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bekas [8.4K]
3 years ago
15

A student takes an 18-question, multiple-choice exam with five choices for each question and guesses on each question. Assume th

e variable is binomial. Part: 0 / 20 of 2 Parts Complete Part 1 of 2 Find the probability of guessing at least 9 out of 18 correctly. Round the answer to at least four decimal places.
Mathematics
1 answer:
Kobotan [32]3 years ago
7 0

Answer:

Probability of getting at least 9 out of 18 correctly is 0.0043

Step-by-step explanation:

We have

n = 18

p = 1/5 for correct answer

q = 4/5 for wrong answer

P( 9 \leq X \leq 18) = 1 - P(0 \leq X\leq 8)

                       = 1 -  0.9957

                         =0.0043

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A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle.
mezya [45]
1. Divide wire b in parts x and b-x. 

2. Bend the b-x piece to form a triangle with side (b-x)/3

There are many ways to find the area of the equilateral triangle. One is by the formula A= \frac{1}{2}sin60^{o}side*side=   \frac{1}{2} \frac{ \sqrt{3} }{2}  (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}
A=\frac{ \sqrt{3} }{36}(b-x)^{2}=\frac{ \sqrt{3} }{36}( b^{2}-2bx+ x^{2}  )=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}

Another way is apply the formula A=1/2*base*altitude,
where the altitude can be found by applying the pythagorean theorem on the triangle with hypothenuse (b-x)/3 and side (b-x)/6

3. Let x be the circumference of the circle.

 2 \pi r=x

so r= \frac{x}{2 \pi }

Area of circle = \pi  r^{2}= \pi  ( \frac{x}{2 \pi } )^{2} = \frac{ \pi }{ 4 \pi ^{2}  }* x^{2} = \frac{1}{4 \pi } x^{2}

4. Let f(x)=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)=\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0

\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0

(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b

x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }

6. So one part is \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) } and the other part is b-\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }

4 0
3 years ago
Read 2 more answers
Joey is staking out a spot for a rectangular pen he is building for his new pig. He has three stakes already plotted on the coor
andreev551 [17]

Answer:2-6

Step-by-step explanation:

36

6 0
3 years ago
Read 2 more answers
Help. IM SO LOST ON THIS! ;(
castortr0y [4]

Answer:

1. 4\frac{2}{3}

Step-by-step explanation:

For question 1, you need to know the ratio between the recipe and the amount in question.

So if in the recipe, it says it uses 3/4 cup of diced ham, and the question uses 1 cup of diced ham, you can divide 1 by 3/4 to get 4/3. This is how many times bigger the amount used in the question than the recipe.

Then it asks for how many cups of potatoes, to do this, you look at the recipe and how many potatoes it uses: 3.5 cups

To solve it then, you just do 3.5 x 4/3 to get 4\frac{2}{3} cups of potatoes

There's your answer.

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3 years ago
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Lelu [443]

Answer:

27 \:  {km}^{2}

Step-by-step explanation:

Given is the shape of a trapezoid.

Therefore,

Area of the trapezoid

=  \frac{1}{2} (2 + 7) \times 6 \\  \\  = 9 \times 3 \\  \\  = 27 \:  {km}^{2}

5 0
3 years ago
the sphere at the right fits snugly inside a cude with 14in edges. what is the approximate volume of the space between the spher
ella [17]

Answer:Find the vol of the sphere:

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V = %284%2F3%29%2Api%2A3%5E3

V = 97.858

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Find the vol between the sphere and cube:

216 - 97.858 = 118.142 cu/in

Step-by-step explanation:

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2 years ago
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