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zhannawk [14.2K]

2 years ago

8

What will be the value of median of a moderately asymmetrical distribution? If the mean and mode are 30 and 24 respectively.

1 answer:

sineoko [7]2 years ago

7 0

The mean, the median and the mode are measures of central tendency, and they are related in some many ways, depending on the type of distribution.

The median of the moderately asymmetrical distribution is 26

The given parameters are:

$Mean = 30$

$Mode = 24$

For a moderately asymmetrical distribution, the mean, median and mode are related by the following equation:

$Mode = 3 * Median - 2 * Mean$

Substitute known values

$30= 3 * Median - 2 * 24$

$30= 3 * Median - 48$

Collect like terms

$3 * Median =30+ 48$

$3 * Median =78$

Divide both sides by 3

$Median =26$

Read more at:

brainly.com/question/15669207

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9(4b-1)=2(b+3) Slove for b plz answer as quick as u can please

Jobisdone [24]

$36b - 9 = 2b + 6 \\ 34b = 15 \\ b = \frac{15}{34}$

3 0

3 years ago

Measure the lengths of the sides of ∆ABC in GeoGebra, and compute the sine and the cosine of ∠A and ∠B. Verify your calculations

marusya05 [52]

Answer:

$Sin \angle A =0.80$

$Cos \angle A=0.60$

$Sin \angle B =0.60$

$Cos \angle B=0.80$

Step-by-step explanation:

Given

I will answer this question using the attached triangle

Solving (a): Sine and Cosine A

In trigonometry:

$Sin \theta =\frac{Opposite}{Hypotenuse}$ and

$Cos \theta =\frac{Adjacent}{Hypotenuse}$

So:

$Sin \angle A =\frac{BC}{BA}$

Substitute values for BC and BA

$Sin \angle A =\frac{8cm}{10cm}$

$Sin \angle A =\frac{8}{10}$

$Sin \angle A =0.80$

$Cos \angle A=\frac{AC}{BA}$

Substitute values for AC and BA

$Cos \angle A=\frac{6cm}{10cm}$

$Cos \angle A=\frac{6}{10}$

$Cos \angle A=0.60$

Solving (b): Sine and Cosine B

In trigonometry:

$Sin \theta =\frac{Opposite}{Hypotenuse}$ and

$Cos \theta =\frac{Adjacent}{Hypotenuse}$

So:

$Sin \angle B =\frac{AC}{BA}$

Substitute values for AC and BA

$Sin \angle B =\frac{6cm}{10cm}$

$Sin \angle B =\frac{6}{10}$

$Sin \angle B =0.60$

$Cos \angle B=\frac{BC}{BA}$

Substitute values for BC and BA

$Cos \angle B=\frac{8cm}{10cm}$

$Cos \angle B=\frac{8}{10}$

$Cos \angle B=0.80$

Using a calculator:

$A = 53^{\circ}$

So:

$Sin(53^{\circ}) =0.7986$

$Sin(53^{\circ}) =0.80$ -- approximated

$Cos(53^{\circ}) = 0.6018$

$Cos(53^{\circ}) = 0.60$ -- approximated

$B = 37^{\circ}$

So:

$Sin(37^{\circ}) = 0.6018$

$Sin(37^{\circ}) = 0.60$ --- approximated

$Cos(37^{\circ}) = 0.7986$

$Cos(37^{\circ}) = 0.80$ --- approximated

8 0

3 years ago

Read 2 more answers

If ef bisects angle ceb,angle cef=7x+31 and angle feb=10x-3

Nastasia [14]

Given : Angle < CEB is bisected by EF.

< CEF = 7x +31.

< FEB = 10x-3.

We need to find the values of x and measure of < FEB, < CEF and < CEB.

Solution: Angle < CEB is bisected into two angles < FEB and < CEF.

Therefore, < FEB = < CEF.

Substituting the values of < FEB and < CEF, we get

10x -3 = 7x +31

Adding 3 on both sides, we get

10x -3+3 = 7x +31+3.

10x = 7x + 34

Subtracting 7x from both sides, we get

10x-7x = 7x-7x +34.

3x = 34.

Dividing both sides by 3, we get

x= 11.33.

Plugging value of x=11.33 in < CEF = 7x +31.

We get

< CEF = 7(11.33) +31 = 79.33+31 = 110.33.

< FEB = < CEF = 110.33 approximately

< CEB = < FEB + < CEF = 110.33 +110.33 = 220.66 approximately

7 0

3 years ago

Lee drove 420 miles and used 15 gallons pf gasoline.How many miles did Lee's car travel per gallon of gasoline?

kondor19780726 [428]

Divide the two numbers. The equation is 420/15=28. So Lee would get 28 miles per gallon of gasoline. I hope this helps!

3 0

3 years ago

When f(n)=17, what is the value of n?

Scilla [17]

N is the same as 17, so it equals 17.

3 0

3 years ago