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2 years ago

kathy and 3 of her friends spent $41.60 on pizza. if they split the bill evenly, how much will each person owe?

Mathematics

1 answer:

MAXImum [283]2 years ago

8 0

Answer:

$10.40

Step-by-step explanation:

you can find this out by dividing the total amount of money owed by how many people are paying, say for instance, you have 6 people paying $40 in total, divide 40 by 6 to get $6.67 (rounded to the nearest .01)

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The average waist size for teenage males is 29 inches with astandard deviation of 2 inches. If waist sizes are normallydistribut

choli [55]

Answer:

The z-score of a teenage with waist size 33 inch is 2

Step-by-step explanation:

Mathematically, the z-score can be calculated using the formula;

z-score = (x- mean)/SD

where mean = 29 inches , SD = 2 inches and x = 33 inch

Plugging these values, we have

z-score = (33-29)/2 = 4/2 = 2

5 0

2 years ago

∠A∠A and ∠B∠B are vertical angles with m∠A=xm∠A=x and m∠B=5x−80m∠B=5x−80 . What is m∠Am∠A ?

iris [78.8K]

The Vertical Angles Theorem states that if two angles are vertical angles, then they are congruent .Given <A and <B are vertical angles so they are equal.

<A=<B.

Measure of <A=x and <B=5x-80

Or 5x-80=x

Adding 80 both sides

5x-80+80=x+80

5x=x+80

Subtracting x both sides

5x-x=x-x+80

4x=80

Dividing both sides by 4

x=20

Measure of <A= x= 20 degrees.

8 0

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).

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2 years ago

When you draw a circle and two of its diameters and connect the end points of the diameters what quadirateral do you get?

sergij07 [2.7K]

A vin diagram I'm pretty sure. or maybe drugs pflucked my brain up. js

3 0

3 years ago

Barbara gets 6 pizza to divide equally among 4 people. How much of a pizza can each person ?

kozerog [31]

Your key word is divide. You'll want to divide your 6 pizzas by your four friends, than most likely convert it to a fraction (unless your teacher accepts decimals.), message if you need more help.

3 0

2 years ago

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