All categories

3 years ago

the average (arithmetic mean) of t and y is 15, and the average of w and x is 15. what is the average of t, w,x,and y

1 answer:

5 0

Oi :)

$\frac{t+y}{2}=15 \\ \\ t+y=15*2 \\ t+y=30$
--------------------------------------------------------------------------------------------------------------------------
$\frac{w+x}{2}=15 \\ \\ w+x=15*2 \\ w+x=30$
---------------------------------------------------------------------------------------------------------------------------
Logo:
$\frac{t+y+w+x}{4}=??? \\ \\ \frac{30+30}{4} \\ \\ \frac{60}{4} \\ \\ 15$
-----------------------------------------------------------------------------------------------------------------------------
Answer = 15

Boa Noite. Bons Estudos !

You might be interested in

A common assumption in modeling drug assimilation is that the blood volume in a person is a single compartment that behaves like

mixas84 [53]

Answer:

a) $\mathbf{\dfrac{dx}{dt} = 30 - 0.015 x}$

b) $\mathbf{x = 2000 - 2000e^{-0.015t}}$

c) the steady state mass of the drug is 2000 mg

d) t ≅ 153.51 minutes

Step-by-step explanation:

From the given information;

At time t= 0

an intravenous line is inserted into a vein (into the tank) that carries a drug solution with a concentration of 500

The inflow rate is 0.06 L/min.

Assume the drug is quickly mixed thoroughly in the blood and that the volume of blood remains constant.

The objective of the question is to calculate the following :

a) Write an initial value problem that models the mass of the drug in the blood for t ≥ 0.

From above information given :

$Rate _{(in)}= 500 \ mg/L \times 0.06 \ L/min = 30 mg/min$

$Rate _{(out)}=\dfrac{x}{4} \ mg/L \times 0.06 \ L/min = 0.015x \ mg/min$

Therefore;

$\dfrac{dx}{dt} = Rate_{(in)} - Rate_{(out)}$

with respect to x(0) = 0

$\mathbf{\dfrac{dx}{dt} = 30 - 0.015 x}$

b) Solve the initial value problem and graph both the mass of the drug and the concentration of the drug.

$\dfrac{dx}{dt} = -0.015(x - 2000)$

$\dfrac{dx}{(x - 2000)} = -0.015 \times dt$

By Using Integration Method:

$ln(x - 2000) = -0.015t + C$

$x -2000 = Ce^{(-0.015t)$

$x = 2000 + Ce^{(-0.015t)}$

However; if x(0) = 0 ;

Then

C = -2000

Therefore

$\mathbf{x = 2000 - 2000e^{-0.015t}}$

c) What is the steady-state mass of the drug in the blood?

the steady-state mass of the drug in the blood when t = infinity

$\mathbf{x = 2000 - 2000e^{-0.015 \times \infty }}$

x = 2000 - 0

x = 2000

Thus; the steady state mass of the drug is 2000 mg

d) After how many minutes does the drug mass reach 90% of its stead-state level?

After 90% of its steady state level; the mas of the drug is 90% × 2000

= 0.9 × 2000

= 1800

Hence;

$\mathbf{1800 = 2000 - 2000e^{(-0.015t)}}$

$0.1 = e^{(-0.015t)$

$ln(0.1) = -0.015t$

$t = -\dfrac{In(0.1)}{0.015}$

t = 153.5056729

t ≅ 153.51 minutes

4 0

3 years ago

Solve the Situation by using an inequality Statement. Choose the range of answers that make the Statement true.

babymother [125]

I going with A hope that helps

7 0

2 years ago

Add. 1/2+(−3/5) <br><br><br> can u help me

attashe74 [19]

Exact form: -1/10

decimal form: -0.1

6 0

3 years ago

Read 2 more answers

Pls help im very desperate i'll defenitely give brainliest

bagirrra123 [75]

Answer:

-25

Step-by-step explanation:

You can take 10÷-⅖=-25

7 0

2 years ago

Read 2 more answers

Anita invested $6000, some at 5% and the rest at 6%. If the yearly income is $337.50, what is the amount invested at each rate?

Pepsi [2]

Let the amount invested with 5% interest be x

Therefore, the amount invested with 6% interest will be (6000-x)

It is given that the total interest earned yearly is $337.5. Thus, the equation of interest will be:

$\frac{5}{100}x+\frac{6}{100}(6000-x)=337.5$

Multiplying both sides by 100 we get:

$5x+6(6000-x)=33750$

$5x+36000-6x=33750$

Subtracting both sides by 36000 we get:

$-x=-2250$

$\therefore x=2250$

Thus, the amount invested at 5% is $2250

Therefore, the amount invested at 6% will be $(6000-2250)=$3750

6 0

3 years ago