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Alchen [17]
2 years ago
14

Idk how to do this without an image honestly

Mathematics
1 answer:
Art [367]2 years ago
6 0

9514 1404 393

Answer:

  (x, y) ⇒ (x -m, y -n)

Step-by-step explanation:

Reverse the arrow and solve for the x and y on the left:

  (x +m, y +n) ⇒ (x, y)

This resolves to two relations:

  x + m ⇒ x

  x ⇒ x - m . . . . . subtract m

and

  y + n ⇒ y

  y ⇒ y - n . . . . . . subtract n

Putting these back together, we have ...

  (x, y) ⇒ (x -m, y -n) . . . . . maps P'Q' to PQ

_____

The original translation was right m and up n. The reverse translation is left m and down n. That is what the rule shows.

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120/50
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What is greater 66mm or 6cm
LuckyWell [14K]
66mm=6cm
So that means that any 66mm is 6cm point 6mm ===> 66cmm=6.6cm
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3 years ago
A simple random sample from a population with a normal distribution of 98 body temperatures has x =98.90 °F and s =0.68°F. Const
Ludmilka [50]

Answer:

0.609 \leq \sigma \leq 0.772  

And the best conclusion would be:

D. This conclusion is safe because 1.40 °F is outside the confidence interval.

Step-by-step explanation:

1) Data given and notation  

s=0.68 represent the sample standard deviation  

\bar x =98.90 represent the sample mean  

n=98 the sample size  

Confidence=90% or 0.90  

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 mean or variance 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.  

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .  

2) Calculating the confidence interval  

The confidence interval for the population variance is given by the following formula:  

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}  

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:  

df=n-1=98-1=97  

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.05,97)" "=CHISQ.INV(0.95,97)". so for this case the critical values are:  

\chi^2_{\alpha/2}=120.990  

\chi^2_{1- \alpha/2}=75.282  

And replacing into the formula for the interval we got:  

\frac{(97)(0.68)^2}{120.990} \leq \sigma \leq \frac{(97)(0.68)^2}{75.282}  

0.371 \leq \sigma^2 \leq 0.596  

Now we just take square root on both sides of the interval and we got:  

0.609 \leq \sigma \leq 0.772  

And the best conclusion would be:

D. This conclusion is safe because 1.40 °F is outside the confidence interval.

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BARSIC [14]

Answer:

y = 2x -8

Step-by-step explanation:

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