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Debora [2.8K]
2 years ago
12

HIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII

Mathematics
1 answer:
Advocard [28]2 years ago
3 0

Answer:

HIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII

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A weight clinic recorded the weight lost (in pounds) by each client of a weight control clinic during the last year, and got the
borishaifa [10]

Answer:

Class interval ____, Frequency _ C/frequency

1 - 10 ____________ 1 _________ 1

11 - 20 ___________ 4 _________ 5

20 - 30 __________ 6 _________ 11

31 - 40 ___________3 _________ 14

41 - 50 ___________ 1 _________ 15

51 - 60 ___________ 1 _________ 16

Step-by-step explanation:

Given the data :

35, 26, 31, 17, 46, 30, 28, 21, 26, 34, 15, 27, 7, 18, 16, 57

Class interval ____, Frequency _ C/frequency

1 - 10 ____________ 1 _________ 1

11 - 20 ___________ 4 _________ 5

20 - 30 __________ 6 _________ 11

31 - 40 ___________3 _________ 14

41 - 50 ___________ 1 _________ 15

51 - 60 ___________ 1 _________ 16

The next to highest frequency group has a frequency of 4 and the highest frequency of 6

Total frequency, n = (1 + 4 + 6 + 3 + 1 + 1) = 16

5 0
3 years ago
Which shows the ratio 16 : 100 as a fraction in simplified form?
Alika [10]

Answer:

16 : 100 = 8 : 50 = 4 : 25

Step-by-step explanation:

You just half each side by the same number

7 0
3 years ago
Need help anyone can help giving brain liest
trapecia [35]

Answer:

14 gel pens/minute

Step-by-step explanation:

divide the amount by time.

210/15 = 14

6 0
3 years ago
Read 2 more answers
Reynaldo is making a model of his school building. The actual building is 28
tankabanditka [31]

Answer:

It's 22.5 inches because the breadth is grater than the length

7 0
3 years ago
Read 2 more answers
If tanA=a <br>then find sin4A-2sin2A/ sin4A+2sin2A​
anygoal [31]

Answer:

The value of the given expression is

\frac{sin4A-2sin2A}{sin4A+2sin2A}=-a^2

Step by step Explanation:

Given that tanA=a

To find the value of \frac{sin4A-2sin2A}{sin4A+2sin2A}

Let us find the value of the expression :

\frac{sin4A-2sin2A}{sin4A+2sin2A}=\frac{2cos2Asin2A-2sin2A}{2cos2Asin2A+2sin2A} ( by using the formula sin2A=2cosAsinA here A=2A)  

=\frac{2sin2A(cos2A-1)}{2sin2A(cos2A+1)}

=\frac{(cos2A-1)}{(cos2A+1)}

=\frac{(-(1-cos2A))}{(1+cos2A)}(using  sin^2A+cos^2A=1  here A=2A)

=\frac{-(sin^2A+cos^2A-(cos^2A-sin^2A))}{sin^2A+cos^2A+(cos^2A-sin^2A)}(using cos2A=cos^2A-sin^2A here A=2A)

=\frac{-(sin^2A+cos^2A-cos^2A+sin^2A)}{sin^2A+cos^2A+(cos^2A-sin^2A)}

=\frac{-(sin^2A+sin^2A)}{cos^2A+cos^2A}

=\frac{-2sin^2A}{2cos^2A}

=-\frac{sin^2A}{cos^2A}

=-tan^2A  ( using tanA=\frac{sinA}{cosA} here A=2A )

 =-a^2 (since tanA=a given )

Therefore \frac{sin4A-2sin2A}{sin4A+2sin2A}=-a^2

6 0
3 years ago
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