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siniylev [52]
4 years ago
10

yesterday, Jack drove 40 1/2 miles. he used 1 1/4 gallons of gasoline. What is the unit rate for miles per gallons

Mathematics
1 answer:
Amanda [17]4 years ago
3 0


From the information we have, Jack drove 401/2 miles and used a total of 1 1/4 gallons of gasoline.

We need to find out how many miles he travelled  per each gallon.

40 1/2 miles can be written as 40.5 miles

1 1/4 gallons  can be written as 1.25 gallons.

So now we form an equation.

40.5 = 1.25

   x = 1

Where x is the number of miles per gallon

We cross multiply the equation

40.5 * 1 = 1.25 * x


40.5 = 1.25x

40.5 / 1.25  = x

 32.4 = x

x = 32 .4

So the unit rate for miles per gallons is 32.4 miles per gallon


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Step-by-step explanation:

Perhaps you're concerned with the sequence ...

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1. This is neither arithmetic nor geometric. Ratios of terms are -4/3, -5/3, -6/3.

The alternating signs mean one factor of the general term is (-1)^(n-1). The divisors of 3 in the term ratios indicate 3^-n is another factor. The increasing multipliers suggest that a factorial is involved.

If we rewrite the sequence factoring out (-1)^(n-1)/3^n, we have ...

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The sequence diverges.

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3 0
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;

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Let X = random variable;

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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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4 years ago
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faltersainse [42]
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3 years ago
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