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3 years ago

Find the equation of the circle that has a diameter with endpoint located at (-5,-3) and (-11,-3).

Mathematics

1 answer:

GREYUIT [131]3 years ago

7 0

Check the picture below.

you can pretty much, without having to use the distance formula, see the diameter is 6 units long, thus the radius is half that or 3 then.

now, without having to use the midpoint formula, if we move from either endpoint inwards 3 units, we'll end up at the center, -5 -3, is just -8, so the center will be at -8, -3.

$\bf \textit{equation of a circle}\\\\ 
(x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2
\qquad 
center~~(\stackrel{-8}{{{ h}}},\stackrel{-3}{{{ k}}})\qquad \qquad 
radius=\stackrel{3}{{{ r}}}
\\\\\\\
[x-(-8)]^2+[y-(-3)]^2=3^2\implies (x+8)^2+(y+3)^2=9$

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You need to rent a bowling lane. On Friday nights, you have two options. Option A is a $20 lane rental plus $3 per game. Option

ExtremeBDS [4]

For 5 games, the cost will be equal for each option.

Step-by-step explanation:

Given,

Lane rental of offer A = $20

Per game charges = $3

Let,

x be the number of games.

A(x) = 3x + 20

Lane rental is $35 for maximum 10 games.

B(x) = 35

For the cost to be same;

A(x) = B(x)

$3x+20=35\\3x=35-20\\3x=15$

Dividing both sides by 3

$\frac{3x}{3}=\frac{15}{3}\\x=5$

For 5 games, the cost will be equal for each option.

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2 years ago

The USA Today reports that the average expenditure on Valentine's Day is $100.89. Do male and female consumers differ in the amo

Aloiza [94]

Answer:

$a)\ \ \bar x_m-\bar x_f=67.03\\\\b)\ \ E=15.7416\\\\c)\ \ CI=[51.2884, \ 82.7716]$

Step-by-step explanation:

a. -Given that:

$n_m=41\ , \ \sigma_m=33, \bar x_m=135.67\\\\n_f=37. \ \ ,\sigma_f=20, \ \ \bar x_f=68.64$

#The point estimator of the difference between the population mean expenditure for males and the population mean expenditure for females is calculated as:

$\bar x_m-\bar x_f\\\\\therefore \bigtriangleup\bar x=135.67-68.64\\\\=67.03$

Hence, the pointer is estimator 67.03

b. The standard error of the point estimator, $\bar x_m-\bar x_f$ is calculated by the following following:

$\sigma_{\bar x_m-\bar x_f}=\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}$

-And the margin of error, E at a 99% confidence can be calculated as:

$E=z_{\alpha/2}\times \sigma_{\bar x_m-\bar x_f}\\\\\\=z_{0.005}\times\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}\\\\\\=2.575\times \sqrt{\frac{33^2}{41}+\frac{20^2}{37}}\\\\\\=15.7416$

Hence, the margin of error is 15.7416

c. The estimator confidence interval is calculated using the following formula:

$\bar x_m-\bar x_f\ \pm z_{\alpha/2}\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}$

#We substitute to solve for the confidence interval using the standard deviation and sample size values in a above:

$CI=\bar x_m-\bar x_f\ \pm z_{\alpha/2}\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}\\\\=(135.67-68.64)\pm 15.7416\\\\=67.03\pm 15.7416\\\\=[51.2884, \ 82.7716]$