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2 years ago

Reflect the point -3, 6 across the y-axis

Mathematics

2 answers:

krek1111 [17]2 years ago

8 0

-3, -6 im pretty sure?!

Charra [1.4K]2 years ago

4 0

Answer is ( 3, 6)

Step by step

The rule for a reflection over the y -axis is (x,y)→(−x,y) .

So

(-3, 6) ➡️ ( 3, 6)

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Based on past experience, the main printer in a university computer centre is operating properly 90% of the time. Suppose inspec

yaroslaw [1]

Answer:

a) 38.74% probability that the main printer is operating properly for exactly 9 inspections.

b) Approximately 100% probability that the main printer is operating properly for at least 3 inspections.

c) The expected number of inspections in which the main printer is operating properly is 9.

Step-by-step explanation:

For each inspection, there are only two possible outcomes. Either it is operating correctly, or it is not. The probability of the printer operating correctly for an inspection is independent of any other inspection, which means that the binomial probability distribution is used 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.

Based on past experience, the main printer in a university computer centre is operating properly 90% of the time.

This means that $p = 0.9$

Suppose inspections are made at 10 randomly selected times.

This means that $n = 10$

A) What is the probability that the main printer is operating properly for exactly 9 inspections.

This is $P(X = 9)$ . So

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

$P(X = 9) = C_{10,9}.(0.9)^{9}.(0.1)^{1} = 0.3874$

38.74% probability that the main printer is operating properly for exactly 9 inspections.

B) What is the probability that the main printer is operating properly for at least 3 inspections?

This is:

$P(X \geq 3) = 1 - P(X < 3)$

In which

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

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

$P(X = 0) = C_{10,0}.(0.9)^{0}.(0.1)^{10} \approx 0$

$P(X = 1) = C_{10,1}.(0.9)^{1}.(0.1)^{9} \approx 0$

$P(X = 2) = C_{10,2}.(0.9)^{2}.(0.1)^{8} \approx 0$

Thus:

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0 + 0 + 0 = 0$

Then:

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0 = 1$

Approximately 100% probability that the main printer is operating properly for at least 3 inspections.

C) What is the expected number of inspections in which the main printer is operating properly?

The expected value for the binomial distribution is given by:

$E(X) = np$

In this question:

$E(X) = 10(0.9) = 9$

3 0

3 years ago

Which ordered pairs make both inequalities true? Select two options. y < 5x + 2 y > One-halfx + 1 On a coordinate plane, 2

Verdich [7]

Answer:

(1, 2)

(2,2)

Step-by-step explanation:

The inequality Given :

y < 5x + 2

y ≥ One-halfx + 1

We can use the values in the options to see which satisfies the inequality :

(-1, 3) ; X = - 1, y = 3

3 < 5(-1) + 2 ; 3 < - 4 (not true)

(0, 2) ; X = 0 ; y = 2

2 < 5(0) + 2 ; 2 < 2 (nor true)

2 > 1/2(0) + 1

2 > 1 (true)

Using (1, 2).; x = 1 ; y = 2

2 < 5(1) + 2 ; 2 < 7 (true)

2 > 1/2(1) + 1 ; 2 > 1.5 (true)

Using (2, 2)

y < 5x + 2

2 < 10 + 2 ; 2 < 12 (true)

y > 1/2x + 1

2 > 1/2(2) + 1 ; 2

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3 years ago

State the x value of the x intercept describe by the following <br> linear equation: (-3x-2y = 9

aivan3 [116]

Answer:

x=-3

Step-by-step explanation:

-3x-2y=9

-3x-2(0)=9

-3x-0=9

-3x=9

x=-3

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3 years ago

Jeffersonville Bank is offering it's customers a record 3.4% compound interest on new savings accounts. Mr. Kirks opens an accou

slega [8]

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3 years ago

Factor the expression using GCF 7+14a

Lana71 [14]

The GCF of 7 and 14a is 7, so the factors are:-

7(1 + 2a)

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3 years ago

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