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Dimas [21]
3 years ago
15

Suppose I have an urn full of balls.10 balls are balck, 2 are blue,2 are red,1 is green. I reach into the urn and pull out a bal

l and record the colour. I then put it back in the urn. If i do this 10 times calculate
a) the probability I obtain exact 4 black balls
b) The probablility I obtain less than 2 red balls

For clarity we are putting the balls back into the urn after the colour is recorded.
Mathematics
1 answer:
Gelneren [198K]3 years ago
6 0

Answer:

Step-by-step explanation:

Assuming that each draw is independent of the last and that each ball is equally likely to be selected makes this a binomial distribution

a.) first find the probability of selecting a black ball, 2/10 or .2

{10\choose{4}}.2^4*.8^6=.08808384

b.) Probability of slecting a red ball: 1/10 or .1

Less than 2 is X=0 + X=1

{10\choose0}*.2^0*.8^{10}+{10\choose1}*.2*.8^9=.3758096384

i will leave you to do the rounding

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A triangle has three sides 35cm 54 cm and 61 cm find its area Also find the smallest of its altitude ​
I am Lyosha [343]

Answer:

Area of given triangle is 939.15cm² and smallest altitude is 30.8cm

<h3>Solution:</h3>

We are given three sides of a triangle, Let the sides be :

  • ( a ) = 35 cm

  • ( b ) = 54 cm

  • ( c ) = 61 cm

We can find the area of the triangle with its three sides using Heron's Formula

  • <u>Heron's </u><u>Formula</u>

Heron's formula was founded by hero of Alexandria, for finding the area of triangle in terms of the length of its sides. Heron's formula can be written as:

\sf{   \pmb { \longrightarrow \:  \sqrt{s(s - a)(s - b)(s - c)} }}

where ( s ) :

\sf \longrightarrow s = \dfrac{a + b + c}{2}

Therefore, for the given triangle first we will calculate ( s )

\begin {aligned}\quad & \quad \longmapsto  \sf  s =  \dfrac{a + b + c}{2}  \\  & \quad \longmapsto  \sf s =  \dfrac{35 + 54 + 61}{2}  \\ & \quad \longmapsto  \sf s =  \dfrac{150}{2}  \\ & \quad \longmapsto  \sf s  = 75cm \end{aligned}

Now, Area of triangle will be:

\begin{aligned}&:\implies \sf\quad \sf \:  A = \sqrt{s(s - a)(s - b)(s - c)} \\ &:\implies \sf\quad \sf \:  A = \sqrt{75(75 - 35)(75 - 54)(75 - 61)}   \\&:\implies \sf\quad \sf \:  A = \sqrt{75 \times 40 \times 21 \times 14}  \\ &:\implies \sf\quad \sf \:  A = \sqrt{5 \times 5 \times 3 \times 3 \times 2 \times 2 \times 7 \times 7 \times 2 \times 2 \times 5}  \\ &:\implies \sf\quad \sf \:  A =5 \times 3 \times 2 \times 7 \times 2 \sqrt{5}  \\ &:\implies \sf\quad \sf \:  A =420 \times 2.23 \\ &:\implies \sf\quad \sf \boxed{ \pmb{ \sf   A =939.15 {cm}^{2} }} \end{aligned}

Also, we have to find the smallest altitude, and the smallest altitude will be on the longest side. So,

\begin{aligned}&:\implies \sf\quad \sf \:  Area =939.15 \\ &:\implies \sf\quad \sf \:   \dfrac{1}{2}  \times b \times h =939.15 \\ &:\implies \sf\quad \sf \:   \dfrac{1}{2} \times 61 \times h  = 939.15 \\&:\implies \sf\quad \sf \:  h =939.15 \times  \dfrac{2}{61}   \\&:\implies \sf\quad \sf \:  h = \dfrac{1818.3}{61}  \\ &:\implies \sf\quad  \boxed{ \pmb{\sf \:  h =30.79 \: (approx)}} \end{aligned}

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143 + 211 + 113 = 467

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if alyssa plans to run 7/8 mile on a trail at one point she sees that she is at mile maker 0.3 How much farther does she have to
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Rewrite 7/8 as a decimal by dividing 7 by 8:

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1-8 is simple. Just subtract 8 from 1. so 1-8 = -7


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