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Tresset [83]
3 years ago
14

A doctor recommends a two-step process to treat a rare form of pancreatic cancer. The first method is successful 80% of the time

. If the first method is successful, the second method is successful 90% of the time. If the first treatment is not a success, the second is 25% of the time. What is the probability that both treatments are unsuccessful?
Mathematics
1 answer:
Rudiy273 years ago
5 0

Answer:

The answer is 15%.

Step-by-step explanation:

The probability that:

  • Both treatments are successful:

       80% x 90% = 72%

  • The first method is a success, but the second one is not:

        80% x (1 - 90%) = 8%

  • The first method is not successful, but the second one is:

        (1 - 80%) x 25% = 5%

  • Both treatments are unsuccessful:

        1 - (72% + 8% + 5%) = 15%

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PLZ HELP ASAP! I WILL MARK BRAINIEST
zmey [24]

Answer:

Step-by-step explanation:

If we look, "y" can be factored out of the entire expression because it goes into them all. This gives y(2x^3 - 4x^2 + 8x - 16)

Since everything inside the brackets is a multiple of 2 we can factor it out.

2y (x^3 - 2x^2 + 8x - 16)

Here to factor (x^3 - 2x^2 + 8x - 16) we could either use a calculator, or notice that when we subsitute 2 into the equation, it equals 0. Therefore (x - 2) must be a factor. The other factor must be (x^2 + 8) because:

(x - 2) (      )

We need to have 1 x^3 so we know the first part of the factor must be x^2

The also need to have -16 when we multiply this out, -2 * something equals -16. Meaning we must have 8 inside of the brackets.

This gives :

(x - 2) (x^2 + 8).

When we expand this out we get (x^3 - 2x^2 + 8x - 16)

Therefore in total we have

2y (x - 2)(x^2 + 8)

There are other methods of factorising (x^3 - 2x^2 + 8x - 16), so use the method which you have been taught in class or the one I used.

3 0
3 years ago
CAN SOMEONE PLEASE HELP ME WITH THIS QUESTION I NEED HELP
lesantik [10]
The answer is 5% . 4000+5% would be 4200
4 0
3 years ago
A tank contains 5,000 L of brine with 13 kg of dissolved salt. Pure water enters the tank at a rate of 50 L/min. The solution is
tresset_1 [31]

Answer:

a) x(t) = 13*e^(^-^\frac{t}{100}^)

b) 10.643 kg

Step-by-step explanation:

Solution:-

- We will first denote the amount of salt in the solution as x ( t ) at any time t.

- We are given that the Pure water enters the tank ( contains zero salt ).

- The volumetric rate of flow in and out of tank is V(flow) = 50 L / min  

- The rate of change of salt in the tank at time ( t ) can be expressed as a ODE considering the ( inflow ) and ( outflow ) of salt from the tank.

- The ODE is mathematically expressed as:

                            \frac{dx}{dt} = ( salt flow in ) - ( salt flow out )

- Since the fresh water ( with zero salt ) flows in then ( salt flow in ) = 0

- The concentration of salt within the tank changes with time ( t ). The amount of salt in the tank at time ( t ) is denoted by x ( t ).

- The volume of water in the tank remains constant ( steady state conditions ). I.e 10 L volume leaves and 10 L is added at every second; hence, the total volume of solution in tank remains 5,000 L.

- So any time ( t ) the concentration of salt in the 5,000 L is:

                             conc = \frac{x(t)}{1000}\frac{kg}{L}

- The amount of salt leaving the tank per unit time can be determined from:

                         salt flow-out = conc * V( flow-out )  

                         salt flow-out = \frac{x(t)}{5000}\frac{kg}{L}*\frac{50 L}{min}\\

                         salt flow-out = \frac{x(t)}{100}\frac{kg}{min}

- The ODE becomes:

                               \frac{dx}{dt} =  0 - \frac{x}{100}

- Separate the variables and integrate both sides:

                       \int {\frac{1}{x} } \, dx = -\int\limits^t_0 {\frac{1}{100} } \, dt  + c\\\\Ln( x ) = -\frac{t}{100} + c\\\\x = C*e^(^-^\frac{t}{100}^)

- We were given the initial conditions for the amount of salt in tank at time t = 0 as x ( 0 ) = 13 kg. Use the initial conditions to evaluate the constant of integration:

                              13 = C*e^0 = C

- The solution to the ODE becomes:

                           x(t) = 13*e^(^-^\frac{t}{100}^)

- We will use the derived solution of the ODE to determine the amount amount of salt in the tank after t = 20 mins:

                           x(20) = 13*e^(^-^\frac{20}{100}^)\\\\x(20) = 13*e^(^-^\frac{1}{5}^)\\\\x(20) = 10.643 kg

- The amount of salt left in the tank after t = 20 mins is x = 10.643 kg

                           

7 0
3 years ago
Simplify:<br><br> -10(t – 8) – 7t<br> A. –17t + 80<br> B. –17t – 80<br> C. 3t + 8<br> D. 3t – 8
Delvig [45]

Answer: A.–17t + 80

Step-by-step explanation:

-10(t –8)–7t

−(10t−80)−7t

−10t+80−7t

(−10t−7t)+80

–17t + 80

Hope this helps, now you know the answer and how to do it. HAVE A BLESSED AND WONDERFUL DAY! As well as a great rest of Black History Month! :-)  

- Cutiepatutie ☺❀❤

5 0
3 years ago
The figures shown are similar. What is the scale factor?
scoray [572]

Answer:

B. 5/9 I TOOK THE TEST

Step-by-step explanation:

6 0
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