How would you draw a counterexample to this?

Answers

Answer 1

Answer: Melanie said:
Every angle bisector in a triangle bisects the opposite side perpendicularly.

A 'counterexample' would show an angle bisector in a triangle that DOESN'T
bisect the opposite side perpendicularly.

See my attached drawing of a counterexample.

Both of the triangles that Melanie examined have equal sides on both sides
of the angle bisector. That's the only way that the angle bisector can bisect
the opposite side perpendicularly. Melanie didn't examine enough different
triangles.

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Use the quadratic formula to solve x^2 + 9x + 10 = 0. Estimate irrational solutions to the nearest tenth. A {–3.6, 16.4}
B {–7.7, –1.3}
C {–25, 16}
D {–15.4, –2.6}

Answers

Basically, just plus your numbers into quadratic formula. A= 1, B= 9, C= 10.
The answer you will get is B.

$x^(2) + 9x + 10 = 0 \nx = \frac{-(9) +/- \sqrt{(9)^(2) - 4(1)(10)}}{2(1)} \nx = (-9 +/- √(81 - 40))/(2) \nx = (-9 +/- √(41))/(2) \nx = (-9 +/- 6.4)/(2) \nx = (-9 + 6.4)/(2) \nx = (-2.6)/(2) \nx = -1.3 \n \nx = (-9 - 6.4)/(2) \nx = (-15.4)/(2) \nx = -7.7$

It is equal to B.

Two landscapers must mow a rectangular lawn that measures 100 feet by 200 feet. Each wants to mow no more than half of the lawn. The first starts by mowing around the outside of the lawn. How wide a strip must the first landscaper mow on each of the four sides in order to mow no more than half of the lawn? The mower has a 24-inch cut. Approximate the required number of trips around the lawn.

Answers

The total area of the complete lawn is (100-ft x 200-ft) = 20,000 ft².
One half of the lawn is 10,000 ft². That's the limit that the first man
must be careful not to exceed, lest he blindly mow a couple of blades
more than his partner does, and become the laughing stock of the whole
company when the word gets around. 10,000 ft² ... no mas !

When you think about it ... massage it and roll it around in your
mind's eye, and then soon give up and make yourself a sketch ...
you realize that if he starts along the length of the field, then with
a 2-ft cut, the lengths of the strips he cuts will line up like this:

First lap:
   (200 - 0) = 200
   (100 - 2) = 98
   (200 - 2) = 198
   (100 - 4) = 96

Second lap:
   (200 - 4) = 196
   (100 - 6) = 94
   (200 - 6) = 194
   (100 - 8) = 92

Third lap:
   (200 - 8) = 192
   (100 - 10) = 90
   (200 - 10) = 190
   (100 - 12) = 88

These are the lengths of each strip. They're 2-ft wide, so the area
of each one is (2 x the length).

I expected to be able to see a pattern developing, but my brain cells
are too fatigued and I don't see it. So I'll just keep going for another
lap, then add up all the areas and see how close he is:

Fourth lap:
   (200 - 12) = 188
   (100 - 14) = 86
   (200 - 14) = 186
   (100 - 16) = 84

So far, after four laps around the yard, the 16 lengths add up to
2,272-ft, for a total area of 4,544-ft². If I kept this up, I'd need to do
at least four more laps ... probably more, because they're getting smaller
all the time, so each lap contributes less area than the last one did.

Hey ! Maybe that's the key to the approximate pattern !

Each lap around the yard mows a 2-ft strip along the length ... twice ...
and a 2-ft strip along the width ... twice. (Approximately.) So the area
that gets mowed around each lap is (2-ft) x (the perimeter of the rectangle),
(approximately), and then the NEXT lap is a rectangle with 4-ft less length
and 4-ft less width.

So now we have rectangles measuring

   (200 x 100), (196 x 96), (192 x 92), (188 x 88), (184 x 84) ... etc.

and the areas of their rectangular strips are
   1200-ft², 1168-ft², 1136-ft², 1104-ft², 1072-ft² ... etc.

==> I see that the areas are decreasing by 32-ft² each lap.
   So the next few laps are
   1040-ft², 1008-ft², 976-ft², 944-ft², 912-ft² ... etc.

How much area do we have now:

   After 9 laps,    Area =   9,648-ft²
   After 10 laps, Area = 10,560-ft².

And there you are ... Somewhere during the 10th lap, he'll need to
stop and call the company surveyor, to come out, measure up, walk
in front of the mower, and put down a yellow chalk-line exactly where
the total becomes 10,000-ft².

There must still be an easier way to do it. For now, however, I'll leave it
there, and go with my answer of: During the 10th lap.

3.30 Measurements of scientific systems are always subject to variation, some more than others. There are many structures for measurement error, and statisticians spend a great deal of time modeling these errors. Suppose the measurement error X of a certain physical quantity is decided by the density function f(x) = k(3 − x2), −1 ≤ x ≤ 1, 0, elsewhere. (a) Determine k that renders f(x) a valid density function. (b) Find the probability that a random error in measurement is less than 1/2. (c) For this particular measurement, it is undesirable if the magnitude of the error (i.e., |x|) exceeds 0.8. What is the probability that this occurs?

Answers

Answer:

a) k should be equal to 3/16 in order for f to be a density function.

b) The probability that the measurement of a random error is less than 1/2 is 0.7734

c) The probability that the magnitude of a random error is more than 0.8 is 0.164

Step-by-step explanation:

a) In order to find k we need to integrate f between -1 and 1 and equalize the result to 1, so that f is a density function.

$1 = k \int\limits^1_(-1) {(3-x^2)} \, dx = k * (3x-(x^3)/(3))|_(x=-1)^(x = 1) = k*[(3-1/3) - (-3 + 1/3)] = 16k/3$

16k/3 = 1

k = 3/16

b) For this probability we have to integrate f between -1 and 0.5 (since f takes the value 0 for lower values than -1)

$P(X < 1/2) = \int\limits^(0.5)_(-1) {(3)/(16)(3-x^2)} \, dx = (3)/(16) [(3x-(x^3)/(3)) |_(x=-1)^(x=0.5)] =(3)/(16) *(1.458333 - (-3+1/3)) = 0.7734$

c) For |x| to be greater than 0.8, either x>0.8 or x < -0.8. We should integrate f between 0.8 and 1, because we want values greater than 0.8, and f is 0 after 1; and between -1 and 0.8.

$P(|X| > 0.8) = \int\limits^(-0.8)_(-1) {(3)/(16)*(3-x^2)} \, dx + \int\limits^(1)_(0.8) {(3)/(16)*(3-x^2)} \, dx =\n (3)/(16) (3x-(x^3)/(3))|_(x=-1)^(x=-0.8) + (3)/(16) (3x-(x^3)/(3))|_(x=0.8)^(x=1) = 0.082 + 0.082 = 0.164$

(a) The value of k that makes f(x) a valid density function is k = 1/6.

(b) The probability that a random error in measurement is less than 1/2 is 3/4.

(a) To make the given function f(x) a valid probability density function, it must satisfy the following conditions:

The function must be non-negative for all x: f(x) ≥ 0.

The total area under the probability density function must equal 1: ∫f(x)dx from -1 to 1 = 1.

Given $f(x) = k(3 - x^2)$ , -1 ≤ x ≤ 1, and f(x) = 0 elsewhere, let's find the value of k that satisfies these conditions.

Non-negativity: The function is non-negative for -1 ≤ x ≤ 1, so we have $k(3 - x^2)$ ≥ 0 for -1 ≤ x ≤ 1. This means that k can be any positive constant.

Total area under the probability density function: To find the value of k, integrate f(x) over the interval [-1, 1] and set it equal to 1:

∫[from -1 to 1] $k(3 - x^2)dx$ = 1

∫[-1, 1] $(3k - kx^2)dx$ = 1

Now, integrate the function:

$[3kx - (kx^3/3)]$ from -1 to 1 = 1

$[(3k(1) - (k(1^3)/3)) - (3k(-1) - (k(-1^3)/3))] = 1$

Simplify:

[3k - k/3 + 3k + k/3] = 1

6k = 1

k = 1/6

So, the value of k that makes f(x) a valid density function is k = 1/6.

(b) To find the probability that a random error in measurement is less than 1/2, you need to calculate the integral of f(x) from -1/2 to 1/2:

P(-1/2 ≤ X ≤ 1/2) = ∫[from -1/2 to 1/2] f(x)dx

P(-1/2 ≤ X ≤ 1/2) = ∫[-1/2, 1/2] (1/6) $(3 - x^2)dx$

Now, integrate the function:

$(1/6) [3x - (x^3/3)]$ from -1/2 to 1/2

$[(1/6)(3(1/2) - ((1/2)^3/3)) - (1/6)(3(-1/2) - ((-1/2)^3/3))]$

Simplify:

(1/6)[(3/2 - 1/24) - (-3/2 + 1/24)]

(1/6)[(9/8) + (9/8)]

(1/6)(18/8)

(3/4)

So, the probability that a randomerror in measurement is less than 1/2 is 3/4.

(c) To find the probability that the magnitude of theerror (|x|) exceeds 0.8, you need to calculate the probability that |X| > 0.8. This is the complement of the probability that |X| ≤ 0.8, which you can calculate as:

P(|X| > 0.8) = 1 - P(|X| ≤ 0.8)

P(|X| > 0.8) = 1 - P(-0.8 ≤ X ≤ 0.8)

We already found P(-0.8 ≤ X ≤ 0.8) in part (b) to be 3/4, so:

P(|X| > 0.8) = 1 - 3/4

P(|X| > 0.8) = 1/4

So, the probability that the magnitude of the error exceeds 0.8 is 1/4.

To Learn more about probability here:

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In the neighborhood park there is a circular grassy picnic area, but half of the grass has died. The city wants to replant the grass, but needs to know the square footage of the damaged area. If the circular picnic area is 150 feet across, what is the square footage of the damaged area? Let π = 3.14.A) 471 sq. feet
B) 8831.25 sq. feet
C) 17662.5 sq. feet
D) 35325 sq. feet

Answers

so, πr^2 is the equation for the area of a circle, if we calculate that, it will equal 17662.5

we then half that to get the area (square feet) of the dead grass, which is
8831.25 square feet which means the answer is B.

150 feet across is diameter
find area
but 1/2 is dead
find area then times 1/2

area=pir^2

150=d
d/2=r=150/2=75=r

area=pi75^2
area=5625pi
times 1/2
area=2812.5pi
pi=3.14 for this problem
area=8831.25

B is answer

The test grades at a large school have an approximately normal distribution with a mean of 50. What is the standard deviation of the data so that 80% of the students are within 12 points (above or below) the mean? Draw a sketch to help reason the procedure and be clever. A.5.875 B.14.5 C.9.375 D.Cannot be determined from the given information. E.10.375

Answers

Answer:

Step-by-step explanation:

The 80% of students within 12 points of the mean are captured by the shaded area, which is two standard deviations away from the mean on either side.

We know that the normal distribution curve is symmetrical, so the two shaded areas have equal areas. Therefore, each shaded area has an area of 40%.

We can use the inverse normal cumulative distribution function (also known as the probit function) to find the z-scores corresponding to the 40th percentile and the 60th percentile. These z-scores are both equal to 1.28.

The z-score is defined as the number of standard deviations away from the mean:

z = (x - μ) / σ

where:

x is the observed value

μ is the mean

σ is the standard deviation

Therefore, the standard deviation is equal to:

σ = (x - μ) / z

From the sketch above, we can see that the x-value corresponding to the 40th percentile is 38 and the x-value corresponding to the 60th percentile is 62.

Substituting these values into the equation above, we can solve for the standard deviation:

σ = (62 - 50) / 1.28 = 9.375

Therefore, the standard deviation of the data so that 80% of the students are within 12 points of the mean is 9.375.

Cleverness:

To solve this problem cleverly, we can use the fact that the normal distribution curve is symmetrical. This means that we can find the standard deviation by calculating the distance between the mean and the 40th percentile, and then multiplying by 2. This is because the 40th percentile is the same distance away from the mean as the 60th percentile.

Final answer:

In a normal distribution with a mean of 50, if 80% of the grades are within 12 points of the mean, the standard deviation can be calculated as approximately 9.375 (option C). We obtained this result by equating the z-score that represents 80% of the data to the ratio of the score range (12 points) to the standard deviation, and solving for the standard deviation.

Explanation:

To determine the standard deviation given 80% of the scores are within 12 points from the mean, we need to understand the properties of a normal distribution curve. In a normal distribution, about 68% of the data falls within one standard deviation of the mean, 95% falls within two standard deviations, and almost all (99.7%) falls within three standard deviations. Now, when we say 80% of the data falls within 12 points of the mean, we are looking at an area of the curve that is equivalent to +/- one standard deviation around the mean.

In a standard normal distribution, an area of approximately 80% correlates to a z-score of about 1.28 (using a z-table or z-score calculator). The z-score is calculated as (X - μ) / σ, where X is the score, μ is the mean and σ is the standard deviation.

In our case, we have X = ±12 (since it's 12 points above and below the mean), and μ = 0 (since we're looking at the difference from the mean). So, 1.28 = 12 / σ. Solving for σ, we get that σ = 12 / 1.28 ≈ 9.375. So the standard deviation, which provides the answer to your question, is 9.375 (option C).

Learn more about Standard Deviation here:

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Two negative integers are 8 units apart on the number line and have a product of 308.Which equation could be used to determine x, the smaller negative integer?

x2 + 8x – 308 = 0
x2 – 8x + 308 = 0
x2 + 8x + 308 = 0
x2 − 8x − 308 = 0

Answers

Calll x the smaller negative number.

The larger number will be x + 8

The product will be x(x+8) = 308

An the equation:

x^2 + 8x = 308
x^2 + 8x - 308 =0

If you solve that equation you will get x = -22 and x = 14.

The result that you are looking is the negative, x = -22.

The larger number is -22 + 8 = -14.

(-22)*(-14) = 308

Answer: x^2 + 8x - 308 =0

that answer is letter A on E2020

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