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3 years ago

7

The histogram below shows the heights, rounded to the nearest inch, of the members of the Oak High School basketball team. A. 1

inch B. 9 inches C. 7 inches D. 5 inches

1 answer:

Elan Coil [88]3 years ago

5 0

Answer:

Range of player's height = 9 inch

Step-by-step explanation:

Given:

Number of players Height range

4 66-67

8 68 -69

14 70-71

9 72-73

1 74-75

Find:

Range of player's height

Computation:

Range = Maximum value - Minimum Value

Range of player's height = Maximum height of a player - height of a player

Range of player's height = 75 inch - 66 inch

Range of player's height = 9 inch

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Answer:

The attached file contains the solution to the question

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2 years ago

Given two 2x3 (2 rows, 3 columns) arrays of integer , x1 and x2, both already initialized , two integer variables , i and j, and

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Answer:

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Step-by-step explanation:

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3 years ago

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3 years ago

Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite

zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt be a gram/L

Let the amount salt in the tank at any time t be Q(t).

$\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}$

Incoming rate = (a g/L)×(1 L/min)

=a g/min

The concentration of salt in the tank at any time t is = $\frac{Q(t)}{10}$ g/L

Outgoing rate = $(\frac{Q(t)}{10} g/L)(1 L/ min)$ $\frac{Q(t)}{10} g/min$

$\frac{dQ}{dt} = a- \frac{Q(t)}{10}$

$\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt$

Integrating both sides

$\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt$

$\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c$ [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

$\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c$

$\Rightarrow -log|10a-15|-2=c$

Therefore ,

$-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2$ .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0 in equation (1) we get

$- log|10a|= -log|10a-15| -2$

$\Rightarrow- log|10a|+log|10a-15|= -2$

$\Rightarrow log|\frac{10a-15}{10a}|= -2$

$\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}$

$\Rightarrow 1-\frac{15}{10a} =e^{-2}$

$\Rightarrow \frac{15}{10a} =1-e^{-2}$

$\Rightarrow \frac{3}{2a} =1-e^{-2}$

$\Rightarrow2a= \frac{3}{1-e^{-2}}$

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Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0

4 years ago

Guys help this is due

lesya [120]

Answer:

x = 45

Step-by-step explanation:

Well here's what we know:

2x + 45 + x = 180

Solve from there:

3x = 135

x = 45

You can plug it back in to check:

2(45) + 45 + 45 = 180

90 + 80 = 180, so it checks out :)

8 0

4 years ago

Read 2 more answers