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3 years ago

What is the distance between (2, 5) and (2, -7)?

Mathematics

1 answer:

MAVERICK [17]3 years ago

4 0

Answer:

Step-by-step explanation:

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I think its letter D because x + 2y fits with the equation and there for -3

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3 years ago

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What is the solution to 3x2 + 4x - 8 = 0?

Ber [7]

Answer:

The answer is c

Step-by-step explanation:

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3 years ago

In the first quadrant, you start at (8, 11)

Anestetic [448]

Answer:

(11,2)

Step-by-step explanation:

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3 years ago

Which equation is equivalent to

marshall27 [118]

Answer:

C) 29x=5

reason:

first, distribute the parentheses
this will make your equation 8-20x=6-3+9x
simplify to get 8-20x=3+9x
add 20x to both sides to get 8=3+29x
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3 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:

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

Hence, the 99% confidence interval is [51.2884,82.7716]

7 0

3 years ago