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3 years ago

Select two ordered pairs that satisfy the function y = –4x – 1.

Mathematics

2 answers:

Nimfa-mama [501]3 years ago

5 0

X = -0.25 and x = -2^-2
I’m not sure if this helps

gogolik [260]3 years ago

3 0

Answer:

(-0.25,0) and (0,-1)

Step-by-step explanation:

I used a graphing calculator.

Put function the the calculator

then ind two points on the function and that is your answer

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If you choose a card at random from a well shuffled deck of 52 cards, what is the probability that the card chosen is not a hear

Vinvika [58]

Sample space ={13 H + 13 D + 13 S + 13 C} = 52 cards

P(getting one heart) = 13/52 = 1/4
P( getting NO heart) = 1-1/4 = 3/4 = 0.75

4 0

3 years ago

You are paid $78 for 612612 hours of work. What is your rate of pay?

sineoko [7]

Not alot i would quit

8 0

3 years ago

10xy(4xy^3-7xy+9y^2)

Thepotemich [5.8K]

Answer:

$40x^{2} y^{4} -70x^{2} y^{2} +90xy^{3}$

Step-by-step explanation:

$10xy(4xy^{3} -7xy+9y^2)$

1. 10xy · 4xy^3 = $40x^2y^4$

2. 10xy · -7xy = $-70x^{2} y^{2}$

3. 10xy · 9y^2 = $90xy^3$

Final step: add up all of those values together to make an equation!

Answer: $40x^2y^4+$ $-70x^2y^2+$ $90xy^3$

Final Answer: $40x^2y^4-70x^2y^2+90xy^3$

3 0

3 years ago

Between 2000 and 2010, the percentage of U.S. Households with cordless phones increased by 13.7% to 91%. What percentage of hous

alina1380 [7]

Answer:

80%

Step-by-step explanation:

$1.137x = 91 \\ x = 80.03518029$

6 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

3 years ago