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3 years ago

A 50-mile trip uses 2 gallons of gas. how much gas would you use on a 110-mile trip?

Mathematics

2 answers:

nikdorinn [45]3 years ago

5 0

We ca use proportion here.

50 mi -----2 gallons
110 mi ---x gallons

x=110*2/50=4.5 gallons

mafiozo [28]3 years ago

3 0

The answer is 4.5!!Hope this helps!!!!!

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Playland Park charges $25 admission plus $2 per ride. Funland Park charges $35 admission plus $1 per ride. For what number of ri

Ket [755]

Answer:

10 rides

Step-by-step explanation:

Make an equation.

25 + 2x = 35 + 1x

Subract 1x from the 2x to get 1.

New Equation: 25 + 1x = 35

Subtract 25 from 35 to get 10.

New Equation: 1x = 10

Divide 1 from both sides.

Answer: 10 rides

8 0

3 years ago

What is 0.3636... round to the nearest tenth?

miskamm [114]

0.36.
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3 years ago

The number of people arriving at a ballpark is random, with a Poisson distributed arrival. If the mean number of arrivals is 10,

Stella [2.4K]

Answer:

a) 3.47% probability that there will be exactly 15 arrivals.

b) 58.31% probability that there are no more than 10 arrivals.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

If the mean number of arrivals is 10

This means that $\mu = 10$

(a) that there will be exactly 15 arrivals?

This is P(X = 15). So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 15) = \frac{e^{-10}*(10)^{15}}{(15)!} = 0.0347$

3.47% probability that there will be exactly 15 arrivals.

(b) no more than 10 arrivals?

This is $P(X \leq 10)$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-10}*(10)^{0}}{(0)!} = 0.000045$

$P(X = 1) = \frac{e^{-10}*(10)^{1}}{(1)!} = 0.00045$

$P(X = 2) = \frac{e^{-10}*(10)^{2}}{(2)!} = 0.0023$

$P(X = 3) = \frac{e^{-10}*(10)^{3}}{(3)!} = 0.0076$

$P(X = 4) = \frac{e^{-10}*(10)^{4}}{(4)!} = 0.0189$

$P(X = 5) = \frac{e^{-10}*(10)^{5}}{(5)!} = 0.0378$

$P(X = 6) = \frac{e^{-10}*(10)^{6}}{(6)!} = 0.0631$

$P(X = 7) = \frac{e^{-10}*(10)^{7}}{(7)!} = 0.0901$

$P(X = 8) = \frac{e^{-10}*(10)^{8}}{(8)!} = 0.1126$

$P(X = 9) = \frac{e^{-10}*(10)^{9}}{(9)!} = 0.1251$

$P(X = 10) = \frac{e^{-10}*(10)^{10}}{(10)!} = 0.1251$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.000045 + 0.00045 + 0.0023 + 0.0076 + 0.0189 + 0.0378 + 0.0631 + 0.0901 + 0.1126 + 0.1251 + 0.1251 = 0.5831$

58.31% probability that there are no more than 10 arrivals.

8 0

3 years ago

WILL GIVE BRANLIEST!!! Pls help! Determine the coordinates of the point on the straight line y=3x+1 that is equidistant from the

iren [92.7K]

Let , coordinate of points are P( h,k ).

Also , k = 3h + 1

Distance of P from origin :

$d=\sqrt{h^2+k^2}$

Distance of P from ( -3, 4 ) :

$d=\sqrt{(h+3)^2+(k-4)^2}$

Now , these distance are equal :

$h^2+(3h+1)^2=(h+3)^2+(3h+1-4)^2\\\\h^2+(3h+1)^2=(h+3)^2+(3h-3)^2$

Solving above equation , we get :

$P=(\dfrac{16}{21},\dfrac{23}{7})$

Hence , this is the required solution.

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3 years ago

What value makes the Inequality n > -5

Crank

Answer:

Any numbers above -5 make this inequality true. Values that will make the inequality true: -4, -3, -2, -1, 0, 1...

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3 years ago

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