Which point on the number line best represents √10

Complete the statement with <, >, or =.<br> 5/5 ____ 1

stiks02 [169]

The answer is =
Any number divided by itself = 1

4 0

4 years ago

The domain of u(x) is the set of all real values except 0 and the domain of v(x) is the set of all real values except 2. What

Oksana_A [137]

Using the domain concept, the restrictions on the domain of (u.v)(x) are given by:

A. u(x) ≠ 0 and v(x) ≠ 2.

<h3>What is the domain of a data-set?</h3>

The domain of a data-set is the set that contains all possible input values for the data-set.

To calculate u(x) x v(x) = (u.v)(x), we calculate the values of u and v and then multiply them, hence the restrictions for each have to be considered, which means that statement A is correct.

Summarizing, u cannot be calculated at x = 0, v cannot be calculated at x = 2, hence uv cannot be calculated for either x = 0 and x = 2.

More can be learned about the domain of a data-set at brainly.com/question/24374080

#SPJ1

3 0

3 years ago

Construct a 99% confidence interval for the population mean, mu. Assume the population has a normal distribution. A group of 1

Zarrin [17]

Answer:

99% confidence interval for the population mean is [19.891 , 24.909].

Step-by-step explanation:

We are given that a group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.

Assuming the population has a normal distribution.

Firstly, the pivotal quantity for 99% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean age of selected students = 22.4 years

s = sample standard deviation = 3.8 years

n = sample of students = 19

$\mu$ = population mean

<em>Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.</em>

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.878 < $t_1_8$ < 2.878) = 0.99 {As the critical value of t at 18 degree of

freedom are -2.878 & 2.878 with P = 0.5%}

P(-2.878 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 2.878) = 0.99

P( $-2.878 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.878 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ < $\mu$ < $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ) = 0.99

<u>99% confidence interval for</u> $\mu$ = [ $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ , $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ]

= [ $22.4 -2.878 \times {\frac{3.8}{\sqrt{19} }$ , $22.4 +2.878 \times {\frac{3.8}{\sqrt{19} }$ ]

= [19.891 , 24.909]

Therefore, 99% confidence interval for the population mean is [19.891 , 24.909].

6 0

4 years ago

What is the value of the expression below when y 8 and z = 8? 9y + 3z

aleksley [76]

Answer:

96

Step-by-step explanation:

8y+3z

9(8)+3(8)

72+24

96

6 0

3 years ago

Week 1 2 3 4 5 Water Level (inches) − 1 1 4 1 3 8 2 1 2 − 1 5 8 − 1 3 4 Between which two weeks did the water level change the m

suter [353]

Answer:

The water level change the most in week 3 and week 4

The change = -370 inches

Step-by-step explanation:

Given - Week 1 2 3 4 5

Water Level (inches) − 1 1 4 1 3 8 2 1 2 -1 5 8 -1 3 4

To find - Between which two weeks did the water level change the most? Calculate the change .

Proof -

The formula for change in water level with respect to week be-

Rate of Change in Water Level = Difference in water level / Difference in weeks

Now,

For week 1 and week 2

Rate of change in water level = $\frac{138 - 114}{2 - 1}$ = $\frac{24}{1}$ = 24 inches

Now,

For week 1 and week 3

Rate of change in water level = $\frac{212 - 114}{3 - 1}$ = $\frac{98}{2}$ = 49 inches

Now,

For week 1 and week 4

Rate of change in water level = $\frac{-158 - 114}{4 - 1}$ = $-\frac{272}{3}$ = -90.67 inches

Now,

For week 1 and week 5

Rate of change in water level = $\frac{-134 - 114}{5 - 1}$ = $-\frac{248}{4}$ = -62 inches

Now,

For week 2 and week 3

Rate of change in water level = $\frac{212 - 138}{3 - 2}$ = $\frac{74}{1}$ = 74 inches

Now,

For week 2 and week 4

Rate of change in water level = $\frac{-158 - 138}{4 - 2}$ = $-\frac{296}{2}$ = -148 inches

Now,

For week 2 and week 5

Rate of change in water level = $\frac{-134 - 138}{5 - 2}$ = $-\frac{272}{3}$ = -90.67 inches

Now,

For week 3 and week 4

Rate of change in water level = $\frac{-158 - 212}{4 - 3}$ = $-\frac{370}{1}$ = -370 inches

Now,

For week 3 and week 5

Rate of change in water level = $\frac{-134 - 212}{5 - 3}$ = $-\frac{346}{2}$ = 173 inches

Now,

For week 4 and week 5

Rate of change in water level = $\frac{-134 + 158}{5 - 4}$ = $\frac{24}{1}$ = 24 inches

∴ we get

The water level change the most in week 3 and week 4

The change = -370 inches

Note :

The highest change means which temperature goes the most change in water level. It can be negative also.

So, We have to take the modulus and then find the highest number.

Here, 370 > 173

So, Highest change occurs in Week 3 and Week 4 but not in Week 3 and Week 5.

7 0

3 years ago