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2 years ago

8

Joe ran 3 miles yesterday and wants to run at least 12 miles this week. Write an inequality that can be used to determine the ad

ditional number of days Joe must run this week if each run is 3 miles. Then solve the inequality.

1 answer:

ivanzaharov [21]2 years ago

8 0

Answer:

3 + x >= 12

not sure what variable your using

Its the > sign with the _ under

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sineoko [7]

The event will have 27 tables, with 2 adults and 3 children in each table.

Step-by-step explanation:

The greatest number of tables the planner can set up is determined by finding the greatest common factor in 54 and 81. This is given by;

54 81----------------divided by 3

3 18 27---------------divided by 3

3 6 9---------------divided by 3

3 2 3---------------no common divisor

The greatest number of table will be given by : 3×3×3=27

The number of adults in each table will be 2

The number of children in each table will be 3

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Greatest common factor : brainly.com/question/13133626

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2 years ago

Determine the equation of the line that has an x-intercept of 6 and passes through the point (6,4)

butalik [34]

Putting a = 6 and b= 3/2 in eqn. (1) the required equation is:-

x /6 + 2.y/3 = 1. or, x + 4y = 6. Answer.

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2 years ago

Read 2 more answers

Question1) Describe three scenarios that involve a real-world linear or exponential function. At least one must be exponential.

elixir [45]

Answer:

1) Let's suppose that you go in a straight line, in a car that moves at a constant speed of 80km/h.

Then the distance from your house (assuming that you start the drive in your house) can be modeled with a linear equation:

D(t) = 80km/h*t

where t is time in hours.

This will be a linear function.

2) Suppose that you have a population of some animal, that grows by 2% each month, and initially, there are 100 individuals of that animal.

Then the first month, the population is 100.

The second month the population increased by a 2%, then it will be:

100 + 100*0.02 = 100*(1.02)

The third month, the population will be 100*(1.02) + 0.2*100*(1.02) = 100*(1.02)^2.

and so on, this is an exponential relation, where the population as a function of the number of months, can be written as:

P(m) = 100*(1.02)^(m - 1)

3) Suppose that you have $100 saved, and each month you can save another $80, let's find a function that says the amount of money that you have saved as a function of the number of months. S(m)

The month number zero (before you started saving) you had $100 saved.

S(0) = $100.

One month after, you have saved $80 more, then you have:

S(1) = $100 + $80

Another month after, you have:

S(2) = $100 + $80 + $80 = $100 + 2*$80

And so on, you already can see the pattern, after m months, you will have:

S(m) = $100 + m*$80 saved.

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2 years ago

Can someone help me do this

DiKsa [7]

The answer is Angle T = angle N

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2 years ago

Two streams flow into a reservoir. Let X and Y be two continuous random variables representing the flow of each stream with join

zlopas [31]

Answer:

c = 0.165

Step-by-step explanation:

Given:

f(x, y) = cx y(1 + y) for 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3,

f(x, y) = 0 otherwise.

Required:

The value of c

To find the value of c, we make use of the property of a joint probability distribution function which states that

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, dy \, dx = 1$

where a and b represent -infinity to +infinity (in other words, the bound of the distribution)

By substituting cx y(1 + y) for f(x, y) and replacing a and b with their respective values, we have

$\int\limits^3_0 \int\limits^3_0 {cxy(1+y)} \, dy \, dx = 1$

Since c is a constant, we can bring it out of the integral sign; to give us

$c\int\limits^3_0 \int\limits^3_0 {xy(1+y)} \, dy \, dx = 1$

Open the bracket

$c\int\limits^3_0 \int\limits^3_0 {xy+xy^{2} } \, dy \, dx = 1$

Integrate with respect to y

$c\int\limits^3_0 {\frac{xy^{2}}{2} +\frac{xy^{3}}{3} } \, dx (0,3}) = 1$

Substitute 0 and 3 for y

$c\int\limits^3_0 {(\frac{x* 3^{2}}{2} +\frac{x * 3^{3}}{3} ) - (\frac{x* 0^{2}}{2} +\frac{x * 0^{3}}{3})} \, dx = 1$

$c\int\limits^3_0 {(\frac{x* 9}{2} +\frac{x * 27}{3} ) - (0 +0) \, dx = 1$

$c\int\limits^3_0 {(\frac{9x}{2} +\frac{27x}{3} ) \, dx = 1$

Add fraction

$c\int\limits^3_0 {(\frac{27x + 54x}{6}) \, dx = 1$

$c\int\limits^3_0 {\frac{81x}{6} \, dx = 1$

Rewrite;

$c\int\limits^3_0 (81x * \frac{1}{6}) \, dx = 1$

The $\frac{1}{6}$ is a constant, so it can be removed from the integral sign to give

$c * \frac{1}{6}\int\limits^3_0 (81x ) \, dx = 1$

$\frac{c}{6}\int\limits^3_0 (81x ) \, dx = 1$

Integrate with respect to x

$\frac{c}{6} * \frac{81x^{2}}{2} (0,3) = 1$

Substitute 0 and 3 for x

$\frac{c}{6} * \frac{81 * 3^{2} - 81 * 0^{2}}{2} = 1$

$\frac{c}{6} * \frac{81 * 9 - 0}{2} = 1$

$\frac{c}{6} * \frac{729}{2} = 1$

$\frac{729c}{12} = 1$

Multiply both sides by $\frac{12}{729}$

$c = \frac{12}{729}$

$c = 0.0165 (Approximately)$

8 0

3 years ago