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1 year ago

12

Find the cumulative frequency

1 answer:

aleksandrvk [35]1 year ago

5 0

We get the cumulative frequency as:

3

7

13

22

29

36

37

We are given the data:

Daily Low(°F) Frequency

35 - 39 3

40 - 44 4

45 - 49 6

50 - 54 9

55 - 59 7

60 - 64 7

65 - 69 1

We have to find the cumulative frequency of the data.

We will do this by adding the data values of previous interval.

Daily Low(°F) Frequency cumulative frequency

35 - 39 3 3

40 - 44 4 7

45 - 49 6 13

50 - 54 9 22

55 - 59 7 29

60 - 64 7 36

65 - 69 1 37

Therefore, we get the cumulative frequency as:

3

7

13

22

29

36

37

Learn more about cumulative frequency here:

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Use linear approximation to approximate √25.3 as follows.

Sophie [7]

The idea is to use the tangent line to $f(x)=\sqrt x$ at $x=25$ in order to approximate $f(25.3)=\sqrt{25.3}$ .

We have

$f(x)=\sqrt x\implies f(25)=\sqrt{25}=5$

$f'(x)=\dfrac1{2\sqrt x}\implies f'(25)=\dfrac1{10}$

so the linear approximation to $f(x)$ is

$L(x)=f(5)+f'(5)(x-5)=5+\dfrac{x-5}{10}=\dfrac x{10}+\dfrac92$

Hence $m=\frac1{10}$ and $b=\frac92$ .

Then

$f(25.3)\approx L(25.3)=\dfrac{25.3}{10}+\dfrac92=\boxed{7.03}$

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3 years ago

Examine the division problem

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Step-by-step explanation:

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2 years ago

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A drama club made $3,200 from a school play. Seventy-five percent of that amount came from ticket sales. Identify the values of

NISA [10]

Answer:

75%

Step-by-step explanation:

For this case, the first thing we will do is name variables:

a: percentage from ticket sales

b: ticket sales

We have then:

a = 75%

b = (0.75) * (3200) = $ 2400

Answer:

a = 75%

b = (0.75) * (3200) = $ 2400

The amount of ticket sales is $ 2400.

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1 year ago

Richard wants to find out his current grade in math dass after taking five quizzes. Which measure of center

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Answer:

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3 years ago

Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

Ludmilka [50]

Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

4 0

3 years ago