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4 years ago

Help! Check answer? My answer x=80

Mathematics

2 answers:

GuDViN [60]4 years ago

8 0

The is the correct answer x=80

liubo4ka [24]4 years ago

7 0

Because their angles are the same,
you can find a scale by doing 25/5 which is 5 then use that like this:

7*5 =35

to find the last side.

Then sum them: 20+25+35=80

80 is correct :)

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Find the cdf F(x) associated with each of the following probability density functions. Sketch the graphs of f(x) and F(x).

wolverine [178]

Answer:

See steps below

Step-by-step explanation:

$\bf f(x)=3(1-x)^2\;(0$

$\bf F(x)=\int_{0}^{x}f(t)dt=\int_{0}^{x}3(1-t)^2dt=3\int_{0}^{x}(1-t)^2=1-(1-x)^3$

The cdf associated with f is

$\bf \boxed{F(x)=1-(1-x)^3}$ for 0<x<1

<h3>See picture 1 </h3>

The median is a point x such that

F(x) = ½

so, the median is

$\bf 1-(1-x)^3=1/2\rightarrow (1-x)^3=1/2\rightarrow \boxed{x=1-\sqrt[3]{2}}$

The 25th percentile equals the 1st quartile and is a point x such

F(x) = ¼

and the 25th percentile is

$\bf 1-(1-x)^3=1/4\rightarrow (1-x)^3=3/4\rightarrow \boxed{x=1-\sqrt[3]{3/4}}$

$\bf f(x)=\frac{1}{x^2}\;(1$

$\bf F(x)=\int_{1}^{x}f(t)dt=\int_{1}^{x}\frac{dt}{t^2}=1-\frac{1}{x}$

The cdf associated with f is

$\bf \boxed{F(x)=1-\frac{1}{x}}$ for x>1

<h3>See picture 2 </h3>

The median is

$\bf 1-\frac{1}{x}=\frac{1}{2}\rightarrow \boxed{x=2}$

The 25th percentile is

$\bf 1-\frac{1}{x}=\frac{1}{4}\rightarrow \boxed{x=4/3}$

f(x) = 1/3 for 0<x<1 or 2<x<4

$\bf F(x)=\int_{0}^{x}\frac{dt}{3}=\frac{x}{3}\;(0$

$\bf F(x)=\frac{1}{3}+\int_{2}^{x}\frac{dt}{3}=\frac{1}{3}+\frac{x-2}{3}=\frac{x-1}{3}\;(2$

The cdf associated with f is

$\bf F(x)=\frac{x}{3}$ for 0<x<1

$\bf F(x)=\frac{x-1}{3}$ for 2<x<4

<h3>See picture 3 </h3>

The median is

$\bf \frac{x-1}{3}=1/2\rightarrow x=1+3/2\rightarrow \boxed{x=5/2}$

The 25th percentile is

$\bf \frac{x}{3}=1/4\rightarrow \boxed{x=3/4}$

4 0

4 years ago

Analyzing a Linear Function Expressed in Different Ways Choose the true statements. A linear function is expressed by a graph of

Irina18 [472]

Answer: The answer is option B and option C.

9 0

4 years ago

A certain radioactive material decays in such a way that the mass in kilograms remaining after t years is given by the function

Angelina_Jolie [31]

The mass of radioactive material remaining after 50 years would be 48.79 kilograms

<h3>How to determine the amount</h3>

It is important to note that half - life is the time it takes for the amount of a substance to reduce by half its original size.

Given the radioactive decay formula as

m(t)=120e−0.018t

Where

t= 50 years

m(t) is the remaining amount

Substitute the value of t

$m(t) = 120e^-^0^.^0^1^8^(^5^0^)$

$m(t) = 120e^-^0^.^9$

Find the exponential value

m(t) = 48.788399

m(t) = 48.79 kilograms to 2 decimal places

Thus, the mass of radioactive material remaining after 50 years would be 48.79 kilograms

Learn more about half-life here:

brainly.com/question/26148784

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7 0

2 years ago

Please help! <3 worth 100 points

shepuryov [24]

Answer:

I cannot see the question! Mind typing it out for me?

Step-by-step explanation:

I could really solve it if you typed it

8 0

3 years ago

Read 2 more answers

49 more than the quotient of an unknown number and 12 is 53. What is the value of the unknown number?

nalin [4]

The unknown number is 48 using the equation: x ÷ 12 +49=53 subtract 49 from both sides and get x ÷ 12 = 4 multiplying by 12 on both side to get rid of the ÷12 and get the final answer of x=48.

6 0

3 years ago