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3 years ago

PLEASE HELP MY SIS IS BREAKING DOWN CRYING

Mathematics

1 answer:

kodGreya [7K]3 years ago

6 0

Answer:

f(1) = 4

Step-by-step explanation:

f(1) means let x=1 and find the y value at that point on the graph

At x=1 there is an open circle and a closed circle. The open circle means that the function does not have a value at that point, so we use the closed circle

f(1) = 4

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Given that the data distribution is approximately normal with the minimum value of 15 and the maximum value of 98 a) Estimate th

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Answer:

a) Mean, median, mode = 56.5

b) $S = 20.75$

Step-by-step explanation:

Normally distributed data have the same value for the mean, median and mode. This value is the addition of the maximum value by the minimum, divided by 2.

Also, the standard deviation in a normally distributed sample can be approximated by the following formula:

$S = \frac{max - min}{4}$

a) Estimate the values of Mean, Median, and Mode.

Highest value = 98

Minimum value = 15

$\mu = \frac{98 + 15}{2} = 56.5$

b) Estimate the value of the standard deviation of these data

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$S = \frac{max - min}{4}$

$S = \frac{98 - 15}{4}$

$S = 20.75$

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3 years ago

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3 years ago

Examination, 152 Candidates offered Agriculture

agasfer [191]

SOLUTION

Given the question in the image, the following are the solution steps to answer the question

STEP 1: Write the given details

$n(G)=136,n(P)=128,n(A)=152,n(A\cap G)=52,n(A\cap P)=60,n(A\cap P\cap G)=30,n(P\cap G)=58$

STEP 2: Get the number of students that takes the subjects only

From the basic stuff, we have Number of candidates who had taken only each of the subjects

$\begin{gathered} n(A\text{ only)=}n(A)-\lbrack n(A\cap P)+n(A\cap G)-n(A\cap P\cap G)_{}\rbrack \\ \Rightarrow152-\lbrack52+60-30\rbrack=152-82=70 \\ \\ n(G\text{ only)=}n(G)-\lbrack n(A\cap G)+n(P\cap G)-n(A\cap P\cap G)\rbrack \\ \Rightarrow136-\lbrack52+58-30\rbrack=56 \\ \\ n(P\text{ only)=}128-\lbrack58+60-30\rbrack=40 \end{gathered}$

STEP 3: Represent the values above on a Venn Diagram after getting the values

STEP 4: Find the total number of candidate that sat for the exam.

$\begin{gathered} We\text{ su}m\text{ all the numbers in the Venn diagram to get the total number of candidates that sat for the exam} \\ We\text{ have:} \\ 70+22+30+30+28+56+40=276 \end{gathered}$

Hence, the total number of candidates that sat for the exam is 276