Ask question

Ask question

All categories

3 years ago

7

T-shirts cost $11 each. Raphael graphs the relationships that give the cost y in dollars of buying x T-shirts. Which ordered pai

r is a point on the graph of the relationship?
A(0, 11)
B(2, 22)
C(3, 14)
D(5, 16)

1 answer:

nikitadnepr [17]3 years ago

7 0

The answer is B(2,22).
Since 1 shirt costs 11 dollars, 2 shirts will cost 22 dollars. Hope this helps!

You might be interested in

A regression by a sample of 10 observations gives: Score=49.5+1.96 Study Hours. The sample mean of study hour is 10.5 and sum of

amm1812

Answer:

0.1463

Step-by-step explanation:

Number of observations = 10

Sample mean = 10.5

Sum of standard deviation = 264.5

X = 14

We are to calculate The leverage statistic for study hour 14 using the data above

= 1/10 + (14-10.5)²/264.5

= 1/10 + 3.5²/264.5

= 0.1 +12.25/264.5

= 0.1+0.04631

= 0.1463 is the leverage statistic for the study

4 0

3 years ago

Binomial Expansion/Pascal's triangle. Please help with all of number 5.

Mandarinka [93]

$\begin{matrix}1\\1&1\\1&2&1\\1&3&3&1\\1&4&6&4&1\end{bmatrix}$

The rows add up to $1,2,4,8,16$ , respectively. (Notice they're all powers of 2)

The sum of the numbers in row $n$ is $2^{n-1}$ .

The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When $n=1$ ,

$(1+x)^1=1+x=\dbinom10+\dbinom11x$

so the base case holds. Assume the claim holds for $n=k$ , so that

$(1+x)^k=\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k$

Use this to show that it holds for $n=k+1$ .

$(1+x)^{k+1}=(1+x)(1+x)^k$
$(1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)$
$(1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}$

Notice that

$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dbinom{k+1}{\ell+1}$

So you can write the expansion for $n=k+1$ as

$(1+x)^{k+1}=1+\dbinom{k+1}1x+\dbinom{k+1}2x^2+\cdots+\dbinom{k+1}{k-1}x^{k-1}+\dbinom{k+1}kx^k+x^{k+1}$

and since $\dbinom{k+1}0=\dbinom{k+1}{k+1}=1$ , you have

$(1+x)^{k+1}=\dbinom{k+1}0+\dbinom{k+1}1x+\cdots+\dbinom{k+1}kx^k+\dbinom{k+1}{k+1}x^{k+1}$

and so the claim holds for $n=k+1$ , thus proving the claim overall that

$(1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n$

Setting $x=1$ gives

$(1+1)^n=\dbinom n0+\dbinom n1+\cdots+\dbinom n{n-1}+\dbinom nn=2^n$

which agrees with the result obtained for part (c).

4 0

2 years ago

Which graph represents the function f(x) equals x(x +2)

joja [24]

Answer:

Step-by-step explanation:

Okay!

A graph representing the function :

f(x)=x(x+2)

I will describe the graph for you!

the points on this graph are : (-3,3),(-2,0),(-1,-1),(0,0),(1,3)

the graph also creates a U shape!

7 0

2 years ago

Help me pleaseeee :)

PilotLPTM [1.2K]

Answer:

first from the top

Step-by-step explanation:

8 0

3 years ago

What is a number in exponential notation that decreases in value as the exponent increases in value

Eva8 [605]

Answer:

Any fraction

Step-by-step explanation:

Example:

(2/3)² = 4/9

2/3 > 4/9

7 0

2 years ago

Read 2 more answers