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Nimfa-mama [501]
3 years ago
7

A tree that is 40 feet tall casts a 30-foot shadow. At the same time, another tree casts a 20-foot shadow. How tall is the secon

d tree?
Mathematics
2 answers:
Olegator [25]3 years ago
7 0
So the tree is 40 ft and then the shadow is 30 so then do 40 -30 = 10 ft difference so then for the second tree do 20 + 10 = 30 ft because the 10 ft difference so the answer is 30 ft tall
charle [14.2K]3 years ago
5 0

Answer:

The second tree is 26.67 foot tall.

Step-by-step explanation:

A tree that is 40 feet tall casts a 30-foot shadow.

\texttt{Ratio of original height to shadow height =}\frac{40}{30}\\\\\texttt{Ratio of original height to shadow height =}\frac{4}{3}

Now we need to find original height of a tree whose shadow is 20 foot.

Original height = Ratio of original height to shadow height x Height of shadow

\texttt{Original height = }\frac{4}{3}\times 20=26.67foot

The second tree is 26.67 foot tall.

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Justin travels at 70 km/h on his motorcycle. If the speed limit is 50 mi./hr., is he speeding? Why or why not?
balu736 [363]

Answer:

Justin is <em>not </em>speeding because there is 1.609 km in a mile. So if you compare it to 70 km/h to 50 mi/h it would 70 km and 43.50 miles per hour. So like I said, no he is not because he is going below 50 mi/h.

6 0
3 years ago
Read 2 more answers
One of the earliest applications of the Poisson distribution was in analyzing incoming calls to a telephone switchboard. Analyst
grandymaker [24]

Answer:

(a) P (X = 0) = 0.0498.

(b) P (X > 5) = 0.084.

(c) P (X = 3) = 0.09.

(d) P (X ≤ 1) = 0.5578

Step-by-step explanation:

Let <em>X</em> = number of telephone calls.

The average number of calls per minute is, <em>λ</em> = 3.0.

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 3.0.

The probability mass function of a Poisson distribution is:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,3...

(a)

Compute the probability of <em>X</em> = 0 as follows:

P(X=0)=\frac{e^{-3}3^{0}}{0!}=\frac{0.0498\times1}{1}=0.0498

Thus, the  probability that there will be no calls during a one-minute interval is 0.0498.

(b)

If the operator is unable to handle the calls in any given minute, then this implies that the operator receives more than 5 calls in a minute.

Compute the probability of <em>X</em> > 5  as follows:

P (X > 5) = 1 - P (X ≤ 5)

              =1-\sum\limits^{5}_{x=0} { \frac{e^{-3}3^{x}}{x!}} \,\\=1-(0.0498+0.1494+0.2240+0.2240+0.1680+0.1008)\\=1-0.9160\\=0.084

Thus, the probability that the operator will be unable to handle the calls in any one-minute period is 0.084.

(c)

The average number of calls in two minutes is, 2 × 3 = 6.

Compute the value of <em>X</em> = 3 as follows:

<em> </em>P(X=3)=\frac{e^{-6}6^{3}}{3!}=\frac{0.0025\times216}{6}=0.09<em />

Thus, the probability that exactly three calls will arrive in a two-minute interval is 0.09.

(d)

The average number of calls in 30 seconds is, 3 ÷ 2 = 1.5.

Compute the probability of <em>X</em> ≤ 1 as follows:

P (X ≤ 1 ) = P (X = 0) + P (X = 1)

             =\frac{e^{-1.5}1.5^{0}}{0!}+\frac{e^{-1.5}1.5^{1}}{1!}\\=0.2231+0.3347\\=0.5578

Thus, the probability that one or fewer calls will arrive in a 30-second interval is 0.5578.

5 0
3 years ago
Let x and y represent two real numbers. Write an algebraic expression to denote the quotient obtained when the product of the tw
sertanlavr [38]
(xy)/(x+y) hope this helps
5 0
3 years ago
What is the midpoint of the segment (1, -1) (4, -6)
Sophie [7]
When looking for the midpoint of a segment defined by two end points, the average of both coordinates are taken. Averaging the 2 x-coordinates give the new x-coordinate, and the same applies for the y-coordinate. This is shown below:

Midpoint = ( (1 + 4)/2 , (-1 + -6)/2 )
Midpoint = (2.5 , -3.5)
6 0
3 years ago
Could u help me make sure its the right answer i can't get this wrong .
Black_prince [1.1K]

Answer:

12x + 6y = 102

5x - 6y = 34

17x = 136

x = 8

y = 1

5 0
3 years ago
Read 2 more answers
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