for the mean is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

The estimated proportion is given by \hat p =\frac{80}{100} =0.8. If we replace the values obtained we got:

0.8 - 2.58\sqrt{\frac{0.8(1-0.8)}{100}}=0.697

0.8 + 2.58\sqrt{\frac{0.8(1-0.8)}{100}}=0.903

The 99% confidence interval would be given by (0.697;0.903)

And the best option is : 0.697 to 0.903

6 0
2 years ago
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