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3 years ago

Find the unit rate. 400 cupcakes in 8 hours

Mathematics

1 answer:

pychu [463]3 years ago

5 0

Answer:

50 per hour

Step-by-step explanation:

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Consider the following sets of sample data: A: $30,500, $27,500, $31,200, $24,000, $27,100, $28,600, $39,100, $36,900, $35,000,

Alecsey [184]

Answer:

$CV=0.2$ ---- dataset 1

$CV = 7.2$ --- dataset 2

Step-by-step explanation:

Given

$A: 30500, 27500, 31200, 24000, 27100,28600, 39100, 36900, 35000, 21400, 37900, 27900, 18700,$ $33100$

$B: 4.29, 4.88, 4.34, 4.17, 4.52, 4.80, 3.28, 3.79, 4.84, 4.77, 3.11$

Required

The coefficient of variation of each

Dataset A

Calculate the mean

$\mu = \frac{\sum x}{n}$

$\mu = \frac{30500+ 27500+31200+24000+ 27100+28600+ 39100+ 36900+ 35000+ 21400+ 37900+ 27900+ 18700+33100}{14}$ $\mu = \frac{418900}{14}$

$\mu = 29921.43$

Next, calculate the standard deviation using:

$\sigma = \sqrt{\frac{\sum(x - \mu)^2}{n}}$

So, we have:

$\sigma= \sqrt{\frac{(30500 - 29921.43)^2 +.................+ (18700- 29921.43)^2 + (33100- 29921.43)^2}{13}}$

$\sigma= \sqrt{\frac{487723571.42857}{14}}$

$\sigma= \sqrt{34837397.959184}$

$\sigma= 5902.32$

So, the coefficient of variation is:

$CV=\frac{\sigma}{\mu}$

$CV=\frac{5902.32}{29921.43}$

$CV=0.2$ --- approximated

Dataset B

Calculate the mean

$\mu = \frac{\sum x}{n}$

$\mu = \frac{4.29+ 4.88+ 4.34+ 4.17+ 4.52+ 4.80+ 3.28+ 3.79+ 4.84+ 4.77+ 3.11}{11}$

$\mu = \frac{46.79}{11}$

$\mu = 4.25$

Next, calculate the standard deviation using:

$\sigma = \sqrt{\frac{\sum(x - \mu)^2}{n}}$

$\sigma = \sqrt{\frac{(4.29 - 4.25)^2 + (4.88- 4.25)^2 +.........+ (3.11- 4.25)^2}{11}}$

$\sigma = \sqrt{\frac{3.859}{11}}$

$\sigma = \sqrt{0.35081818181}$

$\sigma = 0.593$

So, the coefficient of variation is:

$CV=\frac{\sigma}{\mu}$

$CV = \frac{4.25}{0.5903}$

$CV = 7.2$ -- approximated

3 0

3 years ago

You have climbed to the top of a tall tree. When you get to the top, you use your clinometer to discover that the angle between

GuDViN [60]

Answer:

F. 34.64 m

Step-by-step explanation:

The measurements given in the question are;

The angle (of depression) given by the clinometer, θ = 30°

The horizontal distance of the lunchbox from the tree, d = 20 meters

The height (how tall) of the tree = Required

Let the height of the tree be assumed to be perpendicular to the ground, noting that the horizontal distance from the lunchbox to the tree is a straight line, let l represent the line of sight from the top of the tree to the lunchbox, and let h represent the height of the tree we have;

The line of sight to the lunchbox, l, the height of the tree, h, and the horizontal distance of the lunchbox from the base of the tree form a right triangle

The height of the tree is the adjacent leg to the given angle by the clinometer

Using trigonometric ratios, we have;

tan(30°) = d/h

∴ tan(30°) = (20 m)/h

h = (20 m)/tan(30°) ≈ 34.64m

The height of the tree, h ≈ 36.64 m.

5 0

3 years ago

Is my solution correct?

poizon [28]

Answer:

The area of the shaded sector is $51.2\pi \ units^{2}$