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2 years ago

Pls answer thank you

Mathematics

1 answer:

Rashid [163]2 years ago

4 0

Answer:

<h2>ANSWER </h2>

 

 $as \: the \: diagram \: given \: \\ surface \: area \: = (9 \times 17 \times 2 + 4 \times 17 + 4 \times 9 + 4 \times 17 + 9 \times 4) {in}^{2}$ 

 $(153 \times 2+ 68 + 36 + 68 + 36 \\ 306 + 104 + 104 ) {in}^{2}$ 

 $544 \:{ in}^{2}$ 

So the answer here = 544 in^2

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

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

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

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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

g In the sample of 134 South Korean tourists, 57 of them were satisfied with their Jeju experience. In the sample of 150 other c

nasty-shy [4]

Answer:

The test statistic = -0.93

Step-by-step explanation:

The test statistic is given by the formula

z = (X₁ - X₂) ÷ √(σₓ₁² + σₓ₂²)

where X₁ = proportion of data of South Korean tourists = (57/134) = 0.425

X₂ = proportion of other country tourists = (72/150) = 0.48

σₓ₁ = standard error in data 1 = √[p(1-p)/n]

= √(0.425 × 0.575/134) = 0.0427

σₓ₂ = standard error in data 2 = √[p(1-p)/n]

= √(0.48 × 0.52/150) = 0.0408

z = (X₁ - X₂) ÷ √(σₓ₁² + σₓ₂²)

z = (0.425 - 0.48) ÷ √(0.0427² + 0.0408²)

z = -0.055 ÷ 0.0590586996

z = -0.9313

Hope this Helps!!!

6 0

3 years ago