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3 years ago

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Write a function rule for the area of a triangle whose base is 4 ft more than the height. What is the area of the triangle when

its height is 6 ft?
A. A = 0.5h2 + 4h; 42 ft2

B. A = 4h − 0.5h2; 6 ft2

C. A = 0.5h2 + 2h; 30 ft2

D. A = 0.5h2 − 2h; 6 ft2

1 answer:

skelet666 [1.2K]3 years ago

8 0

Full Form of CD?

Compact Disk

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Lawrence found a shirt he liked in the clearance aisle for 15% off the original price. If the original price of the shirt was $3

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Answer:

c

Step-by-step explanation:

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3 years ago

The mean number of automobiles entering a mountain tunnel per two-minute period is one. An excessive number of cars entering the

viva [34]

Answer:

A Poisson model seems reasonable for this problem, since we have the mean during the time interval.

There is a 1.9% probability that the number of autos entering the tunnel during a two-minute period exceeds three.

Step-by-step explanation:

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

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

In which

x is the number of sucesses

e = 2.71828 is the Euler number

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

The mean number of automobiles entering a mountain tunnel per two-minute period is one.

This means that $\mu = 1$ .

For a Poisson model to be reasonable, we only need the mean during the time interval. So yes, a Poisson model seems reasonable for this problem.

Find the probability that the number of autos entering the tunnel during a two-minute period exceeds three.

We want to find $P(X>3)$

Either this number is less or equal to 3, or it exceeds 3. The sum of the probabilities is decimal 1. So:

$P(X \leq 3) + P(X > 3) = 1$

$P(X > 3) = 1 - P(X \leq 3)$

In which

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

$P(X = 0) = \frac{e^{-1}*1^{0}}{(0)!} = 0.3679$

$P(X = 1) = \frac{e^{-1}*1^{1}}{(1)!} = 0.3679$

$P(X = 2) = \frac{e^{-1}*1^{2}}{(2)!} = 0.1839$

$P(X = 3) = \frac{e^{-1}*1^{3}}{(3)!} = 0.0613$

So

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.3679 + 0.3679 + 0.1839 + 0.0613 = 0.981$

Finally

$P(X > 3) = 1 - P(X \leq 3) = 1 - 0.981 = 0.019$

There is a 1.9% probability that the number of autos entering the tunnel during a two-minute period exceeds three.

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3 years ago

A hook in an office storage closet can hold no more than 6 pounds. An order of jumbo paperclips weighs 2 pounds and an

Westkost [7]

Answer:

<u>the first graph:</u> line passing through the points (3, 0) and (0,2), and the shaded region is below the line.

Explanation:

<u>1) Find the expression that represents the situation.</u>

The expression that represents the situation is an inequality:

Number of orders of paper clips: x
Weight of an order of paper clips: 2 lbs
Total weight of x orders of paper clips: 2x

Number of orders of packing tape: y
Weight of an order of packing tape: 3 lbs
Total weight of y orders of packing tape: 3y

Total weight of paper clips and packing tape in a bag: 2x + 3y

Maximum weight hold by the hook: 6 lbs

Hence, the total weight must be less than or equal to (≤) 6 lbs, which is:

2x + 3y ≤ 6

<u>2) Graph of the inequality 2x + 3y ≤ 6</u>

Line:

Border line: 2x + 3y = 6

x-intercept: y = 0 ⇒ 2x = 6 ⇒ x = 6 /2 ⇒ x = 3 ⇒ point (3,0)
y-intercept: x = 0 ⇒ 3y = 6 ⇒ y = 6 /3 ⇒ y = 2 ⇒ point (0,2)

Shaded region:

Symbol ≤ means that the line is included, which is represented with a solid line, and the region is below the line.

Conclusion: the graph is the line passing through the points (3, 0) and (0,2), and the shaded region is below the line, so that is the first graph of the picture.

Note: strictly speaking, you should include the restrictions that the variables x and y cannot be negative, with which the graph would be only on the first quadrant but those constrains are not handled in the problem.

The graph is also attached.

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