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3 years ago

8

Jennifer’s pie shop recorded how many pies it recently sold in each flavor. Lemon meringue pies 2 apple pies 3 peach pies 2 blue

berry pies 4 coconut cream pies 1 considering this data, how many of the next 15 pies sold should you expect to be blueberry pies?

2 answers:

bazaltina [42]3 years ago

7 0

Answer:

correct me if im wrong but im assuming four just cause if you count what you have all ready not couting the 15 pies. You have at leat 12 and not couting the lemon pies. And from there you basically have to just thing and do the math and I got at least 3 to 5 pies that should be blueberry. srry for my spekling it ducks

anyanavicka [17]3 years ago

5 0

Answer:

$2\frac{8}{11}$ will be blueberry

Step-by-step explanation:

2 apple

3 peach

2 blueberry

4 coconut

2 + 3 + 2 + 4 = 11

2 of 11 are blueberry

$\frac{2}{11} =\frac{y}{15}$

y × 11 = 2 × 15

11y = 30

11y ÷ 11 = 30 ÷ 11

y = 2 and 8/11

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Answer:

Step-by-step explanation:

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3 years ago

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis. Verify y

Drupady [299]

Answer:

$V = \frac{\pi^2}{8}$

$V = 1.23245$

Step-by-step explanation:

Given

$y = \cos 2x$

$y = 0; x = 0; x = \frac{\pi}{4}$

Required

Determine the volume of the solid generated

Using the disk method approach, we have:

$V = \pi \int\limits^a_b {R(x)^2} \, dx$

Where

$y = R(x) = \cos 2x$

$a = \frac{\pi}{4}; b =0$

So:

$V = \pi \int\limits^a_b {R(x)^2} \, dx$

Where

$y = R(x) = \cos 2x$

$a = \frac{\pi}{4}; b =0$

So:

$V = \pi \int\limits^a_b {R(x)^2} \, dx$

$V = \pi \int\limits^{\frac{\pi}{4}}_0 {(\cos 2x)^2} \, dx$

$V = \pi \int\limits^{\frac{\pi}{4}}_0 {\cos^2 (2x)} \, dx$

Apply the following half angle trigonometry identity;

$\cos^2(x) = \frac{1}{2}[1 + \cos(2x)]$

So, we have:

$\cos^2(2x) = \frac{1}{2}[1 + \cos(2*2x)]$

$\cos^2(2x) = \frac{1}{2}[1 + \cos(4x)]$

Open bracket

$\cos^2(2x) = \frac{1}{2} + \frac{1}{2}\cos(4x)$

So, we have:

$V = \pi \int\limits^{\frac{\pi}{4}}_0 {\cos^2 (2x)} \, dx$

$V = \pi \int\limits^{\frac{\pi}{4}}_0 {[\frac{1}{2} + \frac{1}{2}\cos(4x)]} \, dx$

Integrate

$V = \pi [\frac{x}{2} + \frac{1}{8}\sin(4x)]\limits^{\frac{\pi}{4}}_0$

Expand

$V = \pi ([\frac{\frac{\pi}{4}}{2} + \frac{1}{8}\sin(4*\frac{\pi}{4})] - [\frac{0}{2} + \frac{1}{8}\sin(4*0)])$

$V = \pi ([\frac{\frac{\pi}{4}}{2} + \frac{1}{8}\sin(4*\frac{\pi}{4})] - [0 + 0])$

$V = \pi ([\frac{\frac{\pi}{4}}{2} + \frac{1}{8}\sin(4*\frac{\pi}{4})])$

$V = \pi ([{\frac{\pi}{8} + \frac{1}{8}\sin(\pi)])$

$\sin \pi = 0$

So:

$V = \pi ([{\frac{\pi}{8} + \frac{1}{8}*0])$

$V = \pi *[{\frac{\pi}{8}]$

$V = \frac{\pi^2}{8}$

or

$V = \frac{3.14^2}{8}$

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3 years ago

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Marianna [84]

Answer:

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Step-by-step explanation:

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3 years ago

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ElenaW [278]

Answer:

Q1 = 61

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Step-by-step explanation:

From the plot attached :

Data obtained is :

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59 0

60 2

61 3

62 3

63 4

64 2

65 1

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Data in raw format :

60,60,61,61,61,62,62,62,63,63,63,63,64,64,65

The first quartile :

Q1 = 1/4(n+1)th term

n = 15

Q1 = 1/4(15+1)th term

Q1 = 1/4 * 16th

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Q3 = 3/4(n+1)th term

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Q3 = 3/4(15+1)th term

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3 years ago

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dalvyx [7]

For the midpoint, you find the differences in x values and y values then divide them by 2. Like so:

Let's do x first:

(-16-0)/2 = (-16)/2 = -8

So our x coordinate is -8.

Now Y:

(-16)/2 = -8

So our Y coordinate is -8 as well.

So your answer is:

(-8,-8)

6 0

4 years ago