Pete was paid $30 for babysitting his cousin. He now has exactly enough money to buy a bike that costs $100.

Helppp pleaseee this is due on 10 minutes with explanation pleasee thank you

Nat2105 [25]

Answer:

Jed's Market

Step-by-step explanation:

Find out the cost of one candy bar for each store:

Smith's:

9.00 ÷ 3 = $3.00

Green's:

10.00 ÷ 4 = $2.50

Jed's:

12.00 ÷ 5 = $2.40

Wan's:

It already tells you the price for each: $2.75

Finding the best deal:

Look at all our costs per 1 candy bar. Which one has the smallest price?
Because $2.40 is the smallest price, Jed's Market has the best buy.

I hope this helps!

8 0

3 years ago

Variance and standard deviation

kirza4 [7]

Answer: 8.3

Step-by-step explanation:

To find the variance, you want to find the mean of the data.

$\frac{3+3+8+1+9+6}{6} =5$

Now that we have the mean, we find the difference between each data point and the mean.

3-5=-2

8-5=3

1-5=-4

9-5=4

6-5=1

With the difference, you square each and find the average.

(-2)²=4

3²=9

(-4)²=16

4²=16

1²=1

$\frac{4+4+9+16+16+1}{6} =\frac{25}{3}$

The variance is 8.3.

3 0

3 years ago

Which will result in a difference of squares?

otez555 [7]

Expanding the given expressions using Foil:

1)(–7x + 4)(–7x + 4) = $7x^{2}-28x-28x+16=49x^{2}-56x+16$

2) (–7x + 4)(4 – 7x)= $-28x+49x^{2}+16-28x=49x^{2}-56x+16.$

3)(–7x + 4)(–7x – 4)= $49x^{2}+28x-28x-16$ = $49x^{2}-16.$

4)(–7x + 4)(7x – 4)= $49x^{2}+28x+28x-16=49x^{2}+56x-16.$

The third option that is (–7x + 4)(–7x – 4) is difference of two squares.

5 0

3 years ago

Read 2 more answers

(a) Find a vector parallel to the line of intersection of the planes −4x+2y−z=1 and 3x−2y+2z=1.

valentinak56 [21]

Find the intersection of the two planes. Do this by solving for z in terms of x and y ; then solve for y in terms of x ; then again for z but only in terms of x.

-4x + 2y - z = 1 ==> z = -4x + 2y - 1

3x - 2y + 2z = 1 ==> z = (1 - 3x + 2y)/2

==> -4x + 2y - 1 = (1 - 3x + 2y)/2

==> -8x + 4y - 2 = 1 - 3x + 2y

==> -5x + 2y = 3

==> y = (3 + 5x)/2

==> z = -4x + 2 (3 + 5x)/2 - 1 = x + 2

So if we take x = t, the line of intersection is parameterized by

r(t) = ⟨t, (3 + 5t )/2, 2 + t⟩

Just to not have to work with fractions, scale this by a factor of 2, so that

r(t) = ⟨2t, 3 + 5t, 4 + 2t⟩

(a) The tangent vector to r(t) is parallel to this line, so you can use

v = dr/dt = d/dt ⟨2t, 3 + 5t, 4 + 2t⟩ = ⟨2, 5, 2⟩

or any scalar multiple of this.

(b) (-1, -1, 1) indeed lies in both planes. Plug in x = -1, y = 1, and z = 1 to both plane equations to see this for yourself. We already found the parameterization for the intersection,

r(t) = ⟨2t, 3 + 5t, 4 + 2t⟩