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tia_tia [17]
3 years ago
10

Which of these lengths CANNOT represent the sides of a triangle? {28, 30, 58} {32, 34, 60} {13, 20, 27} {2, 4, 5}

Mathematics
1 answer:
Nady [450]3 years ago
6 0

Answer:

A.\  {28, 30, 58}

Step-by-step explanation:

Required

Which of the lengths do not form a triangle

To do this, we make use of triangle inequality theorem which states that.

a + b > c

b + c > a

a + c > b

Where a,\ b\ and\ c are the sides of the triangle.

So, we have:

A.\  {28, 30, 58}

28 + 30 > 58 --- False

58 + 30 > 28 --- True

28 + 58 > 30 --- True

B.\ {32, 34, 60}

32 + 34 > 60 -- True

32 + 60 > 34 -- True

34 + 60 > 32 -- True

C.\ {13, 20, 27}

13 + 20 > 27 -- True

13 + 27 > 20 -- True

20 + 27 > 13 --- True

D.\ {2, 4, 5}

2 + 4 > 5 --- True

2 + 5 > 4 -- True

5 + 4 > 2 --- True

From the computations above, only (A) cannot form a triangle

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For a particular diamond mine, 81% of the diamonds fail to qualify as "gemstone grade". A random sample of 92 diamonds is analyz
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Answer:

68.79% probability that more than 79% of the sample diamonds fail to qualify as gemstone grade

Step-by-step explanation:

We use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 92, p = 0.81

So

\mu = E(X) = np = 92*0.81 = 74.52

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{92*0.81*0.19} = 3.76

Find the probability that more than 79% of the sample diamonds fail to qualify as gemstone grade.

This is 1 subtracted by the pvalue of Z when X = 0.79*92 = 72.68. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{72.68 - 74.52}{3.76}

Z = -0.49

Z = -0.49 has a pvalue of 0.3121

1 - 0.3121 = 0.6879

68.79% probability that more than 79% of the sample diamonds fail to qualify as gemstone grade

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3 years ago
Need help with AP CAL
anzhelika [568]

Answer: Choice C

\displaystyle \frac{1}{2}\left(1 - \frac{1}{e^2}\right)

============================================================

Explanation:

The graph is shown below. The base of the 3D solid is the blue region. It spans from x = 0 to x = 1. It's also above the x axis, and below the curve y = e^{-x}

Think of the blue region as the floor of this weirdly shaped 3D room.

We're told that the cross sections are perpendicular to the x axis and each cross section is a square. The side length of each square is e^{-x} where 0 < x < 1

Let's compute the area of each general cross section.

\text{area} = (\text{side})^2\\\\\text{area} = (e^{-x})^2\\\\\text{area} = e^{-2x}\\\\

We'll be integrating infinitely many of these infinitely thin square slabs to find the volume of the 3D shape. Think of it like stacking concrete blocks together, except the blocks are side by side (instead of on top of each other). Or you can think of it like a row of square books of varying sizes. The books are very very thin.

This is what we want to compute

\displaystyle \int_{0}^{1}e^{-2x}dx\\\\

Apply a u-substitution

u = -2x

du/dx = -2

du = -2dx

dx = du/(-2)

dx = -0.5du

Also, don't forget to change the limits of integration

  • If x = 0, then u = -2x = -2(0) = 0
  • If x = 1, then u = -2x = -2(1) = -2

This means,

\displaystyle \int_{0}^{1}e^{-2x}dx = \int_{0}^{-2}e^{u}(-0.5du) = 0.5\int_{-2}^{0}e^{u}du\\\\\\

I used the rule that \displaystyle \int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx which says swapping the limits of integration will have us swap the sign out front.

--------

Furthermore,

\displaystyle 0.5\int_{-2}^{0}e^{u}du = \frac{1}{2}\left[e^u+C\right]_{-2}^{0}\\\\\\= \frac{1}{2}\left[(e^0+C)-(e^{-2}+C)\right]\\\\\\= \frac{1}{2}\left[1 - \frac{1}{e^2}\right]

In short,

\displaystyle \int_{0}^{1}e^{-2x}dx = \frac{1}{2}\left[1 - \frac{1}{e^2}\right]

This points us to choice C as the final answer.

5 0
1 year ago
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