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3 years ago

What is the domain of f(x)= 3squareroot x?

Mathematics

2 answers:

Vinvika [58]3 years ago

7 0

Answer: Domain: x ≥ 0 Interval Notation: x = [0, ∞)

Step-by-step explanation:

f(x) = 3√x

There IS a restriction on x because it is inside the square root symbol.

In order for the radical to be a real number, the expression under the radical must be greater than or equal to 0. The value inside the radical is "x".

So, x ≥ 0 It is already solved for x so no further work is needed.

Andru [333]3 years ago

5 0

Answer:

x is greater than or equal to 0

Step-by-step explanation:

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3 years ago

Is the square root of 10 greater less than or equal to 5

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3 years ago

What number is 432 divisible by?

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3 years ago

The amount of coffee that people drink per day is normally distributed with a mean of 17 ounces and a standard deviation of 4 ou

Setler [38]

Answer:

Step-by-step explanation:

(a)

The distribution of X is Normal Distribution with mean $= \mu =17$ and Variance $= \sigma^{2} = 16 \ i.e., X \sim N (17, 16),$

(b)

The distribution of $\bar{x}$ is Normal Distribution with mean $= \mu =17$ and Variance = $\sigma^{2}/n = 16/15= 1.0667$ .i.e., $\bar{x}\sim N(17,1.0667)$

To find P(15.5 < X < 18):

Case 1: For X from 15.5 to mid value:

Z = (15.5 - 17)/4 = - 0.375

Table of Area Under Standard Normal Curve gives area = 0.1480

Case 2: For X from mid value to 18:

Z = (18 - 17)/4 = 0.25

Table of Area Under Standard Normal Curve gives area = 0.0987

So,

P(15.5 < X< 18) = 0.1480 +0.0987 = 0.2467

So,

Answer is:

0.2467

(d)

$SE = \sigma/\sqrt{n}\\\\= 4/\sqrt{15}$

= 1.0328

To find $P(15.5 < \bar{x}< 18):$

Case 1: For $\bar{x}$ from 15.5 to mid value:

Z = (15.5 - 17)/1.0328 = - 1.4524

Table of Area Under Standard Normal Curve gives area = 0.4265

Case 2: For X from mid value to 18:

Z = (18 - 17)/1.0328 = 0.9682

Table of Area Under Standard Normal Curve gives area = 0.3340

So,

$P(15.5 < \bar{x}< 18) = 0.4265 + 0.3340 = 0.7605$

So,

Answer is:

0.7605

(e)

Correct option:

because Population SD is provided.

(f)

(i)

Q1 is given by:

$- 0.6745 = (\bar{x} - 17)/1.0328$

So,

X = 17 - (0.6745 * 1.0328) = 17 - 0.6966 = 16.3034

So,

Q1 = 16.3034

(ii)

Q3 is given by:

$0.6745 = (\bar{x} - 17)/1.0328$

So,

X = 17 + (0.6745 * 1.0328) = 17 + 0.6966 = 17.6966

So,

Q3= 17.6966

(iii)

IQR = Q3 - Q1 = 17.6966 - 16.3034 = 1.3932

Answers are:

Q1 = 16.3034 ounces

Q3 = 17.6966 Ounces

IQR = 1.3932 Ounces

8 0

3 years ago