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erik [133]
2 years ago
14

A radar antenna, making one revolution every 5 seconds, is located on a ship that is 6 km from a straight shoreline. How fast is

the radar beam moving along the shoreline when the beam makes an angle of 45◦ with the shore?
Mathematics
1 answer:
Anastasy [175]2 years ago
7 0

Answer:

The figure below is showing the situation:

Figure

Now let the point is at a distance x from the shoreline where it is making the angle of 45 degrees.

So we have:

x

=

6

tan

θ

Now we have:

d

θ

d

t

=

2

π

5

So we have to find the velocity that is given by:

d

x

d

t

=

d

d

θ

(

6

tan

θ

)

=

6

sec

2

(

θ

)

d

θ

d

t

=>

d

x

d

t

=

6

sec

2

(

π

4

)

2

π

5

is the required velocity

Step-by-step explanation:

does that answer your question

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Looking at the image attached, we can see the assumed sales of the Chewy Candy Company for its new candy bar in Manila and Seoul. Setting up the given condition for separate branches, we clearly see that during the 6th month, the total sales in Manila first exceeded the cumulative total sales in Seoul.

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The number of hours a group of contestants spends preparing for a quiz show is listed below. What is the frequency table that re
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I'll just show you how to make a frequency table using the above data. 
We will group the data into class intervals and determine the frequency of the group.

<span>8 12 25 32 45 50 62 73 80 99 4 18 9 39 36 67 33
</span>
smallest data value = 4
highest data value = 99
difference = 99 - 4 = 95
number of data = 17

Let us assign a class interval of 20.

Class Interval     Tally                              Frequency
0-20                    8, 12, 4, 18, 9,                     5
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41-60                45, 50, 67                              3
61-80                62, 73, 80                              3
81-100              99                                           1

That is how a frequency table look like. Usually, under the Tally column, tick marks are written instead of the numbers but for easier monitoring, I used the numbers in the data set.
8 0
2 years ago
A rectangle that had the perimeter of 70 and the area of 250 has enlarged; its sides became twice as long. what is the new area
aliya0001 [1]
The perimeter of the original rectangle is:
 P = 2w + 2l = 70
 The area of the original rectangle is:
 A = w * l = 250
 Then, by modifying the length of its sides we have:
 Perimeter:
 P '= 2 (2w) +2 (2l)
 Rewriting:
 P '= 2 (2w + 2l)
 P '= 2P
 P '= 2 (70)
 P '= 140
 Area:
 A '= (2w) * (2l)
 Rewriting:
 A '= (2) * (2) (w) * (l)
 A '= 4 * w * l
 A '= 4 * A
 A '= 4 * 250
 A '= 1000
 Answer:
 
the new area and the new perimeter are:
 
P '= 140
 
A '= 1000
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2 years ago
4 What is the slope of the graph of y = 12x - 19? ​
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Answer:

This means that 12 is the slope of the line y = 12x - 19.

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The mean annual salary for intermediate level executives is about $74000 per year with a standard deviation of $2500. A random s
lidiya [134]

Answer:

11.51% probability that the mean annual salary of the sample is between $71000 and $73500

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 74000, \sigma = 2500, n = 36, s = \frac{2500}{\sqrt{36}} = 416.67

What is the probability that the mean annual salary of the sample is between $71000 and $73500?

This is the pvalue of Z when X = 73500 subtracted by the pvalue of Z when X = 71000. So

X = 73500

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{73500 - 74000}{416.67}

Z = -1.2

Z = -1.2 has a pvalue of 0.1151

X = 71000

Z = \frac{X - \mu}{s}

Z = \frac{71000 - 74000}{416.67}

Z = -7.2

Z = -7.2 has a pvalue of 0.

0.1151 - 0 = 0.1151

11.51% probability that the mean annual salary of the sample is between $71000 and $73500

8 0
2 years ago
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