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3 years ago

PLS HELP FASTTT I NEED TI DUND THE AREA OF EACH SEMICIRCLE

Mathematics

1 answer:

Ne4ueva [31]3 years ago

7 0

Look at the attached picture

Hope it will help you...

Good luck on your assignment

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A certain chalkboard manufacturer determines that their largest blackboard model has a mean length of 5.00 m and a standard devi

Igoryamba

Answer:

23 chalkboards

Step-by-step explanation:

Given:

Mean length = 5 m

Standard deviation = 0.01

Number of units ordered = 1000

Now,

The z factor = $\frac{\textup{x - Mean}}{\textup{standard deviation}}$

The z factor = $\frac{\textup{4.98 - 5}}{\textup{0.01}}$

Z = - 2

Now, the Probability P( length < 4.98 )

Also, From z table the p-value = 0.0228

therefore,

P( length < 4.98 ) = 0.0228

Hence, out of 1000 chalkboard ordered (0.0228 × 1000) = 23 chalkboard are likely to have lengths of under 4.98 m.

7 0

3 years ago

19. Find the volume of a sphere with radius 9.25 cm. Use 3.14 as an approximation for pi. Round the answer to two decimal places

natima [27]

19. The formula for finding the volume of a sphere is 4/3πr^3
The problem says the radius is 9.25 cm.
Just plug in the numbers and solve:
4/3(3.14)(9.25^3)
9.25^3 = ~791.45
4/3(3.14)(791.45)
4/3 • 3.14 = ~4.19
4.19 • 791.45 = 3316.1755
It says to round to 2 decimal places so the answer to this question would be 3316.18.
That means 3316.18 is the volume of the sphere.

20. It's always best to change words into an equation.
-16 - w
w = -23
Plug in the number to the variable:
-16 - (-23)
Double negatives (a negative being subtracted) turn into a plus (positive being added).
-16 + 23 = 7
So the answer is 7.

25. Complementary angles are angles with a sum of 90º.
If angle A is 27º, just subtract that from 90º to get the measure of angle B.
90 - 27 = 63
So angle B is 63º, and the answer is A.

3 0

3 years ago

Use the formula V = s³, where V is the volume and s is the edge length of the cube, to solve this problem.A cube-shaped bin has

KonstantinChe [14]

This is the work shown with the answer. I hope this helps. God bless

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

Can someone please explain this to me I’m so confused and I have a quiz on this tomorrow

Ludmilka [50]

                               x^2 = 16/100
We must solve for x by taking the square root of both sides.
But imagine x^2 = 9. x = 3 is a solution because 3^2 = 9. But x = -3 is also a solution because (-3)^2 = 9. So we need to add the plus and minus sign to account for all of our solutions.
                     √x^2 = √(16/100)
We square root both sides to remove the exponent 2 on x. Then
                         x = ±√(16/100)
Then we have to use a Property of Radicals which is √(x/y) = √(x) / √(y).
Then the square root of 16 is 4 and the square root of 100 is 10. 4/10 is 2/5 simplified.

3 0

3 years ago