All categories

3 years ago

Find the coordinates of the vertices of the image QRST after a reflection over the x axis.

Mathematics

1 answer:

Ivenika [448]3 years ago

4 0

Answer:

t - (-2,-1)

q - (1,-3)

r - (3,3)

s - (0,2)

Step-by-step explanation:

You might be interested in

Find the equation of the line that passes through (-2,-3) and is parallel to 3x+2y=5

Ymorist [56]

Answer:

8,3

Step-by-step explanation:

4 0

3 years ago

Describe how you find an upper bound and a lower bound for the zeros of a polynomial function

Likurg_2 [28]

Answer:

A quick way to calculate upper and lower bands is to halve the degree of accuracy specified, then add this to the rounded value for the upper bound and subtract it from the rounded value for the lower bound.

5 0

3 years ago

HELP!! ILL MARK BRAINLIEST

Sonja [21]

Answer:

Step-by-step explanation:

youre welcome

7 0

3 years ago

Read 2 more answers

Andrew is creating a rectangular dog run in his back yard. The length of the dog run is 18 feet. The perimeter of the dog run mu

ale4655 [162]

Answer: $3\leq x \leq 18$

Step-by-step explanation:

Given, the length of the rectangular dog run is 18 feet.

Formula: Perimeter = 2 ( length + width)

Let width be x.

Then, Perimeter = 2 (18 + x ) feet

The perimeter of the dog run must be at least 42 feet and no more than 72 feet.

That is

$42\leq 2 (x+18)\leq 72$

Divide inequality by 2, we get

$21\leq 18+x\leq 36$

Subtract 18 from each side, we get

$3\leq x \leq 18$

Required inequality : $3\leq x \leq 18$

6 0

3 years ago

For many years businesses have struggled with the rising cost of health care. But recently, the increases have slowed due to les

kaheart [24]

Answer:

The confidence interval would be given by this formula

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

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

The margin of error for this case is given by:

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

And replacing we got:

$ME = 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.0259$

And replacing into the confidence interval formula we got:

$0.52 - 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.4941$

$0.52 + 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.5459$

And the 95% confidence interval would be given (0.4941;0.5459).

Step-by-step explanation:

Data given and notation

n=1000 represent the random sample taken

$\hat p=0.52$ estimated proportion of of U.S. employers were likely to require higher employee contributions for health care coverage

$\alpha=0.05$ represent the significance level (no given, but is assumed)

Solution to the problem

The confidence interval would be given by this formula

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

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

The margin of error for this case is given by:

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

And replacing we got:

$ME = 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.0259$

And replacing into the confidence interval formula we got:

$0.52 - 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.4941$

$0.52 + 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.5459$

And the 95% confidence interval would be given (0.4941;0.5459).

5 0

3 years ago