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Anettt [7]
3 years ago
5

I need help on this question plz

Mathematics
1 answer:
Gelneren [198K]3 years ago
6 0

Hello!

The mean is the average. The median is the middle number of the data set, in terms of quantity. The mode is the number that appears the most. The range is the difference between the largest and smallest numbers in the set.

Therefore, our answers are below.

Range: Difference between largest and smallest data piece.

Median: Middle value of a set of data

Mode: most commonly occurring number(s)

Mean: the sum of the data divided by the quantity of the data items.

I hope this helps!

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How did changes to the voting process affect voter participation?
Jobisdone [24]

Answer:

A) more people voted

Step-by-step explanation:

I did it on edge

4 0
3 years ago
Read 2 more answers
PLEASE I REALLY NEED HELP!
vagabundo [1.1K]

Answer:

\frac{3}{28}

Step-by-step explanation:

Probability (P) is calculated as

P = \frac{requiredoutcome}{count}

The first required outcome is a red sweet from a total of 3 + 5 = 8

P( red) = \frac{3}{8}

There are now 2 red left and a count of 7, since 1 has been eaten, thus

P( second red ) = \frac{2}{7}

P( red and red ) = \frac{3}{8}  × \frac{2}{7} = \frac{6}{56} = \frac{3}{28}

6 0
3 years ago
How many sixteenths sure in 15/16
Feliz [49]

Answer:

15

Step-by-step explanation:

there are 15 sixteenths in 15/16, how you figure this out is the numerator tells you how many are in the fraction, in this case the numerator says 15, so there are 15, sixteenths in, 15/16

8 0
3 years ago
1. if csc β = 7/3 and cot β = - 2√10 / 3, Find sec β
slava [35]

Step-by-step explanation:

1.

\tan \beta  =  \frac{1}{ \cot \beta }  =  -  \frac{3}{2 \sqrt{10} }  =  -  \frac{3 \sqrt{10} }{20}

\csc \beta  \tan \beta  =  \frac{1}{ \cos \beta  }  =  \sec \beta

Therefore,

\sec \beta  = ( \frac{7}{3} )( -  \frac{3 \sqrt{10} }{20} ) =  -  \frac{7 \sqrt{10} }{20}

2.

\csc y =  \frac{1}{ \sin y}  =  -  \frac{ \sqrt{6} }{2}

=  >  \sin y =  -  \frac{ \sqrt{6} }{3}

Use the identity

\cos y =   \sqrt{1 -  \sin ^{2} y}    \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \\ =  \sqrt{1 -  {( -  \frac{ \sqrt{6} }{3}) }^{2} }  =  -  \frac{ \sqrt{3} }{3}

We chose the negative value of the cosine because of the condition where cot y > 0. Otherwise, choosing the positive root will yield a negative cotangent value. Now that we know the sine and cosine of y, we can now solve for the tangent:

\tan \beta  =  \frac{ \sin y}{ \cos y} =( -  \frac{ \sqrt{6} }{3} )( -  \frac{3}{ \sqrt{3} } ) =  \sqrt{2}

3. Recall that sec x = 1/cos x, therefore cos x = 5/6. Solving for sin x,

\sin x =   \sqrt{1 -  \cos ^{2} x} =  \sqrt{ \frac{11}{6} }

Solving for tan x:

\tan x =  \frac{ \sin x}{ \cos x}  =  (\frac{ \sqrt{11} }{ \sqrt{6} } )( \frac{6}{5} ) =  \frac{ \sqrt{66} }{5}

5 0
3 years ago
Find f(-2) for f(x)=3•2^x
8_murik_8 [283]

Answer:

0.75 or 3/4

Step-by-step explanation:

f(-2) = 3•2^-2

f(-2) = 3•0.25

f(-2) = 0.75 or 3/4

Hope this helps! :)

5 0
3 years ago
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