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3 years ago

State whether either extreme in the data set is an outlier.

Mathematics

1 answer:

Drupady [299]3 years ago

5 0

<span>Both extremes in the data set are considered to be outliers because they are outside of the range (89-106)</span>

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Iq scores are known to be normally distributed. the mean iq score is 100 and the standard deviation is 15. what percent of the p

kobusy [5.1K]

The 100%-Norm.Dist(value,mean,sd,true) percent of the population has an iq over 115

According to the statement

we have to find that the what percent of the population has an iq over 115.

And from given information,

the mean iq score is 100 and the standard deviation is 15.

And

Normal distribution, is a probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean.

And by use of this formula, we find the percentage of the population has an iq over 115.

Then the result will comes as 100%.

hence, 100%-Norm.Dist(value,mean,sd,true)

So, The 100%-Norm.Dist(value,mean,sd,true) percent of the population has an iq over 115.

Learn more about Normal distribution here

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2 years ago

3(32 − 4 + 6 )− (82 + + 3) ?<br> simpilyfy<br> plzz

IrinaVladis [17]

Answer:

17 is your answer

Step-by-step explanation:

6 0

3 years ago

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Jerry’s rhombus has a base of 15 inches and a height of 9 inches. Charlene’s rhombus has a base and a height that are 13 the bas

neonofarm [45]

Answer: The area of Charlene's rhombus is nine times smaller than the area of Jerry's rhombus.

Step-by-step explanation:

I will assume that the exercise says " $\frac{1}{3}$ times the base and height of Jerry’s rhombus".

The area of a rhombus can be calculated with the following formula:

$A=b*a$

Where "b" is the length of the base and "a" is the altitude or the height.

Then, you can calculate the area using the formula shown above.

Therefore, you get:

1. Jerry's rhombus:

$A_1=(15\ in)(9\ in)\\\\A_1=135\ in^2$

2. Charlene's rhombus:

$A_2=(\frac{1}{3}*15\ in)(\frac{1}{3}*9\ in)\\\\A_2=(5\ in)(3\ in)\\\\A_2=15\ in^2$

Dividing the area calculated, you get:

$\frac{135\ in^2}{15\ in^2}=9$

Therefore, you can conclude that the area of Charlene's rhombus is nine times smaller than the area of Jerry's rhombus.

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3 years ago

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Lynna [10]

Answer:

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Step-by-step explanation:

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3 years ago

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A candle in the shape of a circular cone has a base of radius r and a height of h that is the same length as the radius. which e

ladessa [460]

Answer:

$\frac{r(1-\sqrt{2})}{-3}$

Step-by-step explanation:

Volume of cone = $\frac{1}{3} \pi r^{2} h$

Since we are given that a circular cone has a base of radius r and a height of h that is the same length as the radius

= $\frac{1}{3} \pi r^{2} \times r$

= $\frac{1}{3} \pi r^{3}$

Surface area of cone including 1 base = $\pi r^{2} +\pi\times r \times \sqrt{r^2+h^2}$

Since r = h

So, area = $\pi r^{2} +\pi\times r \times \sqrt{r^2+r^2}$

= $\pi r^{2} +\pi\times r \times \sqrt{2r^2}$

= $\pi r^{2} +\pi\times r^2 \times \sqrt{2}$

Ratio of volume of cone to its surface area including base :

$\frac{\frac{1}{3} \pi r^{3}}{\pi r^{2} +\pi\times r^2 \times \sqrt{2}}$

$\frac{\frac{1}{3}r}{1+\sqrt{2}}$

$\frac{r}{3(1+\sqrt{2})}$

Rationalizing

$\frac{r}{3(1+\sqrt{2})} \times \frac{1-\sqrt{2}}{1-\sqrt{2}}$

$\frac{r(1-\sqrt{2})}{-3}$

Hence the ratio the ratio of the volume of the candle to its surface area(including the base) is $\frac{r(1-\sqrt{2})}{-3}$

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3 years ago

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