Ask question

Ask question

All categories

2 years ago

9

The distribution of the values of a population is shown below, and a simple Random sample is drawn from the population. Based on

this information alone, can the same will be used to make inferences about the population?

1 answer:

posledela2 years ago

6 0

Answer: C

Step-by-step explanation:

Yes, because the population values appear to be normally distributed

You might be interested in

2r+c+0.5d+8 how many terms does the expression have

Lemur [1.5K]

A <em>term</em> in a math expression is <u>single </u>number or variable and are <u><em>separated </em></u>by a + or - or × or ÷

This expression has 4 terms

2r

1c

.5d

and 8

8 0

2 years ago

5-6x=2+7x step by step plx?

Ivanshal [37]

5-6x=2+7x

+6x +6x

5=2+13x

-2 -2

3=13x

3/3= 13/3

x=4.33

3 0

2 years ago

I'd like some help, thanks!

Otrada [13]

It is likely D I am not positive

7 0

3 years ago

Read 2 more answers

Please help with this I am completely stuck on it

vaieri [72.5K]

Answer:

$f(x)=\sqrt[3]{x-4} , g(x)=6x^{2}\textrm{ or }f(x)=\sqrt[3]{x},g(x)=6x^{2} -4$

Step-by-step explanation:

Given:

The function, $H(x)=\sqrt[3]{6x^{2}-4}$

Solution 1:

Let $f(x)=\sqrt[3]{x}$

If $f(g(x))=H(x)=\sqrt[3]{6x^{2}-4}$ , then,

$\sqrt[3]{g(x)} =\sqrt[3]{6x^{2}-4}\\g(x)=6x^{2}-4$

Solution 2:

Let $f(x)=\sqrt[3]{x-4}$ . Then,

$f(g(x))=H(x)=\sqrt[3]{6x^{2}-4}\\\sqrt[3]{g(x)-4}=\sqrt[3]{6x^{2}-4} \\g(x)-4=6x^{2}-4\\g(x)=6x^{2}$

Similarly, there can be many solutions.

7 0

3 years ago

Find the sum of the geometric series 40 + 40(1.005) + 40(1.005)^2 + ⋯ + 40(1.005)^11.

KiRa [710]

Answer:

The sum is 493.4

Step-by-step explanation:

In order to find the value of the sum, you have to apply the geometric series formula, which is:

$\sum_{i=1}^{n} ar^{i-1} = \frac{a(1-r^{n})}{1-r}$

where i is the starting point, n is the number of terms, a is the first term and r is the common ratio.

The finite geometric series converges to the expression in the right side of the equation. Therefore, you don't need to calculate all the terms. You can use the expression directly.

In this case:

a=40

b= 1.005

n=12 (because the first term is 40 and the last term is 40(1.005)^11 )

Replacing in the formula:

$\frac{a(1-r^{n})}{1-r} = \frac{40(1-1.005^{12})}{1-1.005}$

Solving it:

The sum is 493.4

4 0

2 years ago