What is the interquartile range of this data 0 5 10 15 20

1 answer:

5 0

The interquartile range (IQR) is 15.

To find the IQR, you must:
1) Find the median
2) Split the data set in two at the median.
3) Find the medians from both the groups you made in Step 2
4) Label the smaller one Quartile 1 and the larger one Quartile 3 (Hint: The median is Quartile 2)
5) Subtract: Quartile 3 - Quartile 1

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People in here again and again. I think this is the last question

Lilit [14]

Answer:

2.4 cm

Step-by-step explanation:

We know the length of segment AB and the length of segment BC. To find their midpoints, we can divide their length by 2.

Let's find the midpoint of AB.

Segment AB: 10 cm
Midpoint of AB: 10/2 = 5 cm

Next let's find the midpoint of BC.

Segment BC: 5.2 cm
Midpoint of BC: 5.2/2 = 2.6 cm

In order to find the difference between these midpoints, we can subtract the midpoint of AB by the midpoint of BC.

5 cm - 2.6 cm = 2.4 cm

Therefore, the difference between the midpoints of AB and BC is 2.4 cm.

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3 years ago

Water flows into a tank according to the rate F of t equals the quotient of 6 plus t and the quantity 1 plus t, and at the same

diamong [38]

Answer:

Step-by-step explanation:

The complete question is

Water flows into a tank according to the rate F(t)= (t+6)/(1+t), and at the same time empties out at the rate E(t)= (ln(t+2))/(t+1), with both F(t) and E(t) measured in gallons per minute. How much water, to the nearest galllon, is in the tank at time t=10 minutes.

Let C(t) be the amount of water in the tank at time t. We now that the rate of change of the tank is given by

$\frac{dC}{dt}=[\tex]rate at which water flows in- rate at which water flows out. Then [tex]\frac{dC}{dt}=\frac{t+6}{t+1}-\frac{\ln(t+2)}{(t+1)}[\tex]so, the desired expression is obtained by integrating with respect to t. This leads us to [tex]C(t) = \int \frac{t+1}{t+1}+ \frac{5}{t+1} - \frac{\ln(t+2)}{(t+1)} dt=t+ 5 \ln (|t+1|)-\int \frac{\ln(t+2)}{(t+1)} dt +C$

Unfortunately, the integral $\int \frac{\ln(t+2)}{(t+1)} dt$ cannot be expressed using fundamental functions. So, the problem cannot have an specific function (if you are willing to know the complete answer, the integral of this function uses the polylogarithm function with n=2).

Since you want the exact amount of water at time, you need to give C a value, that is, you need to know an initial condition for the problem. This means, you need to know the amount of water in the tank at time 0

8 0

3 years ago

Rearrange the equation so w is the independent variable.<br> u-5=-4(w-1)

exis [7]

$u-5=-4(w-1) \\u=5-4(w-1)\\u=5-4w+4\\u=-4w+9$