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3 years ago

12

Use matrices to determine the coordinates of the vertices of the rotated figure. Then graph the pre-image and the image of the s

ame coordinate grid. (Pictureprovided)

1 answer:

Sidana [21]3 years ago

5 0

Answer:

10-18=x=9

Step-by-step explanation:

You might be interested in

Students pass a test it they score 50% or more. the marks of a large number of students were sampled and the mean and standard d

Schach [20]

So we are given that the mean is 42% and the sd (standard deviation) is 8%
Assuming our data is normal we can use the 68-95-99 rule

So one thing you should realize is that 42% + 8% is 50% which is passing. That is one standard deviation higher. So we use:

100 - 68 - 13.5 - 2.35 - 0.15 = 16. That means 16% of students passed the test. Which is terrible. They probably need to hit the books more.

Anyways if you have any question feel free to message me!
Hopes this helps!

3 0

3 years ago

Hey can you please help me posted picture of question

Debora [2.8K]

We can use quadratic formula to determine the roots of the given quadratic equation.

The quadratic formula is:

$x= \frac{-b+- \sqrt{ b^{2} -4ac} }{2a}$

b = coefficient of x term = -11
a = coefficient of squared term = 2
c = constant term = 15

Using the values, we get:
$x= \frac{11+- \sqrt{121-4(2)(15)} }{2(2)} \\ \\ 
x= \frac{11+-1}{4} \\ \\ 
x=3, x= 2.5$

So, the correct answer to this question are option B and D

4 0

3 years ago

A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in he

UkoKoshka [18]

Answer:

a) The minimum sample size is 601.

b) The minimum sample size is 2401.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

For this problem, we have that:

We dont know the true proportion, so we use $\pi = 0.5$ , which is when we are are going to need the largest sample size.

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

a. If a 95% confidence interval with a margin of error of no more than 0.04 is desired, give a close estimate of the minimum sample size that will guarantee that the desired margin of error is achieved. (Remember to round up any result, if necessary.)

This is n for which M = 0.04. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.04 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.04\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.04}$

$(\sqrt{n})^2 = (\frac{1.96*0.5}{0.04})^2$

$n = 600.25$

Rounding up

The minimum sample size is 601.

b. If a 95% confidence interval with a margin of error of no more than 0.02 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.

Now we want n for which M = 0.02. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.02 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.02\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.02}$

$(\sqrt{n})^2 = (\frac{1.96*0.5}{0.02})^2$

$n = 2401$

The minimum sample size is 2401.

4 0

3 years ago

What is the estimate answer and the exact answer of 17/10 ÷2 4/5

OLEGan [10]

Both 17/10 and 2 4/5 should be rounded up in preparation for this estimation.

17/10 is close to 2 and 2 4/5 is close to 3. Thus, your estimated answer should be
2/3, or approx. 0.6666....

Exact answer: divide 17/10 by 14/5. LCD is 10, so convert 14/5 to 28/10.

Now divide 17/10 by 28/10. Answer: 17/28 = approx. = 0.607 approx.

These two results are comparable: 0.6666.... and 0.6071 ....

8 0

4 years ago

On a coordinate grid, in which quadrant is the point (-5,4) located

lapo4ka [179]

Quadrant 2 is its location.

3 0

3 years ago

Read 2 more answers