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3 years ago

Find all the points having an x coordinate of 1 whose distance from the point (-2,-3) is 5

1 answer:

6 0

$The\ distance\ between\ A(x_A;\ y_A)\ and\ B(x_B;\ y_B):\\\\d=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}\\\\------------------------------\\\\A(1;\ y);\ B(-2;-3);\ d=5\\\\subtitute\\\\\sqrt{(-2-1)^2+(-3-y)^2}=5\\\sqrt{(-3)^2+(-3-y)^2}=5\\\sqrt{9+(-3-y)^2}=5\Rightarrow9+(-3-y)^2=5^2\\9+(-3-y)^2=25\ \ \ \ |subtract\ 9\ from\ both\ sides\\(-3-y)^2=16\iff-3-y=-\sqrt{16}\ or\ -3-y=\sqrt{16}\\-3-y=-4\ or\ -3-y=4\ \ \ \ |add\ 3\ to\ both\ sides\\-y=-1\ or\ -y=7\ \ \ \ |change\ the\ signs\\y=1\ or\ y=-7$

$Answer:\boxed{(1;\ 1)\ and\ (1;-7).}$

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A random sample of 9 wheels of cheese yielded the following weights in pounds has a sample mean of 20.90 and a sample variance o

Ronch [10]

Answer:

$2.002 \leq \sigma^2 \leq 11.365$

Step-by-step explanation:

1) Data given and notation

s represent the sample standard deviation

$s^2$ represent the sample variance

n=9 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:

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

On this case we need to find the sample standard deviation with the following formula:

$s=sqrt{\frac{\sum_{i=1}^9 (x_i -\bar x)^2}{n-1}}
The sample variance given was [tex]s^2=3.45$

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

$df=n-1=9-1=8$

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(0.05,8)" "=CHISQ.INV(0.95,8)". so for this case the critical values are:

$\chi^2_{\alpha/2}=15.507$

$\chi^2_{1- \alpha/2}=2.732$

And replacing into the formula for the interval we got:

$\frac{(9)(3.45)}{15.507} \leq \sigma \frac{(9)(3.45)}{2.732}$

$2.002 \leq \sigma^2 \leq 11.365$

7 0

3 years ago

Which ratio does not belong with the other three? explain your answer

melomori [17]

Hi I am jazmin...that's what she said :)

4 0

3 years ago

Cherries cost 4 dollars per pound. Mother bought r lbs and father bought 2 lbs. How much more money did mother spend on cherries

vazorg [7]

T= the total cost

T=4r+2

4 0

3 years ago

Read 2 more answers

How do you find answer for 8k=4/9

nasty-shy [4]

Isolate k by dividing both sides by 8

8k=4/9
k=4/72

Reduce if possible

4/72 -> 1/18

8 0

3 years ago

Read 2 more answers

To find the percent of the yellow tint in the new mixture?

Ivanshal [37]

1)We need to know the amount of yellow tint in the old mixture.

amount of yellow tint=30% of 40 liters
30%=30/100=0.3

Amount of yellow tint=0.3(40 liters)=12 liters.

2)We need to find the amount of yellow tint in the new mixture.

amount of yellow tint in the new mixture=12 liters + 9 liters=21 liters.

3) we calculate the amount of the new mixture:
amount of new mixture=40 liters + 9 liters=49 liters

4) we compute the percent of yellow tint in the new mixture:

percent of yellow tint =(amount of yellow tint / amount of mixture)*100

percent of yellow tint=(21 liters / 49 liters)100≈42.9%

Answer: The percent of yellow tint in the new mixture is 42.9%.

note: if this answer is not correct you can try it with 42.8%

6 0

3 years ago