Aslam Farid and majid are going to buy a shop that costs 54 million rupees .if Aslam pays 5 million rupees more than Farid and F

Response:

Detailed explanation:

We are aware that

She made the trip between her residence and the airport twice.

Therefore,

Multiply the distance from her house to the airport by 4.

If point M is located on segment LN, then using the segment addition postulate, we have:

LN is equal to LM plus MN
LN = 22 + 15
Therefore, LN = 37

Answer:

(a) The data point that is considered an outlier is $810.

(b) The distribution regarding expenditures on textbooks among the 34 students is right skewed.

Step-by-step explanation:

The provided data regarding how much 34 college students spent on textbooks is:

S = {120, 130, 130, 140, 150, 150, 160, 170, 180, 210, 220, 230, 240, 250, 260, 280, 280, 290, 290, 290, 310, 320, 320, 370, 380, 390, 410, 440, 450, 470, 510, 530, 620, 810}

(a)

An outlier is defined as a value significantly different from the rest of the dataset, being either excessively high or low.

A widely-used approach for identifying outliers involves:

• Identifying values that fall below Q₁ - 1.5 IQR as outliers.
• Identifying values that exceed Q₃ + 1.5 IQR as outliers.

In our case,

Q₁ = first quartile

Q₃ = third quartile

IQR = Interquartile range = Q₃ - Q₁.

The first quartile value corresponds to more than 25% of the data points. It can be found by examining the median of the first half of the data.

Calculate the first quartile with the set below:

Initial half of data: {120, 130, 130, 140, 150, 150, 160, 170, 180, 210, 220, 230, 240, 250, 260, 280, 280}

This subset contains 17 values.

The median in an odd-sized dataset is the central point.

The central value here is: 180

Thus, the first quartile is Q₁ = 180.

The third quartile value corresponds to more than 75% of the data points.

We determine the third quartile from the second half of the data:

Final half of data: {290, 290, 290, 310, 320, 320, 370, 380, 390, 410, 440, 450, 470, 510, 530, 620, 810}

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The median of an odd dataset serves as the middle value.

The central point is: 390

Hence, the third quartile is Q₃ = 390.

Calculate the interquartile range as follows:

IQR = Q₃ - Q₁

      = 390 - 180

      = 210

Next, we compute the value of [Q₁ - 1.5 IQR]:

And then we compute [Q₃ + 1.5 IQR]:

There are no values below [Q₁ - 1.5 IQR]. However, one value exceeds [Q₃ + 1.5 IQR].

X = 810 > [Q₃ + 1.5 IQR] = 705

Consequently, the outlier identified within the dataset is $810.

(b)

A distribution is classified as positively skewed or right-skewed when the majority of the data congregates on the lower end of the spectrum.

The stem plot indicates that most of the data values are concentrated on the left side, confirming that the distribution is indeed right skewed.

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To address this question, we apply the trinomial theorem. Given a trinomial (a + b + c)ⁿ, we can identify any term using:

[n!/(n-m)!(m-k)!k!] a^(n-m) b^(m-k) c^k

For this scenario, x=a, y=b, and z=c are the variables provided. We already know the exponents associated with each variable. Thus, we set this equal to the trinomial theorem's structure.

n - m = 2
m - k = 5
k = 10

With k known, we can now find m. Once m is determined, n can be found as well. Finally, we solve for the coefficient.

m - 10 = 5
m = 15

n - 15 = 2
n = 17

Thus, the coefficient is calculated to be:

Coefficient = n!/(n-m)!(m-k)!k! = 17!/(17-5)!(15-10)!10! = 408,408

Hi there!

To calculate the mean, we first need to sum all the scores and divide by the total number of tests.

85+93+78+90+88+97+88=619

619/7≈88.43

Mean≈88.43

Next, we will determine the median by arranging the scores in ascending order.

78,85,88,88,90,93,97

To find the median, we look for the middle score, which is 88.

Median: 88

Lastly, we identify the mode, which is the score that occurs most frequently. Here, 88 appears twice while the others appear once, making 88 the mode.

Mode:88

I hope this is useful!