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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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3 years ago

What is the definition of direct control in statistics?

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2 years ago

The manager of a computer retails store is concerned that his suppliers have been giving him laptop computers with lower than av

34kurt

Answer:

Probability that the 50 randomly selected laptops will have a mean replacement time of 3.1 years or less is 0.0092.

Yes. The probability of this data is unlikely to have occurred by chance alone.

Step-by-step explanation:

We are given that the replacement times for the model laptop of concern are normally distributed with a mean of 3.3 years and a standard deviation of 0.6 years.

He then randomly selects records on 50 laptops sold in the past and finds that the mean replacement time is 3.1 years.

Let M = sample mean replacement time

The z-score probability distribution for sample mean is given by;

Z = $\frac{ M-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ ~ N(0,1)

where, $\mu$ = population mean replacement time = 3.3 years

$\sigma$ = standard deviation = 0.6 years

n = sample of laptops = 50

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the 50 randomly selected laptops will have a mean replacement time of 3.1 years or less is given by = P(M $\leq$ 3.1 years)

P(M $\leq$ 3.1 years) = P( $\frac{ M-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ $\leq$ $\frac{ 3.1-3.3}{\frac{0.6}{\sqrt{50} } }} }$ ) = P(Z $\leq$ -2.357) = 1 - P(Z $\leq$ 2.357)

= 1 - 0.99078 = 0.0092 or 0.92%

So, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2.357 in the z table which will lie between x = 2.35 and x = 2.36 which has an area of 0.99078.

Hence, the required probability is 0.0092 or 0.92%.

Now, based on the result above; Yes, the computer store has been given laptops of lower than average quality because the probability of this data is unlikely to have occurred by chance alone as the probability of happening the given event is very low as 0.92%.

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3 years ago

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3 years ago

Evaluate the expression if m = 3. 3m^2 (SHOW YOUR WORK)

iragen [17]

The value of given expression when m = 3 is 27

<h3>Solution:</h3>

Given expression is $3m^2$