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3 years ago

WILL GIVE BRAINLIEST AND 20 POINTS!

1 answer:

defon3 years ago

7 0

Radius of design= 18 in
Area of design= pi*r^2
= 3.14*324
= 1017.36 sq. in

1017.36/144
= 7.065 sq. ft

7.065*35
= 247.275

Therefore, the cost of building the design costs $247.275 to make

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Need help with 8-9.

ANTONII [103]

So it is 5cm to 1m or 5cm to 1000cm
So the scale is the ratio which is 5:1000. this can be reduced to 1:200

10. if the room is 15cm on one side I can say that the ratio to actual size is 15/x. and scale is 1/200 so:
15/x = 1/200
3000 = x is one side

other side:
20/x = 1/200
4000 = x

7 0

3 years ago

Restaurants often slip takeout menus under Meg's apartment door. So far, Meg has collected 18 menus, including 8 for Chinese foo

Kryger [21]

Answer:

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

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

3 years ago

Examine the system of equations.

MrMuchimi

answer:

$x = - 2 \\ y = 2 \\ \\ ( - 2, \: 2)$

explanation:

$y = - \frac{1}{2} x + 1 \\ y = 2x + 6 \\ first \: equation \: add \: \frac{1}{2} from \: both \\ sides \\ \\ second \: equation \: subtract \: 2x \: from \\ both \: sides \\ \\ y + \frac{1}{2} x = 1 \\ y - 2x = 6 \\ solve \: the \: first \: equation \\ \\ y + \frac{1}{2} x = 1 \\ subtract \: \frac{x}{2} from \: both \: sides \\ \\ y = - \frac{1}{2} x + 1 \\ substitute \: - \frac{x}{2} + 1 \: for \: y \: in \: the \\ other \: equation \\ \\ - \frac{1}{2} x + 1 - 2x = 6 \\ add \: \frac{1}{2} x \: to \: 2x \\ \\ - \frac{5}{2} x + 1 = 6 \\ subtract \: 1 \: from \: both \: sides \\ \\ - \frac{5}{2} x = 5 \\ divide \: both \: sides \: by \: - \frac{5}{2} \\ \\ x = - 2 \\ substitute \: the \: value \: of \: x \: into \\ an \: equation \\ \\ y = - \frac{1}{2} ( - 2) + 1 \\ distribute \\ \\ y = 1 + 1 \\ add \\ \\ y = 2 \\ \\ ( - 2, \: 2)$

7 0

3 years ago

A sales manager wanted to determine if increasing sales commissions by 5% would increase employee satisfaction. Her analyst dete

Dafna11 [192]

Answer:

c. We have evidence that there is a difference in population median employee satisfaction rating before and after the commission change.

Step-by-step explanation:

Pvalue is less than the alpha-level:

If the pvalue is less than the alpha-level, we have that there is sufficient evidence of the hypothesis for the population mean/median/standard deviation we are testing.

In this question:

Pvalue less than the alpha-level, so there is evidence that there is a difference in the population median. This means that the correct answer is given by option C.

4 0

3 years ago