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katrin [286]
2 years ago
7

A 99-inch candle burns down in 12 hours. At what rate does the candle burn, in inches per hour?

Mathematics
2 answers:
Verdich [7]2 years ago
7 0
8.25 is the answer because of 99/12
Kitty [74]2 years ago
5 0
8.25 i believe (sorry if i’m wrong )
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Triangles A B C and X Y Z are shown. The length of side A B is 6 and the length of side B C is 5. The length of side X Y is 1.5,
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Answer:

D) a dilation by 1/4 and a translation

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Find an equation of the line. Write the equation in function notation. Through (6,-1); perpendicular to 9y=x-18
docker41 [41]
The first step is to find the slope of the given line by putting its equation in the form y = mx + b.
9y = x - 18
Dividing both sides by 9, gives:
y = (x/9) - 2
The slope of the given line is therefore 1/9.
Let the slope of the perpendicular line be m.
The product of the two slopes must equal -1 for the lines to be perpendicular.
m\times\frac{1}{9}=-1
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3 years ago
A collection of 30 gems, all of which are identical in appearance, are supposedto be genuine diamonds, but actually contain 8 wo
levacccp [35]

Answer:

Step-by-step explanation:

From the given information:

There are 30 collections of gems, of which 8 are worthless;

Thus, the number of the genuine diamonds = 30 - 8 = 22.

Let X = random variable;

X consider the value as 0 (for  2 worthless stone selection),

X = 1200(1 worthless stone  & 1 genuine stone)

X = 2400 (2 genuine stones selected)

However, the numbers of ways of selecting and chosen Gems can be estimated as:

(^n_r) = (^{30}_2) \\ \\ \implies \dfrac{30!}{2!(30-2)!} \\ \\ \implies \dfrac{30!}{2!(28)!}  \\ \\  \implies \dfrac{30*29*28!}{2!(28)!} \\ \\  \implies  \dfrac{30*29}{2*1} \\ \\ \implies 435

Thus;

Pr (X = 0) = \dfrac{(^8_2)}{435}

Pr (X = 0) = \dfrac{\dfrac{8!}{2!(8-2)!}}{435} \\ \\ Pr (X = 0) = \dfrac{\dfrac{8!}{2!(6)!}}{435}  \\ \\ Pr (X = 0) = \dfrac{\dfrac{8*7*6!}{2!(6)!}}{435} \\ \\  Pr (X = 0) = \dfrac{\dfrac{8*7}{2*1}}{435} \\ \\   Pr (X = 0) = 0.0644

P(X =1200) = \dfrac{(^{8}_{1})(^{22}_{1})}{435}

P(X =1200) = \dfrac{ \dfrac{8!}{1!(8-1)!}) ( \dfrac{22!}{1!(22-1)!}) }{435}

P(X =1200) = \dfrac{ (8) ( 22) }{435}

P(X =1200) =0.4046

Pr (X = 2400) = \dfrac{(^{22}_2)}{435}

Pr (X = 2400) = \dfrac{\dfrac{22!}{2!(22-2)!}}{435} \\ \\ Pr (X = 2400) = \dfrac{\dfrac{22!}{2!(20)!}}{435}  \\ \\ Pr (X = 2400) = \dfrac{\dfrac{22*21*20!}{2!(20)!}}{435} \\ \\  Pr (X =2400) = \dfrac{\dfrac{22*21}{2*1}}{435} \\ \\   Pr (X = 2400) = 0.5310

To find E(X):

E(X) = (0 × 0.0644) + (1200 × 0.4046)  + (2400 × 0.5310)

E(X) = 0 + 485.52 + 1274.4

E(X) = 1759.92

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Answer:

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Here, since the age of the crab is given in months and the regression line was computed using the age in months, we just have to replace the value of the age in the formula to estimate the size.

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