All categories

3 years ago

Find the interquartile range using the following data 4 ,4 , 4 ,4 ,5, 6 ,7 , 7 ,7

Mathematics

1 answer:

katen-ka-za [31]3 years ago

7 0

Answer:

Step-by-step explanation:

Ascending order:

4 ,4 , 4 ,4 ,5, 6 ,7 , 7 ,7

9 data: split in

4 ,4 , 4 ,4 | 5 | 6 ,7 , 7 ,7

4 ,4 | 4 ,4 | 5 | 6 ,7 | 7 ,7

(7+7)/2-(4+4)/2

=7-4

You might be interested in

Can you tell me if these are correct, fix them if there not, and answer this one in the form of

bekas [8.4K]

Answer: Here’s some help

Step-by-step explanation:

3 0

2 years ago

Weldon installed a square hot tub in his backyard and wants to enclose it with a fence. The fence will enclose an area that is 4

Katarina [22]

Answer:

fdn

Step-by-step explanation:

sergsrg

6 0

3 years ago

What is the degree of differential equation? ?

Delicious77 [7]

Answer:

Degree = 1

Step-by-step explanation:

Given:

The differential equation is given as:

$\frac{d^2y}{dx^2}+(\frac{dy}{dx})^2+6y=0$

The given differential equation is of the order 2 as the derivative is done 2 times as evident from the first term of the differential equation.

The degree of a differential equation is the exponent of the term which is the order of the differential equation. The terms which represents the differential equation must satisfy the following points:

They must be free from fractional terms.

Shouldn't have derivatives in any fraction.

The highest order term shouldn't be exponential, logarithmic or trigonometric function.

The above differential equation doesn't involve any of the above conditions. The exponent to which the first term is raised is 1.

Therefore, the degree of the given differential equation is 1.

6 0

3 years ago

16051 rounded to the nearest thousand

mr Goodwill [35]

16000 is the answer.

6 0

3 years ago

Read 2 more answers

For what values of x does the curve y^2x^3-15x^2=4 have horizontal tangent lines

enyata [817]

We have the following curve:<span>

$y^2x^3-15x^2=4$

To find the tangent lines to the curve we need to use the concept of derivative, but we can’t solve this problem for $y$ , thus, let's apply implicit differentiation, so:

$\frac{d}{dx}(y^2x^3-15x^2=4) \\ \\
\frac{d}{dx}(y^2x^3)-\frac{d}{dx}(15x^2)=\frac{d}{dx}(4) \\ \\
(y^{2})'x^3+(x^3)'y^2-15(x^2)'=0 \\ \\ 2yy'x^3+3x^2y^2-30x=0 \\ \\
y'=\frac{dy}{dx}=\frac{30x-3x^2y^2}{2x^3y}=\frac{x(30-3xy^2)}{2x^3y} \\ \\
y'=\frac{30-3xy^2}{2x^2y}$

Therefore the horizontal lines occurs when $y'=0$ , then:

$\frac{30-3xy^2}{2x^2y}=0 \\ \\ That \ is, \
when: \\ \\ 30-3xy^2=0 \\ \\ \therefore xy^2=10 \rightarrow y^2=\frac{10}{x}$

If we substitute this in the original equation we have:

$\frac{10}{x}x^3-15x^2=4 \\ \\ \therefore
10x^2-15x^2=4 \\ \\ \therefore x^2=-\frac{4}{5}$

This is an absurd result because it is impossible for a squared number to get a negative number. So the conclusion is that there is no any value of $x$ in which the curve has horizontal tangent lines.</span>

8 0

3 years ago