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VLD [36.1K]
3 years ago
10

URGENT!!! for the given situation identify the independent and D pendant variable‘s and then find a reasonable domain and range

of values listed below will be for statements that relate to the given situation. She’s is the statement that would have to be FALSE given the situation and it’s parameters
Roger has $12 to spend. He wants to buy some apples that cost $.60 apiece.
here are the statements
1) The dependent variable is how many apples he will buy
2) The range will be 0 to 20 apples
3) The domain will be any amount of money less than $12
4) The independent variable is how much money he will spend
Mathematics
1 answer:
sashaice [31]3 years ago
5 0

Answer:

3

Step-by-step explanation:

in order to be correct, it would have to be "any amount of money less than <em>or equal to</em> $12" because if it were less then, that makes the maximum amount he can spend $11.99.

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Find the inverse of the function.<br> f-^1(x)= (x – 5)^3 + 1
Cloud [144]

Answer:

Step-by-step explanation:

To find the inverse of function y=f(x), swap variables and then solve for the original variable.

y=(x-5)^3+1\\ \\ x=(y-5)^3+1\\ \\ x-1=(y-5)^3\\ \\ \sqrt[3]{x-1}=y-5\\ \\ y=\sqrt[3]{x-1}+5\\ \\ f^{-1}(x)=\sqrt[3]{x-1}+5

3 0
3 years ago
One model for the spread of a virusis that the rate of spread is proportional to the product of the fraction of the population P
Darya [45]

Answer:

The differential equation for the model is

\frac{dP}{dt}=kP(1-P)

The model for P is

P(t)=\frac{1}{1-0.99e^{t/447}}

At half day of the 4th day (t=4.488), the population infected reaches 90,000.

Step-by-step explanation:

We can write the rate of spread of the virus as:

\frac{dP}{dt}=kP(1-P)

We know that P(0)=100 and P(3)=100+200=300.

We have to calculate t so that P(t)=0.9*100,000=90,000.

Solving the diferential equation

\frac{dP}{dt}=kP(1-P)\\\\ \int \frac{dP}{P-P^2} =k\int dt\\\\-ln(1-\frac{1}{P})+C_1=kt\\\\1-\frac{1}{P}=Ce^{-kt}\\\\\frac{1}{P}=1-Ce^{-kt}\\\\P=\frac{1}{1-Ce^{-kt}}

P(0)=  \frac{1}{1-Ce^{-kt}}=\frac{1}{1-C}=100\\\\1-C=0.01\\\\C=0.99\\\\\\P(3)=  \frac{1}{1-0.99e^{-3k}}=300\\\\1-0.99e^{-3k}=\frac{1}{300}=0.99e^{-3k}=1-1/300=0.997\\\\e^{-3k}=0.997/0.99=1.007\\\\-3k=ln(1.007)=0.007\\\\k=-0.007/3=-0.00224=-1/447

Then the model for the population infected at time t is:

P(t)=\frac{1}{1-0.99e^{t/447}}

Now, we can calculate t for P(t)=90,000

P(t)=\frac{1}{1-0.99e^{t/447}}=90,000\\\\1-0.99e^{t/447}=1/90,000 \\\\0.99e^{t/447}=1-1/90,000=0.999988889\\\\e^{t/447}=1.010089787\\\\ t/447=ln(1.010089787)\\\\t=447ln(1.010089787)=447*0.010039225=4.487533

At half day of the 4th day (t=4.488), the population infected reaches 90,000.

8 0
4 years ago
Solve each equation by factoring x^2+4x+4=0 ​
makkiz [27]

Answer: x = -2

<u>Step-by-step explanation:</u>

Find 2 numbers whose product is the c-value of 4 and their sum is the b-value of 4.

x² + 4x + 4 = 0

              ∧

           1 + 4 = 5  

           2 + 2 = 4  <em>this works!</em>   <em>so the factors are: </em>

(x + 2)(x + 2) = 0

x + 2 = 0      and      x + 2 = 0

     x = -2      and             x = -2  --> <em>same answer so only need to list it once</em>

6 0
3 years ago
41.687510 to binary form​
kupik [55]

I believe your answer would be 101001.1011000000000000101.

Hope this helps!

Have a great day!

3 0
3 years ago
A headline in a certain newspaper states that​ "most stay at first job less than 2​ years." That headline is based on an online
alukav5142 [94]

Answer:

Mean, μ = 200

Standard deviation, σ = 10

Step-by-step explanation:

Data provided:

a) true percentage of graduates who stay at their first job less than two​ years

p = 50% = 0.5

q = 1 - p = 1 - 0.5 = 0.5

Number of graduates selected in a group, n = 400

Now,

The mean is calculated as:

Mean, μ = p × n

or

μ = 0.5 × 400 = 200

and,

standard deviation, σ = \sqrt{npq}

or

σ = \sqrt{400\times0.5\times0.5}

or

σ = √100

or

σ = 10

7 0
4 years ago
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