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FromTheMoon [43]
3 years ago
12

3/4 of Ana's books are equal to two-thirds of Jake's books find the ratio of the number of Ana's books it to the number of Jake'

s books draw models to help you. Please show your work!!!!!!!
Mathematics
2 answers:
dimaraw [331]3 years ago
8 0
First you want to change the fractions to the same denominator 
3/4 = 9/12 and 2/3 = 8/12
then you put it in ratio form...
the answer is 9:8
Paraphin [41]3 years ago
4 0
The answer would be 9:8
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Simplify:0.3(x-2y)+0.5x-y
astraxan [27]

Answer:

0.3[x-2y] + 0.5x - y


Use the distributive property: a(b-c) = ab - ac


0.3x - 0.6y + 0.5x - y


Now, add/subtract all like terms


0.3x + 0.5x - 0.6y - y


0.8x - 1.6y


If you want, we can simplify that down further to:


0.8(x - 2y)



7 0
3 years ago
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What is the area of the triangle? Use the formula: Area of a triangle = 1/2bh.How do you find the area of a triangle?
velikii [3]

9514 1404 393

Answer:

  4 square inches

Step-by-step explanation:

The only triangle in the figure is 4 in wide and 2 in high. Using these values in the given formula, we find the area to be ...

  A = 1/2bh

  A = (1/2)(4 in)(2 in) = 4 in²

The area of the triangle is 4 square inches.

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2 years ago
Is this correct????????
solniwko [45]
Only use this one pls. Lol

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2 years ago
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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
Use a calculator to find the correlation coefficient of the data set.
ratelena [41]

Answer:

Step-by-step explanation:

Enter the x,y values in the box above. You may enter data in one of the following two formats:

   Each xi,yi couple on separate lines:

   x1,y1

   x2,y2

   x3,y3

   x4,y4

   x5,y5

7 0
3 years ago
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