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2 years ago

Select all the correct answers.

1 answer:

4 0

The function g(x) is created by applying an <em>horizontal</em> translation 4 units left and a reflection over the x-axis. (Correct choices: Third option, fifth option)

<h3>How to determine the characteristics of rigid transformations by comparing two functions</h3>

In this problem we have two functions related to each other because of the existence of <em>rigid</em> transformations. <em>Rigid</em> transformations are transformations applied to <em>geometric</em> loci such that <em>Euclidean</em> distance is conserved at every point of the <em>geometric</em> locus.

Let be f(x) = - 2 · cos (x - 1) + 3, then we use the concept of <em>horizontal</em> translation 4 units in the + x direction:

f'(x) = - 2 · cos (x - 1 + 4) + 3

f'(x) = - 2 · cos (x + 3) + 3 (1)

Now we apply a reflection over the x-axis:

g(x) = - [- 2 · cos (x + 3) + 3]

g(x) = 2 · cos (x + 3) - 3

Therefore, the function g(x) is created by applying an <em>horizontal</em> translation 4 units left and a reflection over the x-axis. (Correct choices: Third option, fifth option)

To learn more on rigid transformations: brainly.com/question/1761538

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A public health organization reports that 40%of baby boys 6-8 months old in the United

Sveta_85 [38]

Using the binomial distribution, the probabilities are given as follows:

0.3675 = 36.75% probability that more than 4 weigh more than 20 pounds.

0.1673 = 16.73% probability that fewer than 3 weigh more than 20 pounds.

Since P(X > 7) < 0.05, it would be unusual if more than 7 of them weigh more than 20 pounds.

<h3>What is the binomial distribution formula?</h3>

The formula is:

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

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

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

The values of the parameters for this problem are:

n = 10, p = 0.4.

The probability that more than 4 weigh more than 20 pounds is:

$P(X > 4) = 1 - P(X \leq 4)$

In which:

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

Then:

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

$P(X = 0) = C_{10,0}.(0.4)^{0}.(0.6)^{10} = 0.0061$

$P(X = 1) = C_{10,1}.(0.4)^{1}.(0.6)^{9} = 0.0403$

$P(X = 2) = C_{10,2}.(0.4)^{2}.(0.6)^{8} = 0.1209$

$P(X = 3) = C_{10,3}.(0.4)^{3}.(0.6)^{7} = 0.2150$

$P(X = 4) = C_{10,4}.(0.4)^{4}.(0.6)^{6} = 0.2502$

Hence:

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0061 + 0.0403 + 0.1209 + 0.2150 + 0.2502 = 0.6325$

$P(X > 4) = 1 - P(X \leq 4) = 1 - 0.6325 = 0.3675$

0.3675 = 36.75% probability that more than 4 weigh more than 20 pounds.

The probability that fewer than 3 weigh more than 20 pounds is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0061 + 0.0403 + 0.1209 = 0.1673

0.1673 = 16.73% probability that fewer than 3 weigh more than 20 pounds.

For more than 7, the probability is:

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10)$

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

$P(X = 8) = C_{10,8}.(0.4)^{8}.(0.6)^{2} = 0.0106$

$P(X = 9) = C_{10,9}.(0.4)^{9}.(0.6)^{1} = 0.0016$

$P(X = 10) = C_{10,10}.(0.4)^{10}.(0.6)^{0} = 0.0001$

Since P(X > 7) < 0.05, it would be unusual if more than 7 of them weigh more than 20 pounds.

More can be learned about the binomial distribution at brainly.com/question/24863377

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2 years ago

Help with practice problem

fenix001 [56]

Answer:

y = 2x - 6

Step-by-step explanation:

For parallel lines, the mx (the slope) stays the same but the y-int changes.

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2 years ago

The greatest common factor of 30.48.60

Elena-2011 [213]

30= 3*10
30= 3*2*5

48= 12*4
48= 3*4*4
48= 3*2*2*2*2

60= 3*20
60= 3*10*2
60= 3*2*5*2

The GCF is 3*2 because all 3 numbers contain a 3 and a 2 in their prime factorizations.

Final answer: 6

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3 years ago

Which statement is true?

Mumz [18]

The answer is D.

|-4| < 4
4 < 4

|-1| < 0
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3 years ago

Please help it is TIMED.

lions [1.4K]

Answer:

message it will be easy er to explain

Step-by-step explanation: