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3 years ago

Which graph correctly represents 1/3y-1/2x>2

Mathematics

2 answers:

Vedmedyk [2.9K]3 years ago

5 0

Move the (1/2) x to the right by adding it to both sides of the inequality ... This is a graph with y intercept of (0,6) and a moderate upward and to the right slope.

BabaBlast [244]3 years ago

3 0

Theses Are The Graphs And Answers

[ A ] [ WRONG ]

[ B ] [ WRONG ]

[ C ] [ WRONG ]

[ D ] [ CORRECT ]

[ E ] [ WRONG ]

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The function f(x)=5(2)^x was replaced with f(x)+k , resulting in the function graphed below. what is the value of k?

xxTIMURxx [149]

Since adding a constant to a function simply moved the graph upwards by that amount, solve f(x) for any value and see how much the graph given is different from that value of y...the simplest way in this case may be to simply find f(0) which is:

5*2^0=5

Clearly the graph at x=0 is at y=-2 so we can see that k is:

5+k=-2 so

k=-7

5 0

3 years ago

The perimeter of a rectangle is 150 cm. the length is 4 cm greater than the width find the width and length

Fudgin [204]

Let's call the width of our rectangle $w$ and the length $l$ . We can say $l = w + 4$ , since the length is equal to 4 cm greater than the width.

Also remember that the perimeter of a rectangle is the sum of two times the width and two times the length, or $P = 2l + 2w$ . To solve this problem, we can substitute in the information we know, as shown below:

$150 = 2(w + 4) + 2w$

$150 = 4w + 8$

$142 = 4w$

$w = \frac{71}{2}$

Now, we can substitute in the width we found into the formula for length, which is $l = w + 4$ :

$l = \frac{71}{2} + 4$

$l = \frac{79}{2}$

The width of our rectangle is $\boxed{\frac{71}{2} \,\, \textrm{cm}}$ cm and the length of our rectangle is $\boxed{\frac{79}{2} \,\, \textrm{cm}}$

8 0

3 years ago

A tree 10 meters high casts a 17.3 meter shadow.

sashaice [31]

Answer:

the angle of elevation of the sun is $\approx 30^o$

Step-by-step explanation:

Let's make a diagram of the tree, ground and inclination of the rays of the sun, so we understand what right angle triangle we have. Please see the attached image.

Notice that our unknown is the angle $\theta$ , and that we know the size of the side opposite to it, and of the side adjacent to it. Therefore the most convenient trigonometric formula to use is the tangent of the angle because:

$tan(\theta)=\frac{opposite}{adjacent} \\tan(\theta)=\frac{10}{17.3}\\\theta=arctan(\frac{10}{17.3})\\\theta\approx 30^o$

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3 years ago

One inch is equivalent to 2.54 centimeters. How many inches are in 50.8 centimeters

Genrish500 [490]

The answer is 20 ^-^.... i think

7 0

3 years ago

Read 2 more answers

The city has an average of 13 days of rainfall for April.

zhenek [66]

Using the Poisson distribution, we have that:

There is a 0.0859 = 8.59% probability of having exactly 10 days of precipitation in the month of April.

There is a 0.00022 = 0.022% probability of having less than three days of precipitation in the month of April.

There is a 0.2364 = 23.64% probability of having more than 15 days of precipitation in the month of April.

<h3>What is the Poisson distribution?</h3>

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:

$P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}$

The parameters are:

x is the number of successes

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

For this problem, the mean is given as follows:

$\mu = 13$

The probability of having exactly 10 days of precipitation in the month of April is P(X = 10), hence:

$P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}$

$P(X = 10) = \frac{e^{-13}13^{10}}{(10)!} = 0.0859$

There is a 0.0859 = 8.59% probability of having exactly 10 days of precipitation in the month of April.

The probability of having less than three days of precipitation in the month of April is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

In which:

$P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-13}13^{0}}{(0)!} \ approx 0$

$P(X = 1) = \frac{e^{-13}13^{1}}{(1)!} = 0.00003$

$P(X = 2) = \frac{e^{-13}13^{2}}{(2)!} = 0.00019$

Then:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0 + 0.00003 + 0.00019 = 0.00022

There is a 0.00022 = 0.022% probability of having less than three days of precipitation in the month of April.

For more than 15 days, the probability is:

P(X > 15) = P(X = 16) + P(X = 17) + ... + P(X = 20)

Applying the formula for each of these values and adding them, we have that P(X > 15) = 0.2364, hence:

There is a 0.2364 = 23.64% probability of having more than 15 days of precipitation in the month of April.

More can be learned about the Poisson distribution at brainly.com/question/13971530

#SPJ1

6 0

2 years ago