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Alekssandra [29.7K]
3 years ago
14

Please show work! will give brainlist!

Mathematics
2 answers:
laila [671]3 years ago
7 0
The answer is a, 21 inches
Akimi4 [234]3 years ago
5 0

Answer:

a) 21 inches :))

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Which of these is the cross section if the slice is made perpendicular to the base?
Vesna [10]

Hi there!

\large\boxed{\text{ C. Rectangle}}

If cross sections were made perpendicular to the base, they would assume the shape of the lateral sides.

Thus, the cross sections would be rectangles. The correction answer is C.

6 0
3 years ago
The paragraph below comes from the rental agreement Susan signed when she opened her account at Super Video.
igor_vitrenko [27]
0283646.8 dhbdvfbe 29376.0
5 0
2 years ago
What single transformation was applied to triangle AAA to get triangle BBB?
UkoKoshka [18]
I assume c, because usually to get to the opposite you reflect the line or figure on the x axis
4 0
4 years ago
You have 15 milliliters of a drink that contains 15% milk. How many millimeters of a drink containing 63% milk needs to be added
Igoryamba

Answer:

4 ml of the 63% milk drink

Step-by-step explanation:

Multiplying 15 ml by 0.15 results in 2.25 ml, the amount of whole milk in the drink.  Let m represent the number of ml of a drink that is 63% milk.

The final amount of milk drink that is to be 45% milk will be 15 ml + m, and the amount of whole milk contained in this drink will be 0.45(15 + m).

Then:

0.15(15 ml) + 0.63(m) = 0.45(15 + m), where m is to be in milliliters.

2.25 + 0.63m = 6.75 + 0.45m

First:  consolidate the m terms on the left.  0.63m less 0.45m yields 18 m; then we have:

2.25 + 18m = 6.75, or

18 m = 4.50, or m = 4 ml.

In conclusion:  adding 4 ml of that 63% milk drink to the initial 15 ml of 15% milk will result in (15 ml + 4 ml) of a 45% milk drink.            

8 0
3 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

5 0
3 years ago
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