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Westkost [7]
3 years ago
8

(Teaser) The total number of sides in two regular polygons is 13, and total number of diagonals is 25. How many sides are in eac

h polygon?
Mathematics
2 answers:
Troyanec [42]3 years ago
6 0

The other student is correct I believe.

NISA [10]3 years ago
5 0

Answer:  

T he two polygons have 8 sides and 5 sides.  

Step-by-step explanation:

A purely algebraic solution seems as if it would be very messy. Use trial and error, along with the formula for the number of diagonals in a n-sided polygon.

number of diagonals in a polygon with n sides: n(n-3)/2

10 sides and 3 sides: 10(7)/2 + 3(0)/2 = 35+0 = 35 nope....

9 sides and 4 sides: 9(6)/2 + 4(1)/2 = 27+2 = 29 nope....

8 sides and 5 sides: 8(5)/2 + 5(2)/2 = 20+5 = 25 yep!

 

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Two chemists working for a chicken fast-food company, have been producing a very popular sauce. Let’s call then Jesse and Mr. Wh
a_sh-v [17]

Answer:

Check the explanation

Step-by-step explanation:

Let \overline{x} and \overline{y} be sample means of white and Jesse denotes are two random variables.

Given that both samples are having normally distributed.

Assume \overline{x} having with mean \mu_{1} and \overline{y} having mean \mu_{2}

Also we have given the variance is constant

A)

We can test hypothesis as

H0: \mu_{1} =  \mu_{2}H1: \mu_{1} > \mu_{2}

For this problem

Test statistic is

T=\frac{\overline {x}-\overline {y}}{s\sqrt{\frac {1}{n1} +\frac{1}{n2}}}

Where

s=\sqrt{\frac{(n1-1)*s1^{2}+(n2-1)*s2^{2}}{n1+n2-2}}

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By calculations we get

s=2.41

T=2.52

Here test statistic is having t-distribution with df=(10+7-2)=15

So p-value is P(t15>2.52)=0.012

Here significance level is 0.05

Since p-value is <0.05 we are rejecting null hypothesis at 95% confidence.

We can conclude that White has significant higher mean than Jesse. This claim we can made at 95% confidence.

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the assumption being that "x" is a plain variable whilst "y" is a function, that matters because the chain rule would be needed for a function, not so for a plain variable.

4x^2+4x+xy=5\implies 8x+4+\stackrel{\textit{product rule}}{\left( 1\cdot y+x\cdot \cfrac{dy}{dx} \right)}=0 \\\\\\ x\cfrac{dy}{dx}=-8x-4-y\implies \cfrac{dy}{dx}=\cfrac{-8x-4-y}{x}

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