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belka [17]
3 years ago
5

The heights of four boys measured 152.0cm, 150.75cm, 149.5cm and 149.25cm. Bradley is the tallest, Calvin is taller than Josh, b

ut shorter than Mark. Josh is the shortest. What is Mark's height?
Mathematics
1 answer:
nataly862011 [7]3 years ago
3 0
I believe the answer would be 150.75 cm as Marks height. I got this by if Bradley is the tallest, he would be 152.0 and if Calvin is taller than Josh, who is the shortest, the only height left would be Marks at 150.75
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X+y=3 x-y=1 find the Find the determinants
Mkey [24]

Answer:

  • D = -2
  • Dx = -4
  • Dy = -2

Step-by-step explanation:

<u>System Determinant</u>

The determinant of the system is the determinant of the matrix of coefficients:

  D=\left|\begin{array}{cc}1&1\\1&-1\end{array}\right|=(1)(-1)-(1)(1)=-2

<u>X Determinant</u>

This is the determinant of the matrix with the x-coefficients replaced by the constants.

  D_x=\left|\begin{array}{cc}3&1\\1&-1\end{array}\right|=(3)(-1)-(1)(1)=-4

<u>Y Determinant</u>

This is the determinant of the matrix with the y-coefficients replaced by the constants.

  D_y=\left|\begin{array}{cc}1&3\\1&1\end{array}\right|=(1)(1)-(3)(1)=-2

_____

This problem didn't ask, but the solution is ...

  x = Dx/D = -4/-2 = 2

  y = Dy/D = -2/-2 = 1

  (x, y) = (2, 1)

3 0
3 years ago
Help, can you tell if im right?
8090 [49]

Answer:

Yes, all of your answers are 100% correct

Step-by-step explanation:

You correctly set the values equal to the numerator or denominator (minutes = minutes) and any value will equal the same value.

4 0
3 years ago
I give brainliest if it is correct
Georgia [21]

Answer:

172 yd²

Step-by-step explanation:

4 0
2 years ago
Suppose X and Y are random variables with joint density function. f(x, y) = 0.1e−(0.5x + 0.2y) if x ≥ 0, y ≥ 0 0 otherwise (a) I
Hatshy [7]

a. f_{X,Y} is a joint density function if its integral over the given support is 1:

\displaystyle\int_{-\infty}^\infty\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\frac1{10}\int_0^\infty\int_0^\infty e^{-x/2-y/5}\,\mathrm dx\,\mathrm dy

=\displaystyle\frac1{10}\left(\int_0^\infty e^{-x/2}\,\mathrm dx\right)\left(\int_0^\infty e^{-y/5}\,\mathrm dy\right)=\frac1{10}\cdot2\cdot5=1

so the answer is yes.

b. We should first find the density of the marginal distribution, f_Y(y):

f_Y(y)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx=\frac1{10}\int_0^\infty e^{-x/2-y/5}\,\mathrm dy

f_Y(y)=\begin{cases}\dfrac15e^{-y/5}&\text{for }y\ge0\\\\0&\text{otherwise}\end{cases}

Then

P(Y\ge8)=\displaystyle\int_8^\infty f_Y(y)\,\mathrm dy=e^{-8/5}

or about 0.2019.

For the other probability, we can use the joint PDF directly:

P(X\le5,Y\le8)=\displaystyle\int_0^5\int_0^8f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=1+e^{-41/10}-e^{-5/2}-e^{-8/5}

which is about 0.7326.

c. We already know the PDF for Y, so we just integrate:

E[Y]=\displaystyle\int_{-\infty}^\infty y\,f_Y(y)\,\mathrm dy=\frac15\int_0^\infty ye^{-y/5}\,\mathrm dy=\boxed5

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3 years ago
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3 years ago
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