How many triangles exist with the given angle measures? 60 60 60

1 answer:

6 0

The sum of the angles in a triangle is 180°, so the angles do form a triangle. Since two of the three angles are congruent, the triangle is an isosceles triangle. An infinite number of isosceles triangles exist with different side lengths. Therefore, more than one triangle exists with the given angle measures.

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Urn 1 contains 3 blue tokens and 2 red tokens; urn 2 contains 2 blue tokens and 4 red tokens. All tokens are indistinguishable.

Dmitry_Shevchenko [17]

Answer:

RR = 0.4

RB = 0.3

BB = 0.22

BR = 0.30

Step-by-step explanation:

P( Urn 1 ) = 2/6 = 1/3

P( Urn 2 ) = 1 - 1/3 = 2/3

Urn 1 contains : 3 blue and 2 red

P( blue | urn 1 ) = 3/5 ( with replacement ) , P( blue | urn 1 ) = 3/4 ( without replacement )

P( red | urn 1 ) = 2 / 5 ( with replacement ) , P(red | urn 1 ) = 1/2 ( without replacement )

Urn 2 contains : 2 blue and 4 red

P ( blue | urn 2 ) = 1/3 ( with replacement ) , P( blue | urn 2 ) = 2/5 ( without replacement )

P ( red | urn 2 ) = 2/3 ( with replacement) , P( red | urn 2 ) = 4/5 ( without replacement )

Determine

i) Possible outcomes when two tokens are drawn from either Urn without replacement

RR = [[ ( 2/5 * 1/3 ) + ( 2/3 * 2/3 ) ] * [( 1/2 * 1/3 ) + ( 4/5 * 2/3 ) ]] = 0.4

RB = [[ (2/5 * 1/3 ) + ( 2/3 * 2/3 ) ] * [ ( 3/4 *1/3 ) + ( 2/5 * 2/3 ) ]] ≈ 0.3

BB = [[ ( 3/5 * 1/3 ) + ( 1/3 * 2/3 ) ] * [ ( 3/4 *1/3 ) + ( 2/5 * 2/3 ) ]] ≈ 0.22

BR = [[ ( 3/5 * 1/3 ) + ( 1/3 * 2/3 ) ] * [ ( 1/2 * 1/3 ) + ( 4/5 * 2/3 ) ]] ≈ 0.30

8 0

3 years ago

A = 1011 + 337 + 337/2 +1011/10 + 337/5 + ... + 1/2021

egoroff_w [7]

The sum of the given series can be found by simplification of the number

of terms in the series.

A is approximately 2020.022

Reasons:

The given sequence is presented as follows;

A = 1011 + 337 + 337/2 + 1011/10 + 337/5 + ... + 1/2021

Therefore;

$\displaystyle A = \mathbf{1011 + \frac{1011}{3} + \frac{1011}{6} + \frac{1011}{10} + \frac{1011}{15} + ...+\frac{1}{2021}}$

The n + 1 th term of the sequence, 1, 3, 6, 10, 15, ..., 2021 is given as follows;

$\displaystyle a_{n+1} = \mathbf{\frac{n^2 + 3 \cdot n + 2}{2}}$

Therefore, for the last term we have;

$\displaystyle 2043231= \frac{n^2 + 3 \cdot n + 2}{2}$

2 × 2043231 = n² + 3·n + 2

Which gives;

n² + 3·n + 2 - 2 × 2043231 = n² + 3·n - 4086460 = 0

Which gives, the number of terms, n = 2020

$\displaystyle \frac{A}{2} = \mathbf{ 1011 \cdot \left(\frac{1}{2} +\frac{1}{6} + \frac{1}{12}+...+\frac{1}{4086460} \right)}$

$\displaystyle \frac{A}{2} = 1011 \cdot \left(1 - \frac{1}{2} +\frac{1}{2} - \frac{1}{3} + \frac{1}{3}- \frac{1}{4} +...+\frac{1}{2021}-\frac{1}{2022} \right)$