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3 years ago

All the common factors of 24, 64, 88

Mathematics

2 answers:

Alenkasestr [34]3 years ago

8 0

All the common factors of 24, 64, 88 are 1, 2, 4, 8

zhuklara [117]3 years ago

8 0

The common factors of 24, 64,88 are (1,2,4,8)

The factors of 24
1,2,3,4,6,8,12,24

The factors of 64
1,2,4,8,16,32,64

The factors of 88
1,2,4,8,11,22,44,88

So if you see the ones that are the same which are 1,2,4 and 8

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Find all of the equilibrium solutions. Enter your answer as a list of ordered pairs (R,W), where R is the number of rabbits and

zloy xaker [14]

Answer:

$(0,0)$ $(4000,0)$ and $(500,79)$

Step-by-step explanation:

Given

See attachment for complete question

Required

Determine the equilibrium solutions

We have:

$\frac{dR}{dt} = 0.09R(1 - 0.00025R) - 0.001RW$

$\frac{dW}{dt} = -0.02W + 0.00004RW$

To solve this, we first equate $\frac{dR}{dt}$ and $\frac{dW}{dt}$ to 0.

So, we have:

$0.09R(1 - 0.00025R) - 0.001RW = 0$

$-0.02W + 0.00004RW = 0$

Factor out R in $0.09R(1 - 0.00025R) - 0.001RW = 0$

$R(0.09(1 - 0.00025R) - 0.001W) = 0$

Split

$R = 0$ or $0.09(1 - 0.00025R) - 0.001W = 0$

$R = 0$ or $0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

Factor out W in $-0.02W + 0.00004RW = 0$

$W(-0.02 + 0.00004R) = 0$

Split

$W = 0$ or $-0.02 + 0.00004R = 0$

Solve for R

$-0.02 + 0.00004R = 0$

$0.00004R = 0.02$

Make R the subject

$R = \frac{0.02}{0.00004}$

$R = 500$

When $R = 500$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 -2.25 * 10^{-5} * 500 - 0.001W = 0$

$0.09 -0.01125 - 0.001W = 0$

$0.07875 - 0.001W = 0$

Collect like terms

$- 0.001W = -0.07875$

Solve for W

$W = \frac{-0.07875}{ - 0.001}$

$W = 78.75$

$W \approx 79$

$(R,W) \to (500,79)$

When $W = 0$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 - 2.25 * 10^{-5}R - 0.001*0 = 0$

$0.09 - 2.25 * 10^{-5}R = 0$

Collect like terms

$- 2.25 * 10^{-5}R = -0.09$

Solve for R

$R = \frac{-0.09}{- 2.25 * 10^{-5}}$

$R = 4000$

So, we have:

$(R,W) \to (4000,0)$

When $R =0$ , we have:

$-0.02W + 0.00004RW = 0$

$-0.02W + 0.00004W*0 = 0$

$-0.02W + 0 = 0$

$-0.02W = 0$

$W=0$

So, we have:

$(R,W) \to (0,0)$

Hence, the points of equilibrium are:

$(0,0)$ $(4000,0)$ and $(500,79)$

4 0

3 years ago

Suppose that taxis in New York are driven an average of 60,000 miles per year with a standard deviation of 11,000 miles. Assume

Gre4nikov [31]

Let n be the number of taxis in NY. The average distance travelled is 60,000 miles, therefore the middle 95% will have the same average as the population, the reason being the mileage is symmetrically distributed about the mean Therefore the total number of miles in one year for the middle 95% is 60,000 * 0.95 * n
The range of miles driven by the middle 95% can be found from the empirical rule that says: For a normal distribution, approximately 95% of the data points lie within the range plus and minus 2 standard deviations of the population mean. In this case the range is (60,000-22,000) to (60,000 + 22,000)

3 0

3 years ago

There is a bag filled with 4 blue and 5 red marbles.

Elan Coil [88]

4/9 (4 out of 9) because you add them all together then count the number of blue marbles then since you put back the first marble, the dominator is still the same

3 0

3 years ago

The amount of carbon-14 present in a paint after t years is given by y equals y Subscript o Baseline e Superscript negative 0.0

Anni [7]

Answer:

16,383.33 years ( approx )

Step-by-step explanation:

Given equation that shows the amount of carbon-14 after t years,

$y = y_0 e^{-0.00012t}$