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Veseljchak [2.6K]

3 years ago

12

Miss Campbell has just bought a rectangular displayboard for her bedroom. It has a length of 90 cm and a width of 50 cm. Miss Ca

mpbell would like to put a border of yellow ribbon around the perimeter of her display board.

1 answer:

AVprozaik [17]3 years ago

4 0

Answer:

Area - l*w

area is 4500 cm 2

perimeter - l + w * 2

perimeter is 280 cm

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3.) Identify all integers.<br> a. 725<br> d. 50<br> e. 10.2<br> C. 806

NeTakaya

Answer:

A, B, C

Step-by-step explanation:

An integer is a whole number (so not a fraction)

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If the ratio of the purple flowers to black flowers is 4 to 8 and there are a total of 108 flowers, how many of the flowers are

tankabanditka [31]

There would be 36 purple flowers.

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3 years ago

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Find the (a) mean, (b) median, (c) mode, and (d) midrange for the data and then (e) answer the given question. Listed below

mafiozo [28]

Answer:

a) $\bar X = 369.62$

b) $Median=175$

c) $Mode =450$

With a frequency of 4

d) $MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5$

<u>e)</u> $s = 621.76$

And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

Step-by-step explanation:

We have the following data set given:

49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000

Part a

The mean can be calculated with this formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

Replacing we got:

$\bar X = 369.62$

Part b

Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:

$Median=175$

Part c

The mode is the most repeated value in the sample and for this case is:

$Mode =450$

With a frequency of 4

Part d

The midrange for this case is defined as:

$MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5$

Part e

For this case we can calculate the deviation given by:

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And replacing we got:

$s = 621.76$

And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

5 0

3 years ago

Please help me with this

sesenic [268]

Ok so the average is 50

average of x number of things is
(add value of all things)/x=average

4 test scores needed
so divide by 4
3 test scores ar known
average is 50
x= missing score

(43+48+42+x)/4=50
times 4 both sides
133+x=200
minus 133 both sides
x=67

a score of 67 is needed

5 0

3 years ago