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2 years ago

Stacey went to six different auto repair shops to get the following estimates to repair her car.

Mathematics

1 answer:

topjm [15]2 years ago

5 0

Answer:

534 is the mean.

Step-by-step explanation:

to calculate the mean you add all the numbers together (for example this set of numbers):

785, 585, 465, 409, 495, 465

=3204

then divide the number of numbers that you added together , (in this case 6)

3204/6

534

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Help i will give brainlist

Arisa [49]

Answer:

3. 2/10 4/20 3/15

4. 12/1 48/4 36/3

Step-by-step explanation:

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3 years ago

Find the interquartile range (IQR) of the data in the dot plot below.

kolbaska11 [484]

Answer:

<em>(IQR) interquartile range: 12 - 9 = 3</em>

Step-by-step explanation:

Median: 10

Lower quartile: 9

Upper quartile: 12

Interquartile Range: 12 - 9 = 3

5 0

3 years ago

Of the 800 participants in a marathon 220 are running to raise money for a cause. How many participants out of 100 are running f

nadezda [96]

Answer:

27.5 /100 are running for a cause

Step-by-step explanation:

We know that 220/800 are running for a cause

Divide the top and bottom by 8 to get the fraction out of 100

220 /8 =27.5

800/8 = 100

27.5 /100 are running for a cause

7 0

2 years ago

I need help finding the answer!!

Whitepunk [10]

Answer:

The second option

Step-by-step explanation:

Here, we need to multiply each part of the matrix by -10.

This will give us: [ -230 380 ]

[ -170 60 ]

So, the answer is the second option.

3 0

1 year ago

Consider the following equation. f(x, y) = y3/x, P(1, 2), u = 1 3 2i + 5 j (a) Find the gradient of f. ∇f(x, y) = Correct: Your

BaLLatris [955]

$f(x,y)=\dfrac{y^3}x$

a. The gradient is

$\nabla f(x,y)=\dfrac{\partial f}{\partial x}\,\vec\imath+\dfrac{\partial f}{\partial y}\,\vec\jmath$

$\boxed{\nabla f(x,y)=-\dfrac{y^3}{x^2}\,\vec\imath+\dfrac{3y^2}x\,\vec\jmath}$

b. The gradient at point P(1, 2) is

$\boxed{\nabla f(1,2)=-8\,\vec\imath+12\,\vec\jmath}$

c. The derivative of $f$ at P in the direction of $\vec u$ is

$D_{\vec u}f(1,2)=\nabla f(1,2)\cdot\dfrac{\vec u}{\|\vec u\|}$

It looks like

$\vec u=\dfrac{13}2\,\vec\imath+5\,\vec\jmath$

so that

$\|\vec u\|=\sqrt{\left(\dfrac{13}2\right)^2+5^2}=\dfrac{\sqrt{269}}2$

Then

$D_{\vec u}f(1,2)=\dfrac{\left(-8\,\vec\imath+12\,\vec\jmath\right)\cdot\left(\frac{13}2\,\vec\imath+5\,\vec\jmath\right)}{\frac{\sqrt{269}}2}$

$\boxed{D_{\vec u}f(1,2)=\dfrac{16}{\sqrt{269}}}$

7 0

3 years ago