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3 years ago

Please Help!!! Calculate the interquartile range for this set of data: {34, 47, 1, 15, 57, 24, 20, 11, 19, 50, 28, 37}

1 answer:

3 0

Let's put it in order.
1, 11, 15, 19, 20, 24, 28, 34, 37, 47, 50, 57
Let's split it in half.
1, 11, 15, 19, 20, 24║28, 34, 37, 47, 50, 57
17 42
42-17
25
The IQR is C) 25.

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Which number line shows the solution set for StartAbsoluteValue 8 minus 2 p EndAbsoluteValue = 6?

Illusion [34]

Answer:

A number line going from negative 1 to positive 7. Points are at positive 1 and positive 7.

Step-by-step explanation:

We have the equation:

|8 - 2*p| = 6

Remember that the equation:

|f(x)| = A

with A > 0

means that:

f(x) = A

f(x) = -A

Then for our case, we can rewrite:

|8 - 2*p| = 6

as:

(8 - 2*p) = 6

(8 - 2*p) = -6

Now we can solve these two equations to find the two possible values of p.

From the first one, we get:

8 - 2*p = 6

8 - 6 = 2*p

2 = 2*p

2/2 = p = 1

so one solution is p = 1

Now from the other equation, we can get the other solution:

(8 -2*p) = -6

8 - 2*p = -6

8 + 6 = 2*p

14 = 2*p

14/2 = p = 7

So the two solutions are p = 1 and p = 7

Then the correct option is:

"A number line going from negative 1 to positive 7. Points are at positive 1 and positive 7."

5 0

3 years ago

Hello , how to do 6(iii)?

Doss [256]

Answer:

$\displaystyle \frac{dS}{dt}=\frac{3}{50r}$

Step-by-step explanation:

Water is being pumped into an inflated rubber sphere at a constant rate of 0.03 cubic meters per second.

So, dV/dt = 0.03.

We want to show that dS/dt is directly proportional to 1/r.

In other words, we want to establish the relationship that dS/dt = k(1/r), where k is some constant.

First, the volume of a sphere V is given by:

$\displaystyle V=\frac{4}{3}\pi r^3$

Therefore:

$\displaystyle \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$

Next, the surface area of a sphere S is given by:

$\displaystyle S=4\pi r^2$

Therefore:

$\displaystyle \frac{dS}{dt}=8\pi r\frac{dr}{dt}$

We can divide both sides by 2:

$\displaystyle \frac{1}{2}\frac{dS}{dt}=4\pi r\frac{dr}{dt}$

We can substitute this into dV/dt. Rewriting:

$\displaystyle \frac{dV}{dt}=r\Big(4\pi r\frac{dr}{dt}\Big)$

So:

$\displaystyle \frac{dV}{dt}=\frac{1}{2}r\frac{dS}{dt}$

Since dV/dt = 0.03 or 3/100:

$\displaystyle \frac{3}{100}=\frac{1}{2}r\frac{dS}{dt}$

Therefore:

$\displaystyle \frac{dS}{dt}=\frac{3}{50r}=\frac{3}{50}\Big(\frac{1}{r}\Big)$

Where k = 3/50.

And we have shown that dS/dt is directly proportional to 1/r.

4 0

3 years ago

Use slope to determine if lines PQ and RS are parallel, perpendicular, or neither P(-3,14) q(2,-1) r(4,8) s(-2,-10)

Vlad1618 [11]

Answer:

Neither

Step-by-step explanation:

The slope of PQ is -3.

The slope of RS is 3.

Parallel lines have the same slope. The slopes for perpendicular lines are the opposite of the reciprocal. PQ and RS are intersecting lines, but not parallel or perpendicular.

4 0

3 years ago

Somebody help me with this?

trasher [3.6K]

Answer:

Zero's

Step-by-step explanation:

4 0

2 years ago

Helppppppppp pleaseeeeee

pishuonlain [190]

Answer:

Yo solo pongo esta respuesta que no dice nada porque busco puntiooooooos

8 0

2 years ago