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AnnyKZ [126]
3 years ago
8

What is the theoretical probability of rolling a 1 or a 5 on a standard six-sided die?

Mathematics
1 answer:
Vaselesa [24]3 years ago
5 0
1/3 probability of rolling a 1 or a 5 
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The functions q and r are defined as follows.<br> HELP PLEASE
dusya [7]

Answer:

-22

Step-by-step explanation:

First do r(3) by replacing x with 3 in r:

2*3^2 + 2 = 2*9 + 2 = 18 + 2 = 20

Then replace r(3) with 20 to do q(20):

-20-2 = -22

7 0
3 years ago
Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br
Lina20 [59]

Answer:

The differential equation for the amount of salt A(t) in the tank at a time  t > 0 is \frac{dA}{dt}=12 - \frac{2A(t)}{500+t}.

Step-by-step explanation:

We are given that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

The concentration of the solution entering is 4 lb/gal.

Firstly, as we know that the rate of change in the amount of salt with respect to time is given by;

\frac{dA}{dt}= \text{R}_i_n - \text{R}_o_u_t

where, \text{R}_i_n = concentration of salt in the inflow \times input rate of brine solution

and \text{R}_o_u_t = concentration of salt in the outflow \times outflow rate of brine solution

So, \text{R}_i_n = 4 lb/gal \times 3 gal/min = 12 lb/gal

Now, the rate of accumulation = Rate of input of solution - Rate of output of solution

                                                = 3 gal/min - 2 gal/min

                                                = 1 gal/min.

It is stated that a large mixing tank initially holds 500 gallons of water, so after t minutes it will hold (500 + t) gallons in the tank.

So, \text{R}_o_u_t = concentration of salt in the outflow \times outflow rate of brine solution

             = \frac{A(t)}{500+t} \text{ lb/gal } \times 2 \text{ gal/min} = \frac{2A(t)}{500+t} \text{ lb/min }

Now, the differential equation for the amount of salt A(t) in the tank at a time  t > 0 is given by;

= \frac{dA}{dt}=12\text{ lb/min } - \frac{2A(t)}{500+t} \text{ lb/min }

or \frac{dA}{dt}=12 - \frac{2A(t)}{500+t}.

4 0
3 years ago
You have 60 CD cases that you would like to store on a shelf that is 24 inches long. if each CD case is 3/8 inch wide, is there
Jet001 [13]
Yes you do.

The shelf is 24 in. long, so to fit, the total width of all the cases must be < 24 in.

If each case is 3/8 inches:
60 * 3/8 = 22.5

22.5 < 24, therefore they will all fit.
5 0
3 years ago
Read 2 more answers
Jenny and Lester are running the 100 meter dash. Jenny completes it in 15.2 seconds and Lester completed it in 15.7 seconds. Wha
matrenka [14]

Jenny won the race as she finished the race much more quickly and has greater average speed than Lester.

\sf \boxed{\sf speed:\frac{distance}{time \ taken} }

Jenny: distance/time = 100/15.2 = 6.58 m/s = 6.6 m/s

Lester: distance/time = 100/15.7 = 6.37 m/s = 6.4 m/s

3 0
2 years ago
A farmer estimated that there were 25 gallons of water left in a tank. If this is an underestimate by 16%, how much water was ac
djyliett [7]

Answer:

The actual water in the tank is 29.76 gallons.

Step-by-step explanation:

It is given that the farmer estimated amount of water is 25 gallons.

It is also given that he underestimate the water in tank by 16%.

Let the actual water in tank be x.

(1-\frac{16}{100})x=25

(\frac{100-16}{100})x=25

84x=25\times 100

x=\frac{2500}{84}

x=29.7619

x\approx 29.76

Therefore the actual water in the tank is 29.76 gallons.

8 0
3 years ago
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