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3 years ago

The image point of A after a translation left 3 units and down 1 unit is the point

Mathematics

1 answer:

trasher [3.6K]3 years ago

8 0

Answer:

(-3, 4)

Step-by-step explanation:

First, add 3 to point B's x coordinate:

-6 + 3 = -3

Then, add 1 to the y coordinate:

3 + 1

= 4

So, the coordinates of pre image point A are (-3, 4)

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A teacher claims that the proportion of students expected to pass an exam is greater than 80%. To test this claim, the teacher a

erica [24]

Answer:

z=-1.591

Step-by-step explanation:

Null Hypotheses,

$H_0 : p=0.8\\H_a: p>0.80$

So we use z-test for one population proportion (right-tailed test)

According to this informaton, we defined that

$\alpha=0.05$

$z_c=1.64$ (critical value)

So our Rejection region is $R={z:z>1.64}$

$z=\frac{\bar{p}-p_0}{\sqrt{p_0(1-p_0)/n}} = \frac{0.755-0.8}{\sqrt{0.8(1-0.8)/200}}=-1.591$

Not rejection.

6 0

3 years ago

How to convert 655.575 to word form?

Leni [432]

Six hundred fifty five thousand five hundred seventy five

6 0

3 years ago

Which of the following sets could be the sides of a right triangle?

Genrish500 [490]

Answer:

None of the above

Step-by-step explanation:

None of the answer choices are even triangles, let alone right triangles. The sum of the two smaller sides isn't larger than the bigger side on any of the triangles, which means that they cannot be closed.

3 0

3 years ago

You estimated the average rental cost of a 2 bedroom apartment in Denton. A random sample of 16 apartments was taken. The sample

Zigmanuir [339]

Answer:

The new sample size required in order to have the same confidence 95% and reduce the margin of erro to $60 is:

n=28

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

$X \sim N(\mu, \sigma=160)$

And the distribution for $\bar X$ is:

$\bar X \sim N(\mu, \frac{160}{\sqrt{n}})$

We know that the margin of error for a confidence interval is given by:

$Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The next step would be find the value of $\z_{\alpha/2}$ , $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$

Using the normal standard table, excel or a calculator we see that:

$z_{\alpha/2}=\pm 1.96$

If we solve for n from formula (1) we got:

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

$n=(\frac{z_{\alpha/2} \sigma}{Me})^2$

And we have everything to replace into the formula:

$n=(\frac{1.96(160)}{78.4})^2 =16$

And this value agrees with the sample size given.

For the case of the problem we ar einterested on Me= $60, and we need to find the new sample size required to mantain the confidence level at 95%. We know that n is given by this formula:

$n=(\frac{z_{\alpha/2} \sigma}{Me})^2$

And now we can replace the new value of Me and see what we got, like this:

$n=(\frac{1.96*160}{60})^2 =27.32$

And if we round up the answer we see that the value of n to ensure the margin of error required $Me=\pm 60$ $ is n=28.

5 0

3 years ago

What is the answer to this problem?<br>2(-7+x)>-2

mina [271]

Isolate the variable by dividing each side by factors that don't contain the variable.

x > 6

8 0

3 years ago