The expression 100n2 – 1 is equivalent to

Consider the following ordered data. 6 9 9 10 11 11 12 13 14 (a) Find the low, Q1, median, Q3, and high. low Q1 median Q3 high (

IrinaVladis [17]

Answer:

Low Q1 Median Q3 High

6 9 11 12.5 14

The interquartile range = 3.5

Step-by-step explanation:

Given that:

Consider the following ordered data. 6 9 9 10 11 11 12 13 14

From the above dataset, the highest value = 14 and the lowest value = 6

The median is the middle number = 11

For Q1, i.e the median of the lower half

we have the ordered data = 6, 9, 9, 10

here , we have to values as the middle number , n order to determine the median, the mean will be the mean average of the two middle numbers.

i.e

median = $\dfrac{9+9}{2}$

median = $\dfrac{18}{2}$

median = 9

Q3, i.e median of the upper half

we have the ordered data = 11 12 13 14

The same use case is applicable here.

Median = $\dfrac{12+13}{2}$

Median = $\dfrac{25}{2}$

Median = 12.5

Low Q1 Median Q3 High

6 9 11 12.5 14

The interquartile range = Q3 - Q1

The interquartile range = 12.5 - 9

The interquartile range = 3.5

7 0

3 years ago

Task: Nonpermissible Values for Rational Expressions [30 points] Write a rational expression that has the nonpermissible values

igor_vitrenko [27]

Answer:

A rational expression that has the nonpermissible values $x = 0$ and $x = 17$ is $f(x) = \frac{4}{x\cdot (x-17)}$ .

Step-by-step explanation:

A rational expression has a nonpermissible value when for a given value of $x$ , the denominator is equal to zero. In addition, we assume that both numerator and denominator are represented by polynomials, such that:

$f(x) = \frac{p(x)}{q(x)}$ (1)

Then, the factorized form of $q(x)$ must be:

$q(x) = x\cdot (x-17)$ (2)

If we know that $p(x) = 4$ , then the rational expression is:

$f(x) = \frac{4}{x\cdot (x-17)}$ (3)

A rational expression that has the nonpermissible values $x = 0$ and $x = 17$ is $f(x) = \frac{4}{x\cdot (x-17)}$ .

8 0

3 years ago

The probability of event A is x, and the probability of event B is y. If the two events are independent, which of these conditio

zzz [600]

If two events are independent, then

$Pr(A\cap B)=Pr(A)\cdot Pr(B)$ .

Use formulas for conditional probabilities:

$Pr(A|B)=\dfrac{Pr(A\cap B)}{Pr(B)},\\ \\ Pr(B|A)=\dfrac{Pr(A\cap B)}{Pr(A)}$ .

For independent events these formulas will be:

$Pr(A|B)=\dfrac{Pr(A\cap B)}{Pr(B)}=\dfrac{Pr(A)\cdot Pr(B)}{Pr(B)}=Pr(A),\\ \\ Pr(B|A)=\dfrac{Pr(A\cap B)}{Pr(A)}=\dfrac{Pr(A)\cdot Pr(B)}{Pr(A)}=Pr(B)$ .

Now in your case $Pr(A)=x,\ Pr(B)=y$ and $Pr(A|B)=x,\ Pr(B|A)=y, Pr(A\cap B)=x\cdot y$ .

This shows that the only correct choice is A.

5 0

3 years ago

Read 2 more answers

The mean of 9 observations was found to be 35. Later on, it was

USPshnik [31]

Answer:

its 42

Step-by-step explanation:

315-18+81=378

Correct mean =378/9=42

3 0

2 years ago

Ayanna plans to bake 3 types of cookies to take to a party. She needs to be sure that she has enough flour. The chocolate cookie

djyliett [7]

The children brought 2+1+114=414 cups of flour and 14+12+34=112 cups of butter.

They have enough flour for

414÷34===174×43173523
batches and they have enough butter for

112÷13===32×3192412
batches, so the butter is the limiting factor. Thus, they can make 4 whole batches of a dozen cookies each.

7 0

2 years ago

Read 2 more answers