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2 years ago

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How will adding the value 75 affect the mean and median of the data set 1, 5, 7, 9 , 9, 10? * A. The median increases less than

the mean increases. B. The mean and the median increase by the same amount. C. The mean increases and the median stays the same. D. The median increases and the mean stays the same.

1 answer:

padilas [110]2 years ago

3 0

The answer is A.
Hope this helps!

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4(8x−1) = 19+32x

Fantom [35]

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Answer:

Let's solve your equation step-by-step.

4 ( 8x − 1 ) = 19 + 32x

Step 1: Simplify both sides of the equation.

4 ( 8x − 1 ) = 19 + 32x

( 4 ) ( 8x ) + ( 4 ) ( −1 ) = 19 + 32x ( Distribute )

32x + −4 = 19 + 32x

32x − 4 = 32x + 19

Step 2: Subtract 32x from both sides.

32x − 4 − 32x = 32x + 19 − 32x

−4 = 19

Step 3: Add 4 to both sides.

−4 + 4 = 19 + 4

0 = 23

Answer:

There are no solutions.

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5 0

2 years ago

Read 2 more answers

I need the answers for 2 and 3

Korolek [52]

2- x= -2 y= -1 so the slope is -1/-2
3- x= 2 y= 3 so the slope is 1 1/2
i’m not sure if this is right but i’m learning this as well

5 0

2 years ago

Mariulka [41]

Answer:

$A = 74.7^\circ$

$B = 42.5^\circ$

$C = 62.8^\circ$

Step-by-step explanation:

Given

$A = (-1,2) \to (x_1,y_1)$

$B = (2,8) \to (x_2,y_2)$

$C = (4,1) \to (x_3,y_3)$

Required

The measure of each angle

First, we calculate the length of the three sides of the triangle.

This is calculated using distance formula

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2$

For AB

$A = (-1,2) \to (x_1,y_1)$

$B = (2,8) \to (x_2,y_2)$

$d = \sqrt{(-1 - 2)^2 + (2 - 8)^2$

$d = \sqrt{(-3)^2 + (-6)^2$

$d = \sqrt{45$

So:

$AB = \sqrt{45$

For BC

$B = (2,8) \to (x_2,y_2)$

$C = (4,1) \to (x_3,y_3)$

$BC = \sqrt{(2 - 4)^2 + (8 - 1)^2$

$BC = \sqrt{(-2)^2 + (7)^2$

$BC = \sqrt{53$

For AC

$A = (-1,2) \to (x_1,y_1)$

$C = (4,1) \to (x_3,y_3)$

$AC = \sqrt{(-1 - 4)^2 + (2 - 1)^2$

$AC = \sqrt{(-5)^2 + (1)^2$

$AC = \sqrt{26$

So, we have:

$AB = \sqrt{45$

$BC = \sqrt{53$

$AC = \sqrt{26$

By representation

$AB \to c$

$BC \to a$

$AC \to b$

So, we have:

$a = \sqrt{53$

$b = \sqrt{26$

$c = \sqrt{45$

By cosine laws, the angles are calculated using:

$a^2 = b^2 + c^2 -2bc \cos A$

$b^2 = a^2 + c^2 -2ac \cos B$

$c^2 = a^2 + b^2 -2ab\ cos C$

$a^2 = b^2 + c^2 -2bc \cos A$

$(\sqrt{53})^2 = (\sqrt{26})^2 +(\sqrt{45})^2 - 2 * (\sqrt{26}) +(\sqrt{45}) * \cos A$

$53 = 26 +45 - 2 * 34.21 * \cos A$

$53 = 26 +45 - 68.42 * \cos A$

Collect like terms

$53 - 26 -45 = - 68.42 * \cos A$

$-18 = - 68.42 * \cos A$

Solve for $\cos A$

$\cos A =\frac{-18}{-68.42}$

$\cos A =0.2631$

Take arc cos of both sides

$A =\cos^{-1}(0.2631)$

$A = 74.7^\circ$

$b^2 = a^2 + c^2 -2ac \cos B$

$(\sqrt{26})^2 = (\sqrt{53})^2 +(\sqrt{45})^2 - 2 * (\sqrt{53}) +(\sqrt{45}) * \cos B$

$26 = 53 +45 -97.67 * \cos B$

Collect like terms

$26 - 53 -45= -97.67 * \cos B$

$-72= -97.67 * \cos B$

Solve for $\cos B$

$\cos B = \frac{-72}{-97.67}$

$\cos B = 0.7372$

Take arc cos of both sides

$B = \cos^{-1}(0.7372)$

$B = 42.5^\circ$

For the third angle, we use:

$A + B + C = 180$ --- angles in a triangle

Make C the subject

$C = 180 - A -B$

$C = 180 - 74.7 -42.5$

$C = 62.8^\circ$

8 0

2 years ago

It x, y, and z are positive intergers and 3x = 4y = 7z, then the least possible value of x + y + z is???

Dmitry [639]

$3x = 4y = 7z \\ \Rightarrow \begin{cases}3x - 4y = 0 \\4y - 7z = 0 \\ 7z - 3x = 0\end{cases} \\ \Leftrightarrow \begin{cases}x = \frac{4}{3} y\\4y - 7z = 0 \\ 7z - 3( \frac{4}{3}y) = 0\end{cases} \\ \Leftrightarrow \begin{cases}x = \frac{4}{3} y\\4y - 7z = 0 \\ 7z - 4y= 0\end{cases} \\ \Leftrightarrow \begin{cases}x = \frac{4}{3} y\\ y \\ z = \frac{4}{7}y \end{cases} \\ x,y,z\: are\: intergers\Rightarrow y=LCM(3,7)=21\Rightarrow x = 28,z = 12 \\ \Rightarrow x + y + z = 61$

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2 years ago

Henry bought 1/4 pounds of screws and 2/5 of backstop build a skateboard ramp. What is the total weight of the screws and nails

nexus9112 [7]

It equals 0.65 pounds

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2 years ago