A parrot is selling for 40% off the original price. the original price was $90.what is these price of the parrot

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Answer 1
Answer: 40 percent off of $90 is $36.


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Consider triangle QRS. The legs each have a length of 10 units. Triangle Q R S is shown. Angle Q S R is 90 degrees. The length of Q X is 10 and the length of S R is 10. What is the length of the hypotenuse of the triangle? 5 units 5 StartRoot 2 EndRoot units 10 units 10 StartRoot 2 EndRoot units
2 times the sum of a number and 8 is equal to the difference of 10 and that number find the number

How many Times greater is the value of 3 in 300 then the value of the 3 in 300

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Simple division:

300/3 = 100.

30/3 = 10.

100/10=10

Therefore the answer is 10.

Solve this equation:
5/8y=-1

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(5)/(8) y = -1

First, change (5)/(8) y to (5y)/(8) / Your problem should look like: (5y)/(8) = -1
Second, multiply both sides by 8. / Your problem should look like: 5y = -8
Third, divide both sides by 5. / Your problem should look like: y =  -(8)/(5)

Answer as fraction: y = - (8)/(5)
Answer as decimal: -1.6



5/8y=−1

Multiply both sides by 8/5.

(8/5)*(5/8y)=(8/5)*(−1)

y=−8/5

Greg is in a car at the top of a roller-coaster ride. The distance, d, of the car from the ground as the car descends is determined by the equation d = 144 – 16t2, where t is the number of seconds it takes the car to travel down to each point on the ride. For which interval of time is Greg’s car moving in the air?

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Answer:

It takes 3 seconds over the interval [0,3]

Step-by-step explanation:

To find when the roller coaster reaches the ground, find when d=0.

0=144-16t^2

To solve divide each term by 16 and factor:

(0)/(16)=(144)/(16) -(-16t^2)/(16)  \n0= 9 - t^2\n0=(3-t)(3+t)

Solve for t by setting each factor to 0.

t-3=0 so t=3

t+3=0 so t=-3

This means the car is in the air from 0 to 3 second.


The time interval will be 0<t<3  or (0, 3)

What is interval?

A interval is a set of real numbers that contains all real numbers lying between any two numbers of the set.

Given:

d = 144 – 16t²

Now,

d>0

144 – 16t²>0

144>16t²

144/16>t²

9>t²

t<±3

Since, timecan't be negative.

So, t<3

Hence, the interval for which Greg’s car moving in the air is,  0<t<3  or (0, 3).

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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.

(c) The probability that the magnitude of the error exceeds 0.8 is 1/4.

(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.

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How many acute angles does a octagon have?

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A regular octagon has zero acute angles. an acute angle has a measure of less than 90°. A regular octagon has a total angle measurement of 1080°, so each angle has a measurement of 135° (Since 1080 / 8 = 135), which is greater than 90°. Therefore, the answer to your question is an octagon has 0 acute angles.
Hope that helped =)

In summary, a regular octagon does not have any acute angles, while an irregular octagon can have some acute angles depending on the specific measurements of its angles.

An octagon is a polygon with eight sides. To determine the number of acute angles in an octagon, we need to understand that an acute angle is an angle that measures less than 90 degrees.

In a regular octagon (where all sides and angles are equal), each interior angle measures 135 degrees. This is because the sum of the interior angles in any octagon is given by the formula (n-2) x 180 degrees, where n is the number of sides.

In this case, (8-2) x 180 = 1080 degrees. Since all angles in a regular octagon are equal, we divide 1080 by 8 to find that each angle measures 135 degrees.

Since an acute angle measures less than 90 degrees, a regular octagon does not have any acute angles.

However, if we consider an irregular octagon (where sides and angles can have different measures), it is possible for some of the angles to be acute. The number of acute angles in an irregular octagon would depend on the specific measurements of the angles.

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Which of equation represents a function? A. x2 + y2 = 9 B. {(4, 2), (4, –2), (9, 3), (9, –3)} C. x = 4. 2x + y = 5

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The answer to your questions is B. {(4, 2), (4, –2), (9, 3), (9, –3)} because you don't have a repeat of a number the X side which is your output. I hope that this is the answer that you were looking for and it has helped you.