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3 years ago

Plz help ..............!!!!!

Mathematics

2 answers:

lana [24]3 years ago

4 0

Answer:

1.8

is the median

Cloud [144]3 years ago

4 0

Answer: 1.8

Step-by-step explanation: 1.8 is the median

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Directions - For the following problem, write a paragraph proof to justify each step you make. All work must be neat,

nlexa [21]

$$ PS = \frac{4}{3} x$

Solution:

Given PRQ is a triangle.

ST is a line parallel to RQ.

$$PT = x, \ PQ = 3x, \ SR=\frac{8}{3}x$

$TQ=PQ-PT$

$TQ=3x -x=2x$

Triangle proportionality theorem,

If a line parallel to one side of a triangle intersects the other two sides, then it divides the two sides proportionally.

$$\frac{PS}{SR} =\frac{PT}{TQ}$

$$\frac{PS}{\frac{8}{3} x} =\frac{x}{2x}$

Do cross multiplication, we get

$$ PS \times 2x=x \times \frac{8}{3} x$

Divide by 2x on both sides, we get

$$ PS = \frac{4}{3} x$

4 0

3 years ago

Joe

lara31 [8.8K]

Answer:

muji chus mero lado madeaghw you mother machinery

6 0

2 years ago

Read 2 more answers

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:

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