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Katena32 [7]
2 years ago
8

the measurements of a photo and it's frame are shown in the diagram. Write a polynomial that represents the width of the photo.

Mathematics
1 answer:
suter [353]2 years ago
4 0

Answer:

The width of the photo is 4w^2+6w+4.

Step-by-step explanation:

From the given figure it is notices that the total width of the frame is

6w^2+8

The photo is covered by a frame border and the width of the border is

w^2-3w+2

To find the width of the photo we have to subtract the width of upper frame border and lower frame border from the total width of frame.

Width of the photo is

\text{Width of the photo}=\text{Width of the frame}-2(\text{Width of the frame border})

\text{Width of the photo}=6w^2+8-2(w^2-3w+2)

\text{Width of the photo}=6w^2+8-2w^2+6w-4

\text{Width of the photo}=4w^2+6w+4

Therefore the width of the photo is 4w^2+6w+4.

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Solve for x. Round to the nearest tenth. Please help
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Step-by-step explanation:

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3 years ago
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Problem #3
AfilCa [17]

Answer:

b = 115.4 feet

Step-by-step explanation:

b^2 = a^2 + c^2 - 2 (a)(b) x cos B

b^2 = 230^2 + 360^2 - 2(230)(360) x cos(38 degrees)

b^2 = 52,900 + 129,600 - 2(230)(360) x cos(38 degrees)

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i hope this helps!! :)

8 0
2 years ago
An honest die is rolled. If the roll comes out even (2, 4, or 6), you will win $1; if the roll comes out odd (1,3, or 5), you wi
jenyasd209 [6]

Answer:

(a) 50%

(b) 47.5%

(c) 2.5%

Step-by-step explanation:

According to the honest coin principle, if the random variable <em>X</em> denotes the number of heads in <em>n</em> tosses of an honest coin (<em>n</em> ≥ 30), then <em>X</em> has an approximately normal distribution with mean, \mu=\frac{n}{2} and standard deviation, \sigma=\frac{\sqrt{n}}{2}.

Here the number of tosses is, <em>n</em> = 2500.

Since <em>n</em> is too large, i.e. <em>n</em> = 2500 > 30, the random variable <em>X</em> follows a normal distribution.

The mean and standard deviation are:

\mu=\frac{n}{2}=\frac{2500}{2}=1250\\\\\sigma=\frac{\sqrt{n}}{2}=\frac{\sqrt{2500}}{2}=25

(a)

To not lose any money the even rolls has to be 1250 or more.

Since, <em>μ</em> = 1250 it implies that the 50th percentile is also 1250.

Thus, the probability that by the end of the evening you will not have lost any money is 50%.

(b)

If the number of "even rolls" is 1250, it implies that the percentile of 1250 is 50th.

Then for number of "even rolls" as 1300,

1300 = 1250 + 2 × 25

        = μ + 2σ

Then P (μ + 2σ) for a normally distributed data is 0.975.

⇒ 1300 is at the 97.5th percentile.

Then the area between 1250 and 1300 is:

Area = 97.5% - 50%

        = 47.5%

Thus, the probability that the number of "even rolls" will fall between 1250 and 1300 is 47.5%.

(c)

To win $100 or more the number of even rolls has to at least 1300.

From part (b) we now 1300 is the 97.5th percentile.

Then the probability that you will win $100 or more is:

P (Win $100 or more) = 100% - 97.5%

                                   = 2.5%.

Thus, the probability that you will win $100 or more is 2.5%.

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Answer:

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Step-by-step explanation:

I've done this before :)

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