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3 years ago

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martin ordered a pizza with a 16-inch diameter. Ricky ordered a pizza with a 20-inch diameter. What is the approximate differenc

e in area of the two pizzas?

1 answer:

ser-zykov [4K]3 years ago

7 0

Answer:

14

Step-by-step explanation:

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This is the 4th question

gavmur [86]

Answer:

a) There are 9 outcomes.

b) **i have attached the tree diagram**

c) 1/9

d) 1/9

Step-by-step explanation:

Both events (choosing a shirt and choosing pants) are not mutually exclusive (disjoint). Meaning that both events can occur at the same time.

a) For a sequence of two events in which the first event can occur <em>m</em> ways and the second event can occur <em>n</em> ways, the events together can occur a total of <em>m</em> • <em>n</em> ways. So for this problem, you would do 3 • 3 which equals 9.

b) **tree diagram is attached**

c) Because there are two different events occurring (choosing a shirt and choosing pants) you need to multiply using the Multiplication Rule for Independent Events : P(A and B) = P(A) * P(B)

The probability of event A occurring (choosing a red shirt) is 1/3 and the probability of event B occurring (choosing brown pants) is also 1/3.

1/3 * 1/3 = 1/9

d) You would do the same thing here as part c.

The probability of event A occurring (choosing a blue shirt) is 1/3 and the probability of event B occurring (choosing blue pants) is again 1/3.

1/3 * 1/3 = 1/9

I hope this was helpful!! :)

4 0

3 years ago

A lawn mowing service charges customers a flat fee plus an hourly rate for their services. The expression 55x + 15 is used to ca

EastWind [94]

Answer:

JK+7]BBUH ,.H,YT MKY,TFY7YKJKKKKKKKKKJKJKHNJ[KL-L;*

Step-by-step explanation:

JK ITS B

6 0

3 years ago

Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D is the trian

Alla [95]

Answer: mass (m) = 4 kg

center of mass coordinate: (15.75,4.5)

Step-by-step explanation: As a surface, a lamina has 2 dimensions (x,y) and a density function.

The region D is shown in the attachment.

From the image of the triangle, lamina is limited at x-axis: 0≤x≤2

At y-axis, it is limited by the lines formed between (0,0) and (2,1) and (2,1) and (0.3):

<u>Points (0,0) and (2,1):</u>

y = $\frac{1-0}{2-0}(x-0)$

y = $\frac{x}{2}$

<u>Points (2,1) and (0,3):</u>

y = $\frac{3-1}{0-2}(x-0) + 3$

y = -x + 3

Now, find total mass, which is given by the formula:

$m = \int\limits^a_b {\int\limits^a_b {\rho(x,y)} \, dA }$

Calculating for the limits above:

$m = \int\limits^2_0 {\int\limits^a_\frac{x}{2} {2(x+y)} \, dy \, dx }$

where a = -x+3

$m = 2.\int\limits^2_0 {\int\limits^a_\frac{x}{2} {(xy+\frac{y^{2}}{2} )} \, dx }$

$m = 2.\int\limits^2_0 {(-x^{2}-\frac{x^{2}}{2}+3x )} \, dx }$

$m = 2.\int\limits^2_0 {(\frac{-3x^{2}}{2}+3x)} \, dx }$

$m = 2.(\frac{-3.2^{2}}{2}+3.2-0)$

m = 2(-4+6)

m = 4

<u>Mass of the lamina that occupies region D is 4.</u>

<u />

Center of mass is the point of gravity of an object if it is in an uniform gravitational field. For the lamina, or any other 2 dimensional object, center of mass is calculated by:

$M_{x} = \int\limits^a_b {\int\limits^a_b {y.\rho(x,y)} \, dA }$

$M_{y} = \int\limits^a_b {\int\limits^a_b {x.\rho(x,y)} \, dA }$

$M_{x}$ and $M_{y}$ are moments of the lamina about x-axis and y-axis, respectively.

Calculating moments:

For moment about x-axis:

$M_{x} = \int\limits^a_b {\int\limits^a_b {y.\rho(x,y)} \, dA }$

$M_{x} = \int\limits^2_0 {\int\limits^a_\frac{x}{2} {2.y.(x+y)} \, dy\, dx }$

$M_{x} = 2\int\limits^2_0 {\int\limits^a_\frac{x}{2} {y.x+y^{2}} \, dy\, dx }$

$M_{x} = 2\int\limits^2_0 { ({\frac{y^{2}x}{2}+\frac{y^{3}}{3})}\, dx }$

$M_{x} = 2\int\limits^2_0 { ({\frac{x(-x+3)^{2}}{2}+\frac{(-x+3)^{3}}{3} -\frac{x^{3}}{8}-\frac{x^{3}}{24} )}\, dx }$

$M_{x} = 2.(\frac{-9.x^{2}}{4}+9x)$

$M_{x} = 2.(\frac{-9.2^{2}}{4}+9.2)$

$M_{x} = 18$

Now to find the x-coordinate:

x = $\frac{M_{y}}{m}$

x = $\frac{63}{4}$

x = 15.75

For moment about the y-axis:

$M_{y} = \int\limits^2_0 {\int\limits^a_\frac{x}{2} {2x.(x+y))} \, dy\,dx }$

$M_{y} = 2.\int\limits^2_0 {\int\limits^a_\frac{x}{2} {x^{2}+yx} \, dy\,dx }$

$M_{y} = 2.\int\limits^2_0 {y.x^{2}+x.{\frac{y^{2}}{2} } } \,dx }$

$M_{y} = 2.\int\limits^2_0 {x^{2}.(-x+3)+\frac{x.(-x+3)^{2}}{2} - {\frac{x^{3}}{2}-\frac{x^{3}}{8} } } \,dx }$

$M_{y} = 2.\int\limits^2_0 {\frac{-9x^3}{8}+\frac{9x}{2} } \,dx }$

$M_{y} = 2.({\frac{-9x^4}{32}+9x^{2})$

$M_{y} = 2.({\frac{-9.2^4}{32}+9.2^{2}-0)$

$M{y} =$ 63

To find y-coordinate:

y = $\frac{M_{x}}{m}$

y = $\frac{18}{4}$

y = 4.5

<u>Center mass coordinates for the lamina are (15.75,4.5)</u>

3 0

4 years ago

The questions are on the attached screenshot file.

Makovka662 [10]

Look at the ss down below

4 0

3 years ago

Read 2 more answers

Jim and Carla are scuba diving. Jim started out 12 feet below the surface. He descended 17 feet, rose 3 feet, and descended 8 mo

Maslowich

Answer:

Jim reached a greater depth by one foot, as he started 12 feet below the surface.

Step-by-step explanation:

Jim

12 feet+17 feet = 29

29-3 feet (rose 3 feet) = 26

26+8= 34 feet

Carla

0 feet (surface)+ 19 feet= 19

19-3 feet= 16

16+17= 33 feet

5 0

3 years ago