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Dima020 [189]
3 years ago
6

ron wants to build a ramp with a length of 14 ft and an angle of elevation of 26° the height of the ramp is about 6.1 ft. what i

s the length of the base of the ramp?
Mathematics
1 answer:
Bezzdna [24]3 years ago
4 0

Answer:

12.58 feet.

Step-by-step explanation:

Ron wants to build a 14 ft. ramp with an angle of the rise of 26° and the height of the ramp is about 6.1 ft.

We have to find out the length of the base of the ramp.

Now, the ramp forms a right triangle with the base and the height and here, the hypotenuse of the triangle is the ramp itself i.e. 14 ft.

If the base length is x feet, then we can write using the trigonometry

\cos 26^{\circ} =  \frac{\textrm{Base}}{\textrm {Hypotenuse}} = \frac{x}{14}

⇒ x = 14 cos 26° = 12.58 feet. (Answer)

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Step-by-step explanation:

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(1) [6pts] Let R be the relation {(0, 1), (1, 1), (1, 2), (2, 0), (2, 2), (3, 0)} defined on the set {0, 1, 2, 3}. Find the foll
goldenfox [79]

Answer:

Following are the solution to the given points:

Step-by-step explanation:

In point 1:

The Reflexive closure:  

Relationship R reflexive closure becomes achieved with both the addition(a,a) to R Therefore, (a,a) is  (0,0),(1,1),(2,2) \ and \ (3,3)

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In point 2:

The Symmetric closure:

R relation symmetrically closes by adding(b,a) to R for each (a,b) of R  Therefore, here (b,a) is:   (0,1),(0,2)\ and \ (0,3)

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8 0
2 years ago
A food company sells salmon to various customers. The mean weight of the salmon is 44 lb with a standard deviation of 3 lbs. The
TiliK225 [7]

Correct question:

A food company sells salmon to various customers. The mean weight of the salmon is 44 lb with a standard deviation of 3 lbs. The company ships them to restaurants in boxes of 9 ​salmon, to grocery stores in cartons of 16 ​salmon, and to discount outlet stores in pallets of 64 salmon. To forecast​ costs, the shipping department needs to estimate the standard deviation of the mean weight of the salmon in each type of shipment. Complete parts​ (a) and​ (b) below.

a. Find the standard deviations of the mean weight of the salmon in each type of shipment.

b. The distribution of the salmon weights turns out to be skewed to the high end. Would the distribution of shipping weights be better characterized by a Normal model for the boxes or pallets?

Answer:

Given:

Mean, u = 44

Sd = 3

The company ships in boxes of 9, cartons of 16 and pallets of 64.

a) For the standard deviations of the mean weight of the salmon in each type of shipment, lets use the formula: \frac{s.d}{\sqrt{u}}

i) For the standard deviation of the mean weight of salmon in boxes of 9, we have:

\frac{s.d}{\sqrt{u}}

= \frac{3}{\sqrt{9}}

= \frac{3}{3} = 1

The standard deviation = 1

ii) For the standard deviation of the mean weight of salmon in cartons of 16, we have:

\frac{s.d}{\sqrt{u}}

= \frac{3}{\sqrt{16}}

= \frac{3}{4} = 0.75

Standard deviation = 0.75

iii) For the standard deviation of the mean weight of salmon in pellets of 64, we have:

\frac{s.d}{\sqrt{u}}

= \frac{3}{\sqrt{64}}

= \frac{3}{8} = 0.375

Standard deviation = 0.375

b) The distribution of shipping weights would be better characterized by a Normal model for the pallets, because regardless of the underlying distribution, the sampling distribution of the mean approaches the Normal model as the sample increases.

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