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Talja [164]
1 year ago
11

Plese HELP !!!

Mathematics
1 answer:
beks73 [17]1 year ago
6 0

Answer:

-9/8

Step-by-step explanation:

\frac{-2a^{-3}b^2}{a} \\ \\ =-\frac{2b^2}{a^4} \\ \\ =-\frac{2(3)^2}{(-2)^4} \\ \\ =-\frac{18}{16} \\ \\ =-\frac{9}{8}

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The school band bought cheese and pepperoni pizzas in the ratio represented in the tape diagram for their
GenaCL600 [577]

Answer:

If they buy 6 cheese pizzas, they buy 2 pepperoni pizzas.

For every 3 cheese pizzas, they bought 1 pepperoni pizza.

If they buy 6 cheese pizzas:

3 * ? = 6

Rearrange to find ?

6 / 3 = ? = 2

You multiply both sides by the same number.

1 * 2 = 2

If they buy 6 cheese pizzas, they buy 2 pepperoni pizzas.

PLEEASEEE GIVE BRAINLIEST

4 0
3 years ago
I need help w/ this please help me!!
artcher [175]

Answer:B

Step-by-step explanation:

B

4 0
2 years ago
Read 2 more answers
S(t) = -120t + 1,080 to determine the salvage value, S(t), in dollars, of a table saw t years after its purchase. How long will
tester [92]
In order for S(t) to equal to 0, it would take 9 years. If you multiply 9 times the -120, you will get -1080, then you add that to the +1,080 and the answer is 0.

4 0
3 years ago
Any help? Much appreciated ​
natulia [17]

Answer:

\frac{6+\sqrt{27} }{4-\sqrt{3} } = \frac{r+s\sqrt{3} }{13} \\We\ may\ multiply\ the\ numerator\ and\ denominator\ with\ (4+\sqrt{3} ).\\Hence,\\\frac{(6+\sqrt{27})(4+\sqrt{3})  }{13} =  \frac{r+s\sqrt{3} }{13} \\Hence,\\24+6\sqrt{3} +4\sqrt{27} +9= r+s\sqrt{3}\\24+6\sqrt{3} +4*3\sqrt{3} } +9= r+s\sqrt{3}\\24+6\sqrt{3} +12\sqrt{3}+9 = r+s\sqrt{3}\\33+18\sqrt{3} = r+s\sqrt{3}\\Hence,\\r=33, s=18

5 0
3 years ago
Let Y1 and Y2 be independent exponentially distributed random variables, each with mean 7. Find P(Y1 > Y2 | Y1 < 2Y2). (En
ArbitrLikvidat [17]

<em>Y</em>₁ and <em>Y</em>₂ are independent, so their joint density is

f_{Y_1,Y_2}(y_1,y_2)=f_{Y_1}(y_1)f_{Y_2}(y_2)=\begin{cases}\frac1{49}e^{-\frac{y_1+y_2}7}&\text{for }y_1\ge0,y_2\ge0\\0&\text{otherwise}\end{cases}

By definition of conditional probability,

P(<em>Y</em>₁ > <em>Y</em>₂ | <em>Y</em>₁ < 2 <em>Y</em>₂) = P((<em>Y</em>₁ > <em>Y</em>₂) and (<em>Y</em>₁ < 2 <em>Y</em>₂)) / P(<em>Y</em>₁ < 2 <em>Y</em>₂)

Use the joint density to compute the component probabilities:

• numerator:

P((Y_1>Y_2)\text{ and }(Y_1

=\displaystyle\frac1{49}\int_0^\infty\int_{\frac{y_1}2}^{y_1}e^{-\frac{y_1+y_2}7}\,\mathrm dy_2\,\mathrm dy_1

=\displaystyle-\frac17\int_0^\infty\int_{-\frac{3y_1}{14}}^{-\frac{2y_1}7}e^u\,\mathrm du\,\mathrm dy_1

=\displaystyle-\frac17\int_0^\infty\left(e^{-\frac{2y_1}7} - e^{-\frac{3y_1}{14}}\right)\,\mathrm dy_1

=\displaystyle-\frac17\left(-\frac72e^{-\frac{2y_1}7} + \frac{14}3 e^{-\frac{3y_1}{14}}\right)\bigg|_0^\infty

=\displaystyle-\frac17\left(\frac72 - \frac{14}3\right)=\frac16

• denominator:

P(Y_1

(I leave the details of the second integral to you)

Then you should end up with

P(<em>Y</em>₁ > <em>Y</em>₂ | <em>Y</em>₁ < 2 <em>Y</em>₂) = (1/6) / (2/3) = 1/4

5 0
2 years ago
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