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3 years ago

What is the median distance jumped 7.48, 7.60, 7.15, 7.44, 7.73, 7.72

1 answer:

3 0

It would be 7.54 because you fist have to put the numbers ordered smallest to biggest amount 7.15, 7.44, 7.48,7.60, 7.72, 7.73 so the numbers that are in the middle are 7.48 and 7.60 you add them up, and the result you divide by 2 which gives you 7.54.

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At a local restaurant, the amount of time that customers have to wait for their food is normally distributed with a mean of 18 m

Colt1911 [192]

Answer:

99.7% of customers have to wait between 8 minutes to 30 minutes for their food.

Step-by-step explanation:

We are given the following in the question:

Mean, μ = 18 minutes

Standard Deviation, σ = 4 minutes

We are given that the distribution of amount of time is a bell shaped distribution that is a normal distribution.

Empirical Formula:

Almost all the data lies within three standard deviation from the mean for a normally distributed data.
About 68% of data lies within one standard deviation from the mean.
About 95% of data lies within two standard deviations of the mean.
About 99.7% of data lies within three standard deviation of the mean.

Thus, 99.7% of the customers have to wait:

$\mu -3\sigma = 18-3(4) = 6\\\mu +3\sigma = 18+3(4) = 30$

Thus, 99.7% of customers have to wait between 8 minutes to 30 minutes for their food.

3 0

3 years ago

US THE DISTRIBUTIVE PROPERTY TO ANSWER

Svetradugi [14.3K]

Answer:

1. -6t - 18

2. 21n - 14

3. -28 + 7t

4. 5m - 30

5. -48 + 8c

6. -15y -24

Step-by-step explanation:

Hope this helps!!!

3 0

3 years ago

Which expression is equivalent to -36 – 8?

balandron [24]

Answer:

-36+-8=-44

Step-by-step explanation:

8 0

3 years ago

What is the soulation to 2x - 8 greater than 12

Masteriza [31]

Answer:

x > 10

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Sketch the equilibrium solutions for the following DE and use them to determine the behavior of the solutions.

GREYUIT [131]

Answer:

$y=\dfrac{1}{1-Ke^{-t}}$

Step-by-step explanation:

Given

The given equation is a differential equation

$\dfrac{dy}{dt}=y-y^2$

$\dfrac{dy}{dt}=-(y^2-y)$

By separating variable

⇒ $\dfrac{dy}{(y^2-y)}=-t$

$\left(\dfrac{1}{y-1}-\dfrac{1}{y}\right)dy=-dt$

Now by taking integration both side

$\int\left(\dfrac{1}{y-1}-\dfrac{1}{y}\right)dy=-\int dt$

⇒ $\ln (y-1)-\ln y=-t+C$

Where C is the constant

$\ln \dfrac{y-1}{y}=-t+C$

$\dfrac{y-1}{y}=e^{-t+c}$

$\dfrac{y-1}{y}=Ke^{-t}$

$y=\dfrac{1}{1-Ke^{-t}}$

from above equation we can say that

When t will increases in positive direction then $e^{-t}$ will decreases it means that ${1-Ke^{-t}}$ will increases, so y will decreases. Similarly in the case of negative t.

4 0

3 years ago