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Mashutka [201]
3 years ago
12

Which best describes the complement of spinning any number less than 3?

Mathematics
1 answer:
tatuchka [14]3 years ago
8 0

Answer:

  • <u>The complement of spinning any number less than 3, is spinning a number equal to or greater than 3.</u>

Explanation:

The complement of a subset is the subset of elements that are not in the given subset.

You must know which numbers the spinner has.

Assuming the spinner has the numbers 1, 2, 3, 4, the complement of spinning any number less than 3, is spinning a number that is not less than 3.

Then, that is spinning a number that is equal to or greater than 3.

The numbers that are equal to  or greater than four, for a spinner that has the numbers 1, 2, 3, and 4 are 3 and 4.

Thus, the complement of spinning any number less than 3 is spinning a three or a four.

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What is Statistical inference?
rjkz [21]

Answer:

<em>Definition 1: The theory, methods, and practice of forming judgments about the parameters of a population and the reliability of statistical relationships, typically on the basis of random sampling.</em>

<em>Definition 2: The use of randomization in sampling allows for the analysis of results using the methods of statistical inference. Statistical inference is based on the laws of probability, and allows analysts to infer conclusions about a given population based on results observed through random sampling. Two of the key terms in statistical inference are parameter and statistic.</em>

Step-by-step explanation:

Hope this helps, have a good day. c;

4 0
3 years ago
Read 2 more answers
What is the rule in the input/output table? A) add 4 B) subtract 4 C) divide by 2 D) multiply by 2 | inputs | 4 6 and 8 | output
MissTica

The rule in the input/out is D multiply by 2

4 0
3 years ago
Read 2 more answers
The equation 3x2 = 6x – 9 has two real solutions<br> True<br> O False
devlian [24]

Answer:

False

Step-by-step explanation:

We first write the equation in the form ax² + bx + c=0 which gives us:

3x² - 6x + 9=0

Given the quadratic formula,

x= [-b ±√(b²- 4ac)]/2a ,the discriminant proves whether the equation has real roots or not.

The discriminant, which is the value under the root sign, may either be positive, negative or zero.

Positive discriminant- the equation has two real roots

Negative discriminant- the equation has no real roots

Zero discriminant - The equation has two repeated roots.

In the provided equation, b²-4ac results into:

(-6)²- (4×3×9)

=36-108

= -72

The result is negative therefore the equation has no real solutions.

3 0
3 years ago
Read 2 more answers
A population of 55 foxes in a wildlife preserve doubles in size every 14 years. The function y equals 55 times 2 Superscript x ​
Pavlova-9 [17]
Since <em>x</em> is the number of 14-year periods, the first thing we do is divide 28 years by 14. 28/14=2.  Now we use that in our function, y=55*2^x.

y=55*2^2=55*4=220
There will be 220 foxes after 28 years.
6 0
3 years ago
Let $f(x) = x^2$ and $g(x) = \sqrt{x}$. Find the area bounded by $f(x)$ and $g(x).$
Anna [14]

Answer:

\large\boxed{1\dfrac{1}{3}\ u^2}

Step-by-step explanation:

Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).

Let's calculate the common points:

x^2=\sqrt{x}\qquad\text{square of both sides}\\\\(x^2)^2=\left(\sqrt{x}\right)^2\\\\x^4=x\qquad\text{subtract}\ x\ \text{from both sides}\\\\x^4-x=0\qquad\text{distribute}\\\\x(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\\\\x^3-1=0\qquad\text{add 1 to both sides}\\\\x^3=1\to x=\sqrt[3]1\to x=1

The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.

We have integrals:

\int\limits_{0}^1(\sqrt{x})dx-\int\limits_{0}^1(x^2)dx=(*)\\\\\int(\sqrt{x})dx=\int\left(x^\frac{1}{2}\right)dx=\dfrac{2}{3}x^\frac{3}{2}=\dfrac{2x\sqrt{x}}{3}\\\\\int(x^2)dx=\dfrac{1}{3}x^3\\\\(*)=\left(\dfrac{2x\sqrt{x}}{2}\right]^1_0-\left(\dfrac{1}{3}x^3\right]^1_0=\dfrac{2(1)\sqrt{1}}{2}-\dfrac{2(0)\sqrt{0}}{2}-\left(\dfrac{1}{3}(1)^3-\dfrac{1}{3}(0)^3\right)\\\\=\dfrac{2(1)(1)}{2}-\dfrac{2(0)(0)}{2}-\dfrac{1}{3}(1)}+\dfrac{1}{3}(0)=2-0-\dfrac{1}{3}+0=1\dfrac{1}{3}

6 0
3 years ago
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