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vladimir1956 [14]
3 years ago
8

Jillian walked 0.5 miles before she started jogging at an average pace of 5 miles per hour. The equation d = 0.5 + 5t can be use

d to relate the total distance, d, in miles to the time, t, that Jillian spent jogging. What are the independent and dependent variables? The independent variable is distance and the dependent variable is time. The independent variable is time and the dependent variable is distance. The independent variable is distance and the dependent variable is speed. The independent variable is speed and the dependent variable is distance.
Mathematics
2 answers:
viktelen [127]3 years ago
3 0
For this case we have the following equation:
 d = 0.5 + 5t
 We will describe the equation step by step.
 0.5: Initial distance Jillian walked
 5: speed with which Jillian begins to jog
 t: independent variable that indicates the number of hours
 y: dependent variable that indicates the distance traveled in miles
 Answer:
 
The variable variable is time and the dependent variable is distance.
Diano4ka-milaya [45]3 years ago
3 0

Answer B The independent variable is time and the dependent variable is distance.

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What value of c makes the equation true? 34 = 6 + 4c
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Step-by-step explanation:

34-6=28

28/4=7

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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.

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Answer:

\bigl(\frac{3}{2},-5\bigr)

or in decimal form

(1.5,-5)

Step-by-step explanation:

The midpoint of the line with endpoints (x_1,y_1) and (x_2,y_2) is \bigl(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\bigr). just take the average between the points

so given the points (1,0) and (2,-10)

x_1=1

y_1=0

x_2=2

y_2=-10

the midpoint is found as follows:

\bigl(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\bigr)

\bigl(\frac{1+2}{2},\frac{0-10}{2}\bigr)

\bigl(\frac{3}{2},\frac{-10}{2}\bigr)

\bigl(\frac{3}{2},-5\bigr)

or in decimal form

(1.5,-5)

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These can be some values of equation
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This question might not get the right answer but this how you set equations.






7 0
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