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Angelina_Jolie [31]
3 years ago
10

A camper wants to know the width of a river. From point A, he walks downstream 60 feet to point B and sights a canoe across the

river. It is determined that \alpha = 34°. About how wide is the river?
A. 34 feet
B. 50 feet
C. 89 feet
D. 40 feet

Mathematics
2 answers:
Anarel [89]3 years ago
6 0
<h2>Hello!</h2>

The answer is:

The correct option is:

D. 40 feet.

<h2>Why?</h2>

To solve the problem and calculate the width of the river, we need to assume that the distance from A to B and the angle formed between that distance and the distance from A to the other point (C) is equal to 90°, meaning that we are working with a right triangle, also, we need to use the given angle which is equal to 34°. So, to solve the problem we can use the following trigonometric relation:

Tan\alpha =\frac{Opposite}{Adjacent}

Where,

alpha is the given angle, 34°

Adjacent is the distance from A to B, which is equal to 60 feet.

Opposite is the distance from A to C which is also equal to the width of the river.

So, substituting and calculating we have:

Tan(34\°) =\frac{Width}{60ft}

Width=60ft*Tan(34\°)=60ft*0.67=40.2ft=40ft

Hence, we have that the correct option is:

D. 40 feet.

Have a nice day!

Veseljchak [2.6K]3 years ago
5 0

Answer: OPTION D

Step-by-tep explanation:

Observe the figure attached.

You can notice that the the width of the river is represented with "x".

To calculate it you need to use this identity:

tan\alpha=\frac{opposite}{adjacent}

In this case:

\alpha=34\°\\opposite=x\\adjacent=60

Now you must substitute values:

 tan(34\°)=\frac{x}{60}

And solve for "x":

60*tan(34\°)=x\\\\x=40.4ft

x≈40ft

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3 years ago
Suppose it is known that the distribution of purchase amounts by customers entering a popular retail store is approximately norm
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Answer:

a. 0.691

b. 0.382

c. 0.933

d. $88.490

e. $58.168

f. 5th percentile: $42.103

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Step-by-step explanation:

We have, for the purchase amounts by customers, a normal distribution with mean $75 and standard deviation of $20.

a. This can be calculated using the z-score:

z=\dfrac{X-\mu}{\sigma}=\dfrac{85-75}{20}=\dfrac{10}{20}=0.5\\\\\\P(X

The probability that a randomly selected customer spends less than $85 at this store is 0.691.

b. We have to calculate the z-scores for both values:

z_1=\dfrac{X_1-\mu}{\sigma}=\dfrac{65-75}{20}=\dfrac{-10}{20}=-0.5\\\\\\z_2=\dfrac{X_2-\mu}{\sigma}=\dfrac{85-75}{20}=\dfrac{10}{20}=0.5\\\\\\\\P(65

The probability that a randomly selected customer spends between $65 and $85 at this store is 0.382.

c. We recalculate the z-score for X=45.

z=\dfrac{X-\mu}{\sigma}=\dfrac{45-75}{20}=\dfrac{-30}{20}=-1.5\\\\\\P(X>45)=P(z>-1.5)=0.933

The probability that a randomly selected customer spends more than $45 at this store is 0.933.

d. In this case, first we have to calculate the z-score that satisfies P(z<z*)=0.75, and then calculate the X* that corresponds to that z-score z*.

Looking in a standard normal distribution table, we have that:

P(z

Then, we can calculate X as:

X^*=\mu+z^*\cdot\sigma=75+0.67449\cdot 20=75+13.4898=88.490

75% of the customers will not spend more than $88.49.

e. In this case, first we have to calculate the z-score that satisfies P(z>z*)=0.8, and then calculate the X* that corresponds to that z-score z*.

Looking in a standard normal distribution table, we have that:

P(z>-0.84162)=0.80

Then, we can calculate X as:

X^*=\mu+z^*\cdot\sigma=75+(-0.84162)\cdot 20=75-16.8324=58.168

80% of the customers will spend more than $58.17.

f. We have to calculate the two points that are equidistant from the mean such that 90% of all customer purchases are between these values.

In terms of the z-score, we can express this as:

P(|z|

The value for z* is ±1.64485.

We can now calculate the values for X as:

X_1=\mu+z_1\cdot\sigma=75+(-1.64485)\cdot 20=75-32.897=42.103\\\\\\X_2=\mu+z_2\cdot\sigma=75+1.64485\cdot 20=75+32.897=107.897

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Quadrilateral ABCD is similiar to quadrilateral EFGH. The lengths of the three longest sides in quadrilateral ABCD are 24 feet,
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The two quadrilaterals are given similar .

In any similar figure sides are in proportion.The second largest side of quadrilateral ABCD is 16 ft .Let the second longest side of quadrilateral EFGH be x ft. These sides will be in proportion to each other .

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To solve for x we cross multiply

12x=(16)(18)

12x=288

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