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2 years ago

Please help! pleaseeeeeeeeeeeeeeeee

Mathematics

1 answer:

Wittaler [7]2 years ago

5 0

From the graph, it looks like the vertex of the parabola is (-4, -9).

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Here are two offers for batteries.

trapecia [35]

OFFER A is cheaper.

Step-by-step explanation:

Given,

OFFER A

Pack of 5

£2.75

Pay for 3 packs get 1 free

And

OFFER A

Pack of 5

£2.75

Pay for 3 packs get 1 free

To find which offer is cheaper.

For OFFER A

For 4 Packets he needs to pay £2.75

He needs to buy 40 batteries.

He needs to buy 40÷(4×5) = 2 packets.

Total cost = £2.75 ×2 = £5.5

For OFFER B

For 4 Packets he needs to pay £2.52

He needs to buy 40 batteries.

He needs to buy 10 packets.

After discount he has to pay = £2.52 -(£2.52 × $\frac{1}{3}$ ) = £1.68

Total cost = £1.68×10 = £16.8

Hence,

OFFER A is cheaper.

7 0

2 years ago

9 - 10x = 2x + 1 - 8x

Furkat [3]

Answer:

$x=2$

Step-by-step explanation:

$9 - 10x = 2x + 1 - 8x$

$9-10x=-6x+1$

Subtract 9 from both sides:

$9-10x-9=-6x+1-9$

$-10x=-6x-8$

Add 6x to both sides:

$-10x+6x=-6x-8+6x$

$-4x=-8$

Divide both sides by -4:

$\frac{-4x}{-4}=\frac{-8}{-4}$

$x=2$

8 0

3 years ago

Read 2 more answers

I. Write the equation of a parabola, in standard form, that goes through these points: (0, 3) (1, 4) (-1, -6) II. Graph the para

Anna11 [10]

Vertex = (1,4)
Axis of symmetry is x = 1, (1,0) or just 1

7 0

3 years ago

Date : n _ 5+ [ 4 +(-7) -{ 6 ×(5+ 1x4)}]

s2008m [1.1K]

Answer:

−28−6^2x4

Step-by-step explanation:

6 0

2 years ago

A source of information randomly generates symbols from a four letter alphabet {w, x, y, z }. The probability of each symbol is

koban [17]

The expected length of code for one encoded symbol is

$\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha\ell_\alpha$

where $p_\alpha$ is the probability of picking the letter $\alpha$ , and $\ell_\alpha$ is the length of code needed to encode $\alpha$ . $p_\alpha$ is given to us, and we have

$\begin{cases}\ell_w=1\\\ell_x=2\\\ell_y=\ell_z=3\end{cases}$

so that we expect a contribution of

$\dfrac12+\dfrac24+\dfrac{2\cdot3}8=\dfrac{11}8=1.375$

bits to the code per encoded letter. For a string of length $n$ , we would then expect $E[L]=1.375n$ .

By definition of variance, we have

$\mathrm{Var}[L]=E\left[(L-E[L])^2\right]=E[L^2]-E[L]^2$

For a string consisting of one letter, we have

$\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha{\ell_\alpha}^2=\dfrac12+\dfrac{2^2}4+\dfrac{2\cdot3^2}8=\dfrac{15}4$

so that the variance for the length such a string is

$\dfrac{15}4-\left(\dfrac{11}8\right)^2=\dfrac{119}{64}\approx1.859$

"squared" bits per encoded letter. For a string of length $n$ , we would get $\mathrm{Var}[L]=1.859n$ .

5 0

2 years ago