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2 years ago

Madeline usually makes 85% of her shots in basketball. if she attempts 20, how many will she likely make?

Mathematics

2 answers:

Bad White [126]2 years ago

8 0

Answer:

Step-by-step explanation:

She will make 17 shots because:

$\frac{85 x 20}{100}$ = 17.

So, the answer is 17. Hope this helps!

kherson [118]2 years ago

6 0

Answer:

Step-by-step explanation:

If you divide 20 by 100 then multiply by 85 you get the number shes likely to make

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The number of hurricanes hitting the coast of Florida annually has a Poisson distribution with a mean of 0.8. Answer the followi

slava [35]

Answer:

a) $P(X>2)= 1-P(X \leq 2) = 1-[P(X=0)+P(X=1)+P(X=2)]$

And we can find the individual probabilities like this:

$P(X=0) = \frac{e^{-0.8} 0.8^0}{0!}= 0.4493$

$P(X=1) = \frac{e^{-0.8} 0.8^1}{1!}= 0.3595$

$P(X=2) = \frac{e^{-0.8} 0.8^2}{2!}= 0.1438$

And replacing we got:

$P(X>2)= 1-P(X \leq 2) = 1-[0.4493+0.3595+0.1438]=0.0474$

b) $P(X=1) = \frac{e^{-0.8} 0.8^1}{1!}= 0.3595$

Step-by-step explanation:

Let X the random variable that represent the number of hurricanes hitting the coast of Florida annualle. We know that $X \sim Poisson(\lambda=0.8)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda=0.8$

$E(X)=\mu =\lambda=0.8$

Part a

For this case we want this probability: $P(X>2)$

And for this case we can use the complement rule like this:

$P(X>2)= 1-P(X \leq 2) = 1-[P(X=0)+P(X=1)+P(X=2)]$

And we can find the individual probabilities like this:

$P(X=0) = \frac{e^{-0.8} 0.8^0}{0!}= 0.4493$

$P(X=1) = \frac{e^{-0.8} 0.8^1}{1!}= 0.3595$

$P(X=2) = \frac{e^{-0.8} 0.8^2}{2!}= 0.1438$

And replacing we got:

$P(X>2)= 1-P(X \leq 2) = 1-[0.4493+0.3595+0.1438]=0.0474$

Part b

Using the probability mass function we have:

$P(X=1) = \frac{e^{-0.8} 0.8^1}{1!}= 0.3595$

3 0

2 years ago

Tiffany is monitoring the decay of two radioactive compounds in test tubes at her lab. Compound A is continuously decaying at a

trapecia [35]

Answer:

30e-0.12t

40e-0.18t

Step-by-step explanation:

6 0

2 years ago

A point in the figure is selected at random. Find the probability that the point will be in the part that is NOT shaded.

Dovator [93]

Answer:

Step-by-step explanation:

So the question really is what is the probability that the point is in the square? You cannot pick something surrounding the circles because you do not know anything about that region.

This is solved by comparing areas. There are 2 full circles there. 1 circle has an area of pi*r^2

Total shaded area = 2 * pi * r^2.

Let r = 3 This is a completely random choice. No matter what number you choose for r, the answer will come out the same. When you get along a little further in math, you will find that you can just use letters.

Total shaded area = 2 * 3.14 * 3^2

Total shaded area = 2 * 3.14 * 9

Total shaded area = 56.55

Total area of the square.

s = 2*r

s = 2*3

s = 6

Area of the square = 6^2 = 36

Total area of both regions = 56.55 + 36 = 92.55

Answer

% = (area of square) * 100% / Total area of both regions

% = 3600 / 92

% = 39%

Obviously the answer I get is not offered. The closest answer is B so I will choose that.

==========================

This is how this question would be done without using any value for r.

Area of shaded region = 2 pi r^2 = 6.28 r^2

Area of square = d^2 where d = 2r

Area of square = 4 r^2

Total area = 6.28 r^2 + 4r^2 = 10.28 r^2

Area of square to total = (4r^2/10.28 r^2 ) * 100%

Area of square to total (as a %) = 4/10.28 * 100 = 39% which gives the same answer.

7 0

2 years ago

Read 2 more answers

Which equation is y = 3(x – 2)2 – (x – 5)2 rewritten in vertex form? Y = 3 (x minus seven-halves) squared minus StartFraction 27

densk [106]

Answer: y = 2 (x minus one-half) squared minus StartFraction 27 Over 2 EndFraction

$y=2((x-\dfrac{1}{2})^2)-\dfrac{27}{2}$

Step-by-step explanation:

Vertex form of equation : $f (x) = a(x - h)^2 + k,$ where (h, k) is the vertex of the parabola.

$y=3(x-2)^2-(x-5)^2\\\\=3(x^2+4-4x)-(x^2+25-10x)\\\\=3x^2+12-12x-x^2-25+10x\\\\=2x^2-2x-13\\\\=2(x^2-x-\dfrac{13}{2})\\\\=2(x^2-x+\dfrac{1}{4}-\dfrac{1}{4}-\dfrac{13}{2})\\\\=2((x-\dfrac{1}{2})^2-\dfrac{1+26}{4})\\\\=2((x-\dfrac{1}{2})^2-\dfrac{27}{4})=2((x-\dfrac{1}{2})^2)-\dfrac{27}{2}$