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2 years ago

Given 4 coordinate points A(2, 0), B(4, 0), C(5, -2), D(1, -2). If the image is rotated by 180° about the origin O, then D' =

Mathematics

2 answers:

Jet001 [13]2 years ago

5 0

Answer:

Option 2nd is correct

(-1, 2)

Step-by-step explanation:

The rule of rotation by 180 degree about the origin is given by:

$(x, y) \rightarrow (-x, -y)$

As per the statement:

Given 4 coordinate points

A(2, 0), B(4, 0), C(5, -2), D(1, -2).

Apply the rule of rotation on D we have;

$D(1, -2) \rightarrow (-1, -(-2))=(-1, 2)=D'$

Therefore, the image of D' is, (-1, 2)

nekit [7.7K]2 years ago

3 0

The answer to your question is b

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Dave’s Automatic Door, referred to in Exercise 29, installs automatic garage door openers. Based on a sample, following are the

HACTEHA [7]

The question is not complete and the full question says;

Calculate the (a) range, (b) arithmetic mean, (c) mean deviation, and (d) interpret the values. Dave’s Automatic Door installs automatic garage door openers. The following list indicates the number of minutes needed to install a sample of 10 door openers: 28, 32, 24, 46, 44, 40, 54, 38, 32, and 42.

Answer:

A) Range = 30 minutes

B) Mean = 38

C) Mean Deviation = 7.2

D) This is well written in the explanation.

Step-by-step explanation:

A) In statistics, Range = Largest value - Smallest value. From the question, the highest time is 54 minutes while the smallest time is 24 minutes.

Thus; Range = 54 - 24 = 30 minutes

B) In statistics,

Mean = Σx/n

Where n is the number of times occurring and Σx is the sum of all the times occurring

Thus,

Σx = 28 + 32 + 24 + 46 + 44 + 40 + 54 + 38 + 32 + 42 = 380

n = 10

Thus, Mean(x') = 380/10 = 38

C) Mean deviation is given as;

M.D = [Σ(x-x')]/n

Thus, Σ(x-x') = (28-38) + (32-38) + (24-38) + (46-38) + (44-38) + (40-38) + (54-38) + (38-38) + (32-38) + (42-38) = 72

So, M.D = 72/10 = 7.2

D) The range of the times is 30 minutes.

The average time required to open one door is 38 minutes.

The number of minutes the time deviates on average from the mean of 38 minutes is 7.2 minutes

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2 years ago

A study is being conducted in which the health of two independent groups of ten policyholders is being monitored over a one-year

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Answer:

46.91% probability that at least nine participants complete the study in one of the two groups, but not in both groups

Step-by-step explanation:

We use two binomial trials to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Probability of at least nine participants finishing the study in a group.

0.2 probability of a students dropping out. So 1 - 0.2 = 0.8 probability of a student finishing the study. This means that $p = 0.8$ .

10 students, so $n = 10$

We have to find:

$P(X \geq 9) = P(X = 9) + P(X = 10)$

Then

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 9) = C_{10,9}.(0.8)^{9}.(0.2)^{1} = 0.2684$

$P(X = 10) = C_{10,10}.(0.8)^{10}.(0.2)^{0} = 0.1074$

$P(X \geq 9) = P(X = 9) + P(X = 10) = 0.2684 + 0.1074 = 0.3758$

0.3758 probability that at least nine participants complete the study in a group.

Calculate the probability that at least nine participants complete the study in one of the two groups, but not in both groups?

0.3758 probability that at least nine participants complete the study in a group. This means that $p = 0.3758$

Two groups, so $n = 2$

We have to find P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{2,1}.(0.3758)^{1}.(0.6242)^{1} = 0.4691$

46.91% probability that at least nine participants complete the study in one of the two groups, but not in both groups

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Hey there! Welcome to Brainly!

Let's see how many marbles we have in total.

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