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3 years ago

Determine whether the following statement is true or false.

Mathematics

1 answer:

shutvik [7]3 years ago

8 0

Answer:

Step-by-step explanation:

In constructing a confidence interval about the mean, the central limit theorem is usually applied. This makes it possible to use the normal distribution. As the number of samples is increasing, the distribution tends to be normal. This would require using the z distribution. In the case where the sample size is small, we assume a normal distribution and use the t distribution. Therefore, the given statement is true.

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The length of a square is 4x-3. Find the area in terms of x

prisoha [69]

Answer:

The area of square in terms of x = 16x² -24x + 9

Step-by-step explanation:

It is stated in the question that the length of a square = 4x - 3

Formula to find the area of the square is

<h2 /><h2>Area = length²</h2><h2 />

Area = ( 4x - 3 )²

<em>by using ( a - b )² = a² + b² - 2ab</em>

here a = 4x and b = -3

So,

(4x)² + (-3)² - 2(4x)(3)

16x² + 9 - 2(12x)

16x² + 9 - 24x

16x² -24x + 9

The area of square in terms of x = 16x² -24x + 9

6 0

3 years ago

I need help with this ?

GenaCL600 [577]

1)Shade 56 of the cubes
2)0.056

4 0

3 years ago

Please help with the attached question. Thanks

Mademuasel [1]

Answer:

Choice A) $F(x) = 3\sqrt{x + 1}$ .

Step-by-step explanation:

What are the changes that would bring $G(x)$ to $F(x)$ ?

Translate $G(x)$ to the left by $1$ unit, and
Stretch $G(x)$ vertically (by a factor greater than $1$ .)

$G(x) = \sqrt{x}$ . The choices of $F(x)$ listed here are related to $G(x)$ :

Choice A) $F(x) = 3\;G(x+1)$ ;
Choice B) $F(x) = 3\;G(x-1)$ ;
Choice C) $F(x) = -3\;G(x+1)$ ;
Choice D) $F(x) = -3\;G(x-1)$ .

The expression in the braces (for example $x$ as in $G(x)$ ) is the independent variable.

To shift a function on a cartesian plane to the left by $a$ units, add $a$ to its independent variable. Think about how $(x-a)$ , which is to the left of $x$ , will yield the same function value.

Conversely, to shift a function on a cartesian plane to the right by $a$ units, subtract $a$ from its independent variable.

For example, $G(x+1)$ is $1$ unit to the left of $G(x)$ . Conversely, $G(x-1)$ is $1$ unit to the right of $G(x)$ . The new function is to the left of $G(x)$ . Meaning that $F(x)$ should should add $1$ to (rather than subtract $1$ from) the independent variable of $G(x)$ . That rules out choice B) and D).

Multiplying a function by a number that is greater than one will stretch its graph vertically.
Multiplying a function by a number that is between zero and one will compress its graph vertically.
Multiplying a function by a number that is between $-1$ and zero will flip its graph about the $x$ -axis. Doing so will also compress the graph vertically.
Multiplying a function by a number that is less than $-1$ will flip its graph about the $x$ -axis. Doing so will also stretch the graph vertically.

The graph of $G(x)$ is stretched vertically. However, similarly to the graph of this graph $G(x)$ , the graph of $F(x)$ increases as $x$ increases. In other words, the graph of $G(x)$ isn't flipped about the $x$ -axis. $G(x)$ should have been multiplied by a number that is greater than one. That rules out choice C) and D).

Overall, only choice A) meets the requirements.

Since the plot in the question also came with a couple of gridlines, see if the points $(x, y)$ 's that are on the graph of $F(x)$ fit into the expression $y = F(x) = 3\sqrt{x - 1}$ .