A restaurant listed its prices for a medium pizza based on the number of toppings in the table below

Saul decides to use the IQR to measure the spread of the data. Saul calculates the IQR of the data set to be 27

german

Answer:

n = 78

Step-by-step explanation:

IQR is the interquartile range of the data set.

So Saul had already arranged the data in order, that is the first step when calculating the interquartile range.

25, 30, 50, 50, 50, 50, 56, n., 250

Next, is to find the median so we could easily separate the data into quartiles.

The median of the data 25, 30, 50, 50, 50, 50, 56, n., 250 = 50

Then, arrange the data into the first quartile and the third quartile

first quartile = 25, 30, 50, 50

third quartile = 50, 56, n., 250

first quartile median = 30 + 50 = 80/2 = 40

third quartile = 56 + n /2

The interquartile range formula is the first quartile subtracted from the third quartile:

IQR = Q3 – Q1.

Since 27 is the interquartile range, n=

27 = (56 + n /2) - 40

Use 2 to multiply through to eliminate the denominator

54 = (56 + n) - 80

n = 78

5 0

3 years ago

Evaluate the integral of the quantity x divided by the quantity x to the fourth plus sixteen, dx . (2 points) one eighth times t

Anika [276]

Answer:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

Step-by-step explanation:

Given

$\int\limits {\frac{x}{x^4 + 16}} \, dx$

Required

Solve

Let

$u = \frac{x^2}{4}$

Differentiate

$du = 2 * \frac{x^{2-1}}{4}\ dx$

$du = 2 * \frac{x}{4}\ dx$

$du = \frac{x}{2}\ dx$

Make dx the subject

$dx = \frac{2}{x}\ du$

The given integral becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{x}{x^4 + 16}} \, * \frac{2}{x}\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{1}{x^4 + 16}} \, * \frac{2}{1}\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

Recall that: $u = \frac{x^2}{4}$

Make $x^2$ the subject

$x^2= 4u$

Square both sides

$x^4= (4u)^2$

$x^4= 16u^2$

Substitute $16u^2$ for $x^4$ in $\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16u^2 + 16}} \,\ du$

Simplify

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16}* \frac{1}{8u^2 + 8}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{2}{16}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

In standard integration

$\int\limits {\frac{1}{u^2 + 1}} \,\ du = arctan(u)$

So, the expression becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(u)$

Recall that: $u = \frac{x^2}{4}$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

4 0

3 years ago

Solve for x and y 9/x-4/y=8

borishaifa [10]

You can solve 9/x-4/y=8 for x or for y but NOT at the same time.

Solving for x: 9/x-4/y=8 Mult all 3 terms by xy to eliminate the fractions.

9(xy)/x - 4xy/y = 8xy => 9y - 4x = 8xy, or 9y = 4x + 8xy = 4x(1+2y)

then 9y = x [ 4(1+2y) ]
9y
therefore x = ------------
4(1+2y)

Solve for y using a similar approach.

7 0

3 years ago

Find the area of the following shape. (8 points) what is the answer?

Ivenika [448]

Answer:

72 square units

Step-by-step explanation:

Identify the height
Identify the length
Multiply those two together
That is your answer

H = 9 units

L = 8 units

A = H × L

A = 9 × 8

A = 72 square units

8 0

2 years ago

Why there must be at least two lines on any given plane

Elanso [62]

Part of tue reason is to balance the plane so you arent flying sideways.. :D not many happy people on a sideways plane ;) I am an idiot haha oh welp go ahead and delete it haha

3 0

3 years ago

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