Hi i need help on geometry

nata0808 [166]

Try and find the common denominator which is 6 since 2×3 is 6 then once u get ur denominator u have to find the numerator which would be 4/6+3/6 And that would equal 7/6 and if you would u would keep it like that but if u would have to change it into a mixed fraction it would be 1 and 1/6

Hoped this helped

3 0

3 years ago

A bowl contained 250 skittles. The result of 25 fruit-flavored candies pulled at random is given below. Based on this informatio

Anastaziya [24]

OK so you have 250 skittles. But you only pull out 25. You pull out 5 red skittles, 7 yellow skittles,8 orange skittles,2 green skittles and 3 purple skittles. So how many red skittles are in the whole box. The way i would look at it is theres 5 red skittles in every 25 skittles. So whats 250 divided by 25 that would be 10. So you can imagine you have 10 groups of skittles. In each group theres 5 skittles. So all together there would be 50 red skittles. Sooo based on the infromation there would be 50 red skittles expected in the box. I hope this helped pls mark me as brainliest!!

7 0

3 years ago

4. Six workers are hired to weed a field by

user100 [1]

Answer:

594 square feet

Step-by-step explanation:

Area of a rectangle is length times width. We can find the area of each workers plot by multiplying 9 and 11 to get 99 sq feet. There is six workers, and therefore six plots of land, so we'd take our 99 sq feet and count that six times to get 594 square feet.

4 0

3 years ago

N independent random samples of 10 men and 10 women in a coed basketball league, the numbers of points scored per season are giv

faltersainse [42]

Answer:

Step-by-step explanation:

Hello!

a)

To compare both datasets I've made a box plot for them.

The yellow one belongs to the points scored by the men and the green one is for the points scored by the women.

Women: The box seems symmetrical, the median and the mean (black square) are almost the same. The whiskers are almost the same length, you could say that in general the points scored by the women have a symmetrical distribution.

Men: The box looks a little skewed to the right, the median is closer to the 1st quantile and the mean is greater than the median. The whiskers are the same length, in general the distribution seems almost symmetrical.

b)

As you can see in the box plot the median and mean of the points scored by the women are almost the same. Using the data I've calculated both values:

Me= 24.50

X[bar]= 23.80

The median divides the sample in halves (50-50), it shows you where the middle of the distribution is.

The average shows you the value that centers the distribution, meaning, it is the value around which you'll find most of the data set. It summarizes best the sample information and therefore is a better measure of center.

c.

As you can see in the box plots, there are no outliers in both distributions.

An outlier is an observation that is significantly distant from the rest of the data set. They usually represent experimental errors (such as a measurement) or atypical observations. Some statistical measurements, such as the sample mean, are severely affected by this type of values and their presence tends to cause misleading results on a statistical analysis.

Considering the 1st quartile (Q₁), the 3rd quartile (Q₃) and the interquartile range IQR, any value X is considered an outlier if:

X < Q₁ - 1.5 IQR

X > Q₃ + 1.5 IQR

Or extreme outliers if:

X < Q₁ - 3 IQR

X > Q₃ + 3 IQR

For the women data set:

Q₁= 17; Q₃= 30; IQR= 30 - 17= 13

Lower outliers: X < Q₁ - 1.5 IQR= 17-1.5*13= -2.5 ⇒ The minimum value recorded is 03, so there are no outliers.

Upper outliers: X > Q₃ + 3 IQR= 30 + 1.5*13= 49.5 ⇒ The maximum value registered is 41, so there are no outliers.

I hope this helps!