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3 years ago

How do I find the Q1, Q3, and IQR of a median that is a decimal?

1 answer:

3 0

Step 1: Put the numbers in order.
1,2,5,6,7,9,12,15,18,19,27Step 2: Find the median (How to find a median).
1,2,5,6,7,9,12,15,18,19,27Step 3: Place parentheses around the numbers above and below the median.
Not necessary statistically–but it makes Q1 and Q3 easier to spot.
(1,2,5,6,7),9,(12,15,18,19,27)Step 4: Find Q1 and Q3
Q1 can be thought of as a median in the lower half of the data. Q3 can be thought of as a median for the upper half of data.
(1,2,5,6,7), 9, ( 12,15,18,19,27). Q1=5 and Q3=18.Step 5: Subtract Q1 from Q3 to find the interquartile range.
18-5=13.
Hope this helped:)
-BRIEMODEE:)

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prisoha [69]

<h3>Given:</h3>

h= 19 ft
r= 12 ft

h= Height
r= Radius

The volume of the given cone.

<h3>Solution:</h3>

$\large\boxed{Formula:V= \frac{1}{3}\pi{r}^{2}h}$

$\large\boxed{\red \pi \red = \red 3 \red . \red 1 \red 4}$

Let's solve!

Substitute the values according to the formula.

$V=\frac{1}{3}×3.14×{12}^{2}×19$

$\large\boxed{V=2863.68 \: {ft}^{3}}$

Therefore, the volume of the given cone is 2863.68 cubic feets.

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2 years ago

Find the size of the final unknown exterior angle in a polygon whose other exterior angles are:

soldi70 [24.7K]

Answer:

=72

Step-by-step explanation:

57+68+32+59+72=288

360-288=72

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3 years ago

PLEASE HELP (30 POINTS) Solve the rational equation x/3 = x^2/x + 5 , and check for extraneous solutions.

anygoal [31]

Option C: $x=0$ and $x=\frac{5}{2}$ are the solutions.

Explanation:

The equation is $\frac{x}{3} =\frac{x^{2} }{x+5}$

We shall determine the value of x, by simplifying the equation.

$\begin{aligned} x(x+5) &=3 x^{2} \\ x^{2}+5 x &=3 x^{2} \\ 2 x^{2}-5 x &=0 \\ x(2 x-5) &=0 \end{aligned}$

Thus, $x=0$ and $x=\frac{5}{2}$ are the solutions.

Now, let us check whether the solutions are extraneous solutions.

Let us substitute $x=0$ in the original equation to check whether both sides of the equation are equal.

$\begin{aligned}&\frac{0}{3}=\frac{0^{2}}{0+5}\\&0=\frac{0}{5}\\&0=0\end{aligned}$

Thus, both sides of the equation are equal.

Hence $x=0$ is a true solution.

Now, Let us substitute $x=\frac{5}{2}$ in the original equation to check whether both sides of the equation are equal.

$\begin{aligned}\frac{\left(\frac{5}{2}\right)}{3} &=\frac{\left(\frac{5}{2}\right)^{2}}{\left(\frac{5}{2}\right)+5} \\\frac{5}{6} &=\frac{\left(\frac{25}{4}\right)}{\left(\frac{15}{2}\right)} \\\frac{5}{6} &=\frac{5}{6}\end{aligned}$

Thus, both sides of the equation are equal.

Hence, $x=\frac{5}{2}$ is a true solution.

Thus, solutions are not extraneous.

Hence, Option C is the correct answer.

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3 years ago

If 14 songs were 9.38 dollars how much would one be

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Answer:

each song would be 67 cents (0.67)

Step-by-step explanation:

9.38/14=0.67

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2 years ago