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marin [14]
3 years ago
5

At sea, the distance D to the horizon is directly proportional to the square root of the elevation E of the observer. If a perso

n who is 36 feet above the water can see 7.4 miles, find how far a person 49 feet above the water can see.
Mathematics
1 answer:
Alla [95]3 years ago
7 0
<span>The problem says that the distance D to the horizon is directly proportional to the square root of the elevation E of the observer, so:
</span>
 The person who is 36 feet above the water ⇒ √36=6 

 A person 49 feet above the water ⇒ √49=7
<span>
 So, by proportions, you have:

 (7/6)x7.4= 8.63 miles

 Therefore, the answer is: </span><span>a person 49 feet above the water can see 8.63 miles</span><span> </span>
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Matt earns $5 for each lawn he does. He wants to earn at least $100. Which inequality represents his situation?
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Answer:

$100 is less than or equal to x

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3 years ago
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12% of children are nearsighted, but this condition often is not detected until they go to kindergarten. A school district tests
Klio2033 [76]

Answer:

There is a 34.60% probability that 0 or 1 of them is nearsighted.

Step-by-step explanation:

For each children, there are only two possible outcomes. Either they are nearsighted, or they are not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem, we have that:

12% of children are nearsighted. This means that p = 0.12.

A school district tests all incoming kindergarteners' vision. In a class of 18 kindergarten students, what is the probability that 0 or 1 of them is nearsighted?

There are 18 students, so n = 18

This probability is:

P = P(X = 0) + P(X = 1)

So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{18,0}.(0.12)^{0}.(0.88)^{18} = 0.1002

P(X = 1) = C_{18,1}.(0.12)^{1}.(0.88)^{17} = 0.2458

So

P = P(X = 0) + P(X = 1) = 0.1002 + 0.2458 = 0.3460

There is a 34.60% probability that 0 or 1 of them is nearsighted.

5 0
3 years ago
6463/4 show your work first to answer get brainlist
masya89 [10]

Answer:

1 6 1 3 r.11

Step-by-step explanation:

  1 6 1 3 r.11

4\sqrt{6463\\}

   -4

_________________\\____________

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    -24

____________

        63

       -4

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          23

          -12

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7 0
3 years ago
A rectangular prism has a height of 12 cm and a square base with sides measuring 5 cm. A pyramid with the same base and the heig
Ksju [112]
The answer is 200 cm³


The volume of the rectangular prism (V1) is:
V1 = l · w · h                       (l - length,  w - width,  h - height)
It is given:
h = 12 cm
w = l = 5 cm (since it has a square base which all sides are the same size).
Thus: V1 = 12 · 5 · 5 = 300 cm³

The volume of pyramid (V2) is:
V2 = 1/3 · l · w · h                   (l - length,  w - width,  h - height)
It is given:
h = 12 cm
w = l = 5 cm (since it has a square base which all sides are the same size).
V2 = 1/3 · 12 · 5 · 5 = 1/3 · 300 = 100 cm³


The volume of the space outside the pyramid but inside the prism (V) is a difference between the volume of the rectangular prism (V1) and the volume of the pyramid (V2): 
V = V1 - V2 = 300 cm³ - 100 cm³ = 200 cm³
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3 years ago
When 1,250
SSSSS [86.1K]

Answer:

2

Step-by-step explanation:

1250/5 = 250/5 = 50/ 5 = 10/5 = 2

So we pulled sqrt 5, sqrt 5, sqrt 5, sqrt 5

sqrt 5 * sqrt 5 = 5 (do twice)

25 sqrt 2

3 0
2 years ago
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