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Marina CMI [18]

2 years ago

10

Which value of y makes the equation true. 72=y over6

1 answer:

Lina20 [59]2 years ago

6 0

Answer: y = 432

Step-by-step explanation:

72 = y/6

y = 6 x 72

y = 432

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After tax it will cost $12.99.

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A swimming pool is open for 7 1/2 hours during a day. The pool keeps one lifeguard on duty at a time, and each lifeguard's shift

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3 years ago

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Can you prove that DEF = HGF.!? Justify your answer.

frosja888 [35]

Step-by-step explanation:

$In \: \triangle DEF \: \& \: \triangle HGF \\ EF \: \cong \: GF...(given) \\ \angle EFD \cong \angle GFH \\ (vertically \: opposite \: angles) \\ \ DF \: \cong \: HF...(given) \\ \therefore \: \triangle DEF \cong \triangle HGF \\ (By\: SAS \:criterion\: of \:congruence)$

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2 years ago

The photo shows a small coin. The scale from the actual coin to the photo is 8 mm to 2 cm In the photo, the distance across the

alekssr [168]

Answer:

1.3cm

Step-by-step explanation:

The scale from the actual coin to the photo is 8 mm to 2 cm In the photo,

converting 8mm to cm = 0.8 cm

using ratio and proportion,

if 2cm = 3.25 cm

then 0.8cm = ? less

if less then more devides.

thus ; (0.8cm ÷2cm ) × 3.25cm =

( 0.4) × 3.25cm = 1.3cm

therefore the distance across the actual coin is 1.3cm.

3 0

2 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