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Leokris [45]
2 years ago
9

Please help! Giving brainliest

Mathematics
1 answer:
lianna [129]2 years ago
6 0

Answer:

5y^2-12y+21-73/2y+3

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Let f(x) = cxe−x2 if x ≥ 0 and f(x) = 0 if x < 0.
sergij07 [2.7K]

(a) If <em>f(x)</em> is to be a proper density function, then its integral over the given support must evaulate to 1:

\displaystyle\int_{-\infty}^\infty f(x)\,\mathrm dx = \int_0^\infty cxe^{-x^2}\,\mathrm dx=1

For the integral, substitute <em>u</em> = <em>x</em> ² and d<em>u</em> = 2<em>x</em> d<em>x</em>. Then as <em>x</em> → 0, <em>u</em> → 0; as <em>x</em> → ∞, <em>u</em> → ∞:

\displaystyle\frac12\int_0^\infty ce^{-u}\,\mathrm du=\frac c2\left(\lim_{u\to\infty}(-e^{-u})-(-1)\right)=1

which reduces to

<em>c</em> / 2 (0 + 1) = 1   →   <em>c</em> = 2

(b) Find the probability P(1 < <em>X </em>< 3) by integrating the density function over [1, 3] (I'll omit the steps because it's the same process as in (a)):

\displaystyle\int_1^3 2xe^{-x^2}\,\mathrm dx = \boxed{\frac{e^8-1}{e^9}} \approx 0.3678

5 0
2 years ago
8 3/4 miles in 5/8 write as a rate?
dlinn [17]
Yes I Belive so yeah
8 0
3 years ago
HELP ME FAST PLEASE
antoniya [11.8K]
B. 28
42 \div 1.5 = 28
7 0
3 years ago
Solve the following system of equations and show all work. Please help!!! y = 2x2 y = –3x −1
Mice21 [21]
We can solve these equations by using substitution method where you substitute one equation to the other and solve for the value of the other variable. We do as follows:

y = 2x^2
y = –3x −1

2x^2 = –3x −1
2x^2 +3x +1 = 0

x1= -0.5
x2 = -1
y1 = 0.5
y2 = 2
6 0
2 years ago
Read 2 more answers
According to a​ study, 68​% of all males between the ages of 18 and 24 live at home. ​ (Unmarried college students living in a d
Marat540 [252]

Answer:

a) P(X≥143​)=0

b) This contradicts the study as getting a sample with this proportion is almost impossible (if the proportion of 68% is true).

Step-by-step explanation:

If we use the normal approximation to the binomial distribution we have the following parameters (mean and standard deviation):

\mu=np=0.68*206=140.1\\\\ \sigma=\sqrt{np(1-p)}=\sqrt{206*0.68*0.32}=\sqrt{44.8256}=6.7

Then, we can calculate the probability of X being equal or more than 143 using the z-score:

z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{143-140.1}{6.7/\sqrt{206}}=\dfrac{2.9}{0.4668}=6.2124\\\\\\P(x\geq143)=P(z>6.2124)=0

This contradicts the study as getting a sample with this proportion is almost impossible (if the proportion of 68% is true).

6 0
3 years ago
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