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3 years ago

I need help with a MathsWatch question!

Mathematics

2 answers:

KIM [24]3 years ago

6 0

To find the area of a circle, you do π × diameter (in this case 10). They have told you to use 3.142 as π, so you do 3.142 × 10 = 31.42. Because it's a semi-circle, you need to halve 31.42 to get 16.21, which is the answer

yuradex [85]3 years ago

6 0

A = $\pi$ x r²
A = 3.142 x 5²
A = 9818.75

Semicircle
9818.75 / 2 = 4909.38

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Evaluate the expression

Marrrta [24]

The answer would be 16

11 + (8-3)^2 - 20 = 16

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2 years ago

Arrange the list of numbers in increasing order. Three fractions, arranged from left to right, separated by commas. Begin fracti

Lelechka [254]

The right answer for the question that is being asked and shown above is that: "D. Three fractions, arranged from left to right, separated by commas. Begin fraction four over seven end fraction. Begin fraction three over five end fraction. Begin fraction five over nine end fraction."

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3 years ago

Read 2 more answers

Which equation has a graph that is a parabola with a vertex at (5, 3)?

yanalaym [24]

Answer:

Its A because when you do (x-5)=0 you add 5 to get x=5 and thats your x value and the 3 for your k is your y value

6 0

3 years ago

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

2.50c+6a=6344.50 <br> Find each variable

Ksju [112]

Answer:

c=2.4a+2537.8

a=-2.5/6c+1057.4

Step-by-step explanation:

solve for c:

2.50c+6a=6344.50

-6a -6a

2.50c=-6a+6344.50

/2.50 /2.50 /2.50

c=2.4a+2537.8

solve for a:

2.50c+6a=6344.50

-2.5c -2.5c

6a=-2.5c+6344.50

/6 /6 /6

a=-2.5/6c+1057.4

6 0

3 years ago