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saw5 [17]
3 years ago
9

Tell whether -2 is a solution of the inequality n-5<8

Mathematics
1 answer:
ruslelena [56]3 years ago
6 0

Answer:

-2 is a solution

Step-by-step explanation:

-2-5<8

-7<8

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(Please help a sister out)!
MrRissso [65]

Answer:

Distributive

Step-by-step explanation:

when you distribute 3 to x, 3y, and 5, you get 3x+9y+15, which is equivalent to the other expression when you simplify it

5 0
3 years ago
Make n the subject of the formula: M=3n
natta225 [31]
Divide each side by 3. ----- n=M/3 .
4 0
3 years ago
A bag contains tickets numbered 1 through 5. Use the probability distribution to determine the probability of drawing an odd num
Makovka662 [10]

Answer:

The probability of drawing an odd numbered ticket is 60%.

Step-by-step explanation:

Odd numbered tickets:

Probability of one is 1/5 plus half of 1/5.

P(X = 1) = \frac{1}{5} + \frac{1}{10} = \frac{3}{10}

Probability of 3 is half of 1/5.

P(X = 3) = \frac{1}{10}

Probability of 5 is 1/5. So

P(X = 5) = \frac{1}{5}

Probability of drawing an odd numbered ticket:

p = P(X = 1) + P(X = 3) + P(X = 5) = \frac{3}{10} + \frac{1}{10} + \frac{1}{5} = \frac{4}{10} + \frac{2}{10} = \frac{6}{10} = 0.6

0.6*100% = 60%

The probability of drawing an odd numbered ticket is 60%.

3 0
2 years ago
Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains liters of a dye solution with a
Alja [10]

Answer:

t = 460.52 min

Step-by-step explanation:

Here is the complete question

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.

Solution

Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.

inflow = 0 (since the incoming water contains no dye)

outflow = concentration × rate of water inflow

Concentration = Quantity/volume = Q/200

outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.

So, Q' = inflow - outflow = 0 - Q/100

Q' = -Q/100 This is our differential equation. We solve it as follows

Q'/Q = -1/100

∫Q'/Q = ∫-1/100

㏑Q  = -t/100 + c

Q(t) = e^{(-t/100 + c)} = e^{(-t/100)}e^{c}  = Ae^{(-t/100)}\\Q(t) = Ae^{(-t/100)}

when t = 0, Q = 200 L × 1 g/L = 200 g

Q(0) = 200 = Ae^{(-0/100)} = Ae^{(0)} = A\\A = 200.\\So, Q(t) = 200e^{(-t/100)}

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

2 = 200e^{(-t/100)}\\\frac{2}{200} =  e^{(-t/100)}

㏑0.01 = -t/100

t = -100㏑0.01

t = 460.52 min

6 0
3 years ago
A school concert has students tickets for $7 and nonstudents tickets for $12. If the school made $1973 selling 214 tickets? How
bearhunter [10]

Answer:

119 student tickets and 95 non-student tickets

Step-by-step explanation:

I did a trial-and-error solution.

I started with 107 students and 107 non-students:

(107*$7)+(107*$12) = $2033

It was too high, which tells us that there should be more who paid for the cheaper price.

I ended up with 119 and 95:

(119*$7)+(95*$12) = $1973

8 0
2 years ago
Read 2 more answers
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