If a 75 ft flagpole casts a shadow that is 43 long, what is the angle of elevation of the sun from the tip of the shadow?

A department store buys 200 shirts at a cost of $1,400 and sells them at a selling price of $10 each. Find the percent markup.

Dima020 [189]

The answer is 200 -1400 +10=19

5 0

2 years ago

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6x - 5y = 15, x=y+3 what does x and y equal

Natasha_Volkova [10]

Answer:

Step-by-step explanation:

6x - 5y = 15....... (1)

x=y+3..........(2)

Putting (2) in (1)

6(y + 3) - 5y = 15

6y + 18 - 5y = 15

6y - 5y = 15 - 18

y = -3

And x = y + 3

x = -3 + 3

x = 0

3 0

3 years ago

I posted this 2 hours ago can someone pls help?

nikdorinn [45]

Answer:

0.314 gallon is required to cover a circular region with a radius of 40 feet

Step-by-step explanation:

Given

a gallon of paint will cover about 400 square feet in a single coat

To Find:

Number of gallons required to cover a circular region with a radius of 40 feet = ?

Solution:

Step 1: Finding Area of the circular region

The area of the circular region, A = $\pi r^2$

where r is the radius

Now substituting r = 40 , we get

A = $\pi \times 40$

$A = 125.6 ft^2$

Step 2: Finding the number of gallons of paint required

To cover 400 sq feet = 1 gallon

so to cover 1 square feet = $\frac{1}{400}$ gallon

Now to cover 125.6 ft we need

$125.6 \times \frac{1}{400}$

$\frac{125.6}{400}$

=> 0.314 gallon

4 0

3 years ago

What are the endpoint coordinates for the midsegment of △BCD that is parallel to BC¯¯¯¯¯?

wariber [46]

Answer:

The perfect person does not ex---False

Thank you so much man, great help for us people facing brain farts.

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Calculate the sample standard deviation and sample variance for the following frequency distribution of hourly wages for a sampl

lidiya [134]

Answer:

(a) The sample variance is 16.51

(a) The sample standard deviation is 4.06

Step-by-step explanation:

Given

$\begin{array}{cc}{Class} & {Frequency} & 8.26 - 10.00 & 20 &10.01-11.75 & 38 &11.76 - 13.50& 36 & 13.51-15.25 &25&15.26-17.00 &27 &\ \end{array}$

Solving (a); The sample variance.

First, calculate the class midpoints.

This is the mean of the intervals.

i.e.

$x_1 = \frac{8.26+10.00}{2} = \frac{18.26}{2} = 9.13$

$x_2 = \frac{10.01+11.75}{2} = \frac{21.76}{2} = 10.88$

$x_3 = \frac{11.76+13.50}{2} = \frac{25.26}{2} = 12.63$

$x_4 = \frac{13.51+15.25}{2} = \frac{28.76}{2} = 14.38$

$x_5 = \frac{15.26+17.00}{2} = \frac{32.26}{2} = 16.13$

So, the table becomes:

$\begin{array}{ccc}{Class} & {Frequency} & {x} & 8.26 - 10.00 & 20&9.13 &10.01-11.75 & 38 &10.88&11.76 - 13.50& 36 &12.63& 13.51-15.25 &25&14.38&15.26-17.00 &27 &16.13\ \end{array}$

Next, calculate the mean

$\bar x = \frac{\sum fx}{\sum f}$

$\bar x = \frac{20*9.13 + 38 * 10.88+36*12.63+25*14.38+27*16.13}{20+38+36+25+27}$

$\bar x = \frac{1845.73}{146}$

$\bar x = 12.64$

Next, the sample variance is:

$\sigma^2 = \frac{\sum f(x - \bar x)^2}{\sum f - 1}$

So, we have:

$\sigma^2 = \frac{20*(9.13-12.63)^2 + 38 * (10.88-12.63)^2 +...........+27 * (16.13 -12.63)^2}{20+38+36+25+27-1}$

$\sigma^2 = \frac{2393.6875}{145}$

$\sigma^2 = 16.51$

The sample standard deviation is:

$\sigma = \sqrt{\sigma^2}$

$\sigma = \sqrt{16.51}$

$\sigma = 4.06$

6 0

2 years ago