Help please.............

The scheduled arrival time for a daily flight from Boston to New York is 9:35 am. Historical data show that the arrival time fol

Lady_Fox [76]

Answer:

The mean is 11.5 minutes and the standard deviation is of 6.64 minutes

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The mean is:

$M = \frac{a+b}{2}$

The standard deviation is:

$S = \sqrt{\frac{(b-a)^{2}}{12}}$

Arrival time of 9:18 am and a late arrival time of 9:41 am.

9:41 is 23 minutes from 9:18. So the time is uniformily distributed between 0 and 23 minutes, so a = 0, b = 23.

Mean:

$M = \frac{a+b}{2} = \frac{0+23}{2} = 11.5$

Standard deviation:

$S = \sqrt{\frac{(b-a)^{2}}{12}} = \sqrt{\frac{(23 - 0)^{2}}{12}} = 6.64$

The mean is 11.5 minutes and the standard deviation is of 6.64 minutes

8 0

3 years ago

Please help and explain the answer as best as you can so i can understand it better! (6th grader)

qwelly [4]

Answer:

The rug will fit. The rug is smaller than the parallelogram.

Step-by-step explanation:

This question is testing to see if you can figure out how the rug may fit in the room.

The rug does fit. Here's how we know.

The only given information from the question is...

Length and Width of the <em>Rectangular Rug</em>
Area of the Parallelogram

Notice that you can use the Length and the Width to get the area of the Rectangular Rug. We need to find the area of the rectangle so we can compare it to the parallelogram.

The area of the Parallelogram is listed at : 108 sqft

The area of the rectangular rug is : 60 sqft

<em>Area of rectangle = Length * Width</em>
<em>6 * 10 = 60</em>

<em />

The question now asks us if the rug will fit in her room. Well, 60 sqft is less than 108 sqft. So, yes, it will fit.

3 0

3 years ago

Michael ran 655 feet in 19 seconds jasmine ran 1 mile in 4 minutes and 24 seconds in miles per hour and to the nearest tenth how

Natalka [10]

Answer:

Step-by-step explanation:

<h3>Step 1: Convert miles to feet</h3>

First, we need to convert the miles to feet keeping in mind that:

1 mile = 5,280 feet

655 ft 19 sec

5,280 ft 4 min 24 sec

<h3>Step 2: Divide the distance by the time to find the unit rate</h3>

3,447 feet per minute

1,245 feet per minute

<h3>Step 3: Convert minutes to hours</h3>

206,820 feet per hour

74,700 feet per hour

<h3>Step 4: Convert feet to miles</h3>

39.17 miles per hour

14.15 miles per hour

To find how much faster Michael ran, we need one more step.

<h3>Step 5: Subtract smaller amount from larger amount</h3>

39.17 - 14.15 = 25

I'm always happy to help :)

4 0

3 years ago

Find the length of the curve given by ~r(t) = 1 2 cos(t 2 )~i + 1 2 sin(t 2 ) ~j + 2 5 t 5/2 ~k between t = 0 and t = 1. Simplif

xxMikexx [17]

Answer:

The length of the curve is

L ≈ 0.59501

Step-by-step explanation:

The length of a curve on an interval a ≤ t ≤ b is given as

L = Integral from a to b of √[(x')² + (y' )² + (z')²]

Where x' = dx/dt

y' = dy/dt

z' = dz/dt

Given the function r(t) = (1/2)cos(t²)i + (1/2)sin(t²)j + (2/5)t^(5/2)

We can write

x = (1/2)cos(t²)

y = (1/2)sin(t²)

z = (2/5)t^(5/2)

x' = -tsin(t²)

y' = tcos(t²)

z' = t^(3/2)

(x')² + (y')² + (z')² = [-tsin(t²)]² + [tcos(t²)]² + [t^(3/2)]²

= t²(-sin²(t²) + cos²(t²) + 1 )

................................................

But cos²(t²) + sin²(t²) = 1

=> cos²(t²) = 1 - sin²(t²)

................................................

So, we have

(x')² + (y')² + (z')² = t²[2cos²(t²)]

√[(x')² + (y')² + (z')²] = √[2t²cos²(t²)]

= (√2)tcos(t²)

Now,

L = integral of (√2)tcos(t²) from 0 to 1

= (1/√2)sin(t²) from 0 to 1

= (1/√2)[sin(1) - sin(0)]

= (1/√2)sin(1)

≈ 0.59501

8 0

3 years ago

How can you use an equation or a graph to decide if a function is linear?

vredina [299]

An equation can be determined as linear if it is first order in a single variable, i.e. y=3x, but not y=3x^2.
A graph will be linear if the slope is constant

8 0

3 years ago

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Help please.............​

Help please.............