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3 years ago

Jeff company issues a

1 answer:

6 0

<span>Jeff Company issues a promissory note to David Company to get extended time on an account payable. David records this transaction by debiting </span>

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What is X if 2 times the square root of 2X-1 =4

svp [43]

Remove the radical by raising each side to the index of the radical.
x = 5/2

5 0

3 years ago

A circle has a radius of 6. An arc in this circle has a central angle of 330

sashaice [31]

Answer: $Arc\ lenght\approx34.55\ units$

Step-by-step explanation:

<h3> The complete exercise is: " A circle has a radius of 6. An arc in this circle has a central angle of 330 degrees. What is the arc length?"</h3><h3></h3>

To solve this exercise you need to use the following formula to find the Arc lenght:

$Arc\ lenght=2\pi r(\frac{C}{360})$

Where "C" is the central angle of the arc (in degrees) and "r" is the radius.

In this case, after analize the information given in the exercise, you can identify that the radius and the central angle in degrees, are:

$r=6\ units\\\\C=330$

Therefore, knowing these values, you can substitute them into the formula:

$Arc\ lenght=2\pi (6\ units)(\frac{330}{360})$

And finally,you must evaluate in order to find the Arc lenght.

You get that this is:

$Arc\ lenght=2\pi (6\ units)(\frac{11}{12})\\\\Arc\ lenght=11\pi \ units\\\\Arc\ lenght\approx34.55\ units$

3 0

2 years ago

Read 2 more answers

The USA won 15 Wimbledon championships. Sweden and Switzerland both won 7. Australia won 6, German won 4, and Spain won 2. Write

mihalych1998 [28]

Well you add the total of championships. 15+7+7+6+4+2=41
The ratio is 6:41
This means that Australia won 6 of the 41 championships.

5 0

3 years ago

Mr. Barresi is ordering pizzas for a team party. He knows that 7 pizzas will feed 25 students. If 300 students are attending the

liubo4ka [24]

Divide 300 by 25 and multiply with 7.

300/25 x 7 = X

7 0

3 years ago

Read 2 more answers

A cellular phone company monitors monthly phone usage. The following data represent the monthly phone use in minutes of one part

Sergeu [11.5K]

Answer:

The standard deviation increased but there was no change in the interquantile range

Step-by-step explanation:

We are given the following data in the question:

320, 411, 348, 537, 420, 449, 462, 403, 454, 517, 515, 358, 438, 541, 387, 368, 502, 437, 431, 428.

n = 20

a) Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{8726}{20} = 436.3$

Sum of squares of differences =

13525.69 + 640.09 + 7796.89 + 10140.49 + 265.69 + 161.29 + 660.49 + 1108.89 + 313.29 + 6512.49 + 6193.69 + 6130.89 + 2.89 + 10962.09 + 2430.49 + 4664.89 + 4316.49 + 0.49 + 28.09 + 68.89 = 75924.2

$S.D = \sqrt{\frac{75924.2}{19}} = 63.21$

Sorted Data = 320, 348, 358, 368, 387, 403, 411, 420, 428, 431, 437, 438, 449, 454, 462, 502, 515, 517, 537, 541

$IQR = Q_3 - Q_1\\Q_3 = \text{upper median},\\Q_1 = \text{ lower median}$

$Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}$

$Median = \frac{431 + 437}{2} = 434$

$Q_1 = \frac{387 + 403}{2} = 395\\\\Q_3 = \frac{462 + 502}{2} = 482$

IQR = $482 - 395 = 87$

b) After changing the observation

0, 411, 348, 537, 420, 449, 462, 403, 454, 517, 515, 358, 438, 541, 387, 368, 502, 437, 431, 428

$Mean =\displaystyle\frac{8406}{20} = 420.3$

Sum of squares of differences =

176652.09 + 86.49 + 5227.29 + 13618.89 + 0.09 + 823.69 + 1738.89 + 299.29 + 1135.69 + 9350.89 + 8968.09 + 3881.29 + 313.29 + 14568.49 + 1108.89 + 2735.29 + 6674.89 + 278.89 + 114.49 + 59.29 = 247636.2

$S.D = \sqrt{\frac{247636.2}{19}} = 114.16$

Sorted Data = 0, 348, 358, 368, 387, 403, 411, 420, 428, 431, 437, 438, 449, 454, 462, 502, 515, 517, 537, 541

$Median = \frac{431 + 437}{2} = 434$

$Q_1 = \frac{387 + 403}{2} = 395\\\\Q_3 = \frac{462 + 502}{2} = 482$

IQR = $482 - 395 = 87$

Thus. the standard deviation increased but there was no change in the interquantile range.

5 0

3 years ago