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3 years ago

How many times does 9 go into 999999

Mathematics

2 answers:

Kipish [7]3 years ago

7 0

999999 divided by nine is 111,111!

Hope this helps!! :)

max2010maxim [7]3 years ago

5 0

To figure out your question, just divide 999,999 by 9.

999,999/9 = 111,111

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5.3, CD r- 8.8, and AD - 14.3, find AB to the nearest tenth.<br>If BC

pychu [463]

5.3/8.8=x/14.3
AB is x,
x= 2.43

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Which inequality is represented on the line?<br> -11-10

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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

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

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

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

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

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

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

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

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

$\Rightarrow a = 1.73$

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

8 0

3 years ago

(1 point) A pond contains 100 fish, of which 40 are carp. If 20 fish are caught from the lake, what are the mean and variance of

marta [7]

Answer:

E(X) = 6

Var(X) = 3.394

Step-by-step explanation:

Let X represent the number of carp caught out of the 20 fishes caught. Now, if we are to assume that each

of the (100, 20) ways to catch the 20 fishes will be equally likely.

Thus, we can say that X fulfills a hypergeometric

distribution with parameters as follows;

n = 20, N = 100, k = 30

Formula for expected mean value in hypergeometric distribution is;

E(X) = nk/N

E(X) = (20 × 30)/100

E(X) = 6

Formula for variance is;

Var(X) = (nk/N) × [((n - 1)(k - 1)/(N-1))) + (1 - nk/N)]

Var(X) = ((20 × 30)/100) × [((20 - 1)(30 - 1)/(100 - 1)) + (1 - (20 × 30/100)]

Var(X) = 6 × 0.5657

Var(X) = 3.394

5 0

3 years ago