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sweet [91]
3 years ago
13

A cat is watching a bird in a tree nearby. The tree is approximately 20 ft from the cat (ground distance). If the cat’s line of

sight makes a 25° with the ground when he has his eye on the bird, how high up is the bird in the tree?
A. Draw a picture


B. Solve the problem, Solve to the nearest ft.

Mathematics
1 answer:
Masja [62]3 years ago
6 0

Answer:

9.3 feet

Step-by-step explanation:

See attachment for the picture.

Let f represent how high up the bird is in the tree.

This side length of the right triangle is opposite the given angle which is 25 degrees.

The side length adjacent to the given angle is 20 ft.

We use the tangent ratio to obtain:

\tan 25\degree=\frac{opposite}{adjacent}

We substitute the values to obtain:

\tan 25\degree=\frac{f}{20}

f=20\tan 25\degree

f=9.3 ft

Therefore the bird is 9.3 feet high up in the tree.

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John has two jobs. For daytime work at a jewelry store he is paid
djyliett [7]

Given Information:

John's mean monthly commission = μ = $10,000

Standard deviation of monthly commission = σ =  $2,000

Answer:

P(9,000 < X < 11,000) = 0.383\\\\P(9,000 < X < 11,000) = 38.3 \%

The probability that John's commission from the jewelry store is  between $9,000 and $11,000 is 38.3%

Step-by-step explanation:

What is Normal Distribution?

We are given a Normal Distribution, which is a continuous probability distribution and is symmetrical around the mean. The shape of this distribution is like a bell curve and most of the data is clustered around the mean. The area under this bell shaped curve represents the probability.  

We want to find out the probability that John's commission from the jewelry store is  between $9,000 and $11,000?

P(9,000 < X < 11,000) = P( \frac{x - \mu}{\sigma} < Z < \frac{x - \mu}{\sigma} )\\\\P(9,000 < X < 11,000) = P( \frac{9,000 - 10,000}{2,000} < Z < \frac{11,000 - 10,000}{2,000} )\\\\P(9,000 < X < 11,000) = P( \frac{-1,000}{2,000} < Z < \frac{1,000}{2,000} )\\\\P(9,000 < X < 11,000) = P( -0.5 < Z < 0.5 )\\\\P(9,000 < X < 11,000) = P( Z < 0.5 ) - P( Z < -0.5 ) \\\\

The z-score corresponding to 0.50 is 0.6915

The z-score corresponding to -0.50 is 0.3085

P(9,000 < X < 11,000) = 0.6915 - 0.3085 \\\\P(9,000 < X < 11,000) = 0.383\\\\P(9,000 < X < 11,000) = 38.3 \%

Therefore, the probability that John's commission from the jewelry store is  between $9,000 and $11,000 is 38.3%

How to use z-table?

Step 1:

In the z-table, find the two-digit number on the left side corresponding to your z-score. (e.g 1.4, 2.2, 0.5 etc.)

Step 2:

Then look up at the top of z-table to find the remaining decimal point in the range of 0.00 to 0.09. (e.g. if you are looking for 0.50 then go for 0.00 column)

Step 3:

Finally, find the corresponding probability from the z-table at the intersection of step 1 and step 2.

4 0
3 years ago
If NO=17 and NP=5x-6, what is the value of X. Also how do you find X.
Anarel [89]
Are NO and NP congruent?
7 0
2 years ago
What is the outlier of 29, 2, 28, 30, 26, 31
Vladimir [108]

Answer:

2

Step-by-step explanation:

Given the data : 29, 2, 28, 30, 26, 31

Outlier ;

Lower :Q1 - (1.5 * IQR)

Upper : Q3 + (1.5 * IQR)

Q1 = Lower quartile ; Q3 = upper quartile ; IQR = Interquartile range

Using calculator :

Q1 = 26

Q3 = 30

IQR = (Q3 - Q1) = 30 - 26 = 4

Lower : 26 - (1.5 * 4) = 20

Upper : 30 + (1.5 * 4) = 36

Hence, the number in the given data which falls outside the range is 2

4 0
3 years ago
What is the solution set of the given equation?<br> 4y - 6y + 9y = -2
maw [93]
4y - 6y + 9y = -2
      -2y + 9y = -2
               7y = -2
                7      7
                 y = ⁻²/₇
8 0
3 years ago
Read 2 more answers
The side lengths of a square are each 4p. By adding 3 to the length and subtracting 3 from the width, a rectangle is made. What
AysviL [449]

Answer:

area will be

16p {}^{2}  - 9

Step-by-step explanation:

after addition of 3, the length will be =》(4p+3)

after subtracting 3, the width will be =》(4p-3)

area = length × width

  • (4p+3) (4p-3)
  • 16p^2 - 9

the area will be

16p {}^{2}   - 9

<em>i</em><em> </em><em>hope</em><em> </em><em>it</em><em> </em><em>helped</em><em>.</em><em>.</em><em>.</em><em>.</em>

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3 years ago
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