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3 years ago

Describe the diagram

Mathematics

2 answers:

Degger [83]3 years ago

6 0

Answer:

The diagram represents 3 sets assuming they refer to 3 activities A, B and C.

The shaded point on A shows those engage in activities A only.

The shaded point on B shows those engage in activities B only.

The shaded point on C shows those engage in activities C only.

Incidentally the numbers of people engage in activities A only and B only are the same number.

The colour grade for the following is the same with that engage in only activity B;

The intersection of A and B

The intersection of B and C

The intersection of A and C

Therefore it means the numerical value is the same with that of B only;

A n B= B n C = A n C = n( B only); n on the right represent intersection and on the left number.

A n B n C = no of people engaged in all 3 activities.

A n B n C = n( A only) = n( B only) ;

From the colour grade

Note: n on the right represent intersection and on the left

Tasya [4]3 years ago

3 0

Answer:

The diagram reprents the 3 sets and the universal set . The shaded portion clearly can be seen as :

$A \cap B \cap C + A - (A \cap B) - (A \cap C) + A \cap B \cap C + C - (C \cap B) - ( C \cap A) + (A \cap B \cap C)$

Shaded region = $3(A \cap B \cap C) + A + C - (A \cap B) - 2(A \cap C) - (C \cap B)$

Try to understand some what complex

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In a school,2/3 of the students study a language.

Cloud [144]

Answer:

The ratio is 4/11

Step-by-step explanation:

Let

x ----> the total number of students in a school

we know that

1) 2/3 of the students study a language.

(2/3)x ----> number of students studying a language

and

x-(2/3)x=(1/3)x ---->number of students who do not study a language

2) Of those who study a language 2/5 were studying french

(2/3)x(2/5)=(4/15)x ----> number of students studying French

and

(2/3)x(3/5)=(6/15)x ----> number of students studying a language but not studying French

3) we know that

The number of students who do not study French is equal to the number of students who do not study a language plus the number of students who study a language but do not study French.

(1/3)x+(6/15)x=(11/15)x ----> number of students who not study French

(4/15)x ----> number of students who study French

4 ) Find the ratio of students who study french and who do not

(4/15)x/(11/15)x=4/11

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3 years ago

What is the sum of the measures of all the interior angles of a regular octagon?

Nitella [24]

Answer:

Step-by-step explanation:

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4 years ago

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What volume of this pyramid with Height 8ft and Width 6Ft on one side and 6ft on the other side.

Fiesta28 [93]

$\bf \textit{volume of a square pyramid}\\\\
V=\cfrac{1}{3}Bh\qquad 
\begin{cases}
B=base=width\cdot length\\\\
h=height\\
----------\\
width=6\\
length=6\\
h=8
\end{cases}$

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3 years ago

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crimeas [40]

Answer:

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2 years ago

Randomly selected 110 student cars have ages with a mean of 8 years and a standard deviation of 3.6 years, while randomly select

monitta

Answer:

1. Yes, there is sufficient evidence to support the claim that student cars are older than faculty cars.

2. The 98% confidence interval for the difference between the two population means is [1.432 years, 3.968 years].

Step-by-step explanation:

We are given that randomly selected 110 student cars to have ages with a mean of 8 years and a standard deviation of 3.6 years, while randomly selected 75 faculty cars to have ages with a mean of 5.3 years and a standard deviation of 3.7 years.

Let $\mu_1$ = mean age of student cars.

$\mu_2$ = mean age of faculty cars.

So, Null Hypothesis, $H_0$ : $\mu_1 \leq \mu_2$ {means that the student cars are younger than or equal to faculty cars}

Alternate Hypothesis, $H_A$ : $\mu_1>\mu_2$ {means that the student cars are older than faculty cars}

(1) The test statistics that will be used here is Two-sample t-test statistics because we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)} {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t_n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean age of student cars = 8 years

$\bar X_2$ = sample mean age of faculty cars = 5.3 years

$s_1$ = sample standard deviation of student cars = 3.6 years

$s_2$ = sample standard deviation of student cars = 3.7 years

$n_1$ = sample of student cars = 110

$n_2$ = sample of faculty cars = 75

Also, $s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(110-1)\times 3.6^{2}+(75-1)\times 3.7^{2} }{110+75-2} }$ = 3.641

So, the test statistics = $\frac{(8-5.3)-(0)} {3.641 \times \sqrt{\frac{1}{110}+\frac{1}{75} } }$ ~ $t_1_8_3$

= 4.952

The value of t-test statistics is 4.952.

Since the value of our test statistics is more than the critical value of t, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we support the claim that student cars are older than faculty cars.

(2) The 98% confidence interval for the difference between the two population means ( $\mu_1-\mu_2$ ) is given by;

98% C.I. for ( $\mu_1-\mu_2$ ) = $(\bar X_1-\bar X_2) \pm (t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} })$

= $(8-5.3) \pm (2.326 \times 3.641 \times \sqrt{\frac{1}{110}+\frac{1}{75} })$

= $[2.7 \pm 1.268]$

= [1.432, 3.968]

Here, the critical value of t at a 1% level of significance is 2.326.

Hence, the 98% confidence interval for the difference between the two population means is [1.432 years, 3.968 years].

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3 years ago