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Tpy6a [65]
1 year ago
10

each transformation is performed on the line with the equation y=2x-1. write the equation of the new line. vertical translation

down 3 units, slope increased by 4, slope divided in half, and shifted up 1 unit.
Mathematics
1 answer:
Phoenix [80]1 year ago
6 0

Answer:

y2= 2x-4

y3=6x-1

y4= x-1

y5=2x

Step-by-step explanation:

for y=2x-1

1) for a vertical translation down of 3 units

y2= y-3 =(2x-1)-3= 2x-4

y2= 2x-4

2) for a slope increased by 4

y3= y+ 4x = 2x-1 +4x = 6x-1

y3=6x-1

3) for sloped divided in half. slope of y : m=2 → slope of y4=2/2 =1

y4= x-1

4) shifted up (vertical translation) of  1 unit

y5= y+1 = 2x-1+1=2x

y5=2x

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Answer:

Step-by-step explanation:

From the information given, we would write the hypothesis.

For the null hypothesis,

H0 : µ = 70

For the null hypothesis,

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This is a right tailed test because of the symbol of greater than.

The decision rule is to reject the null hypothesis if the level of significance is greater than the p value and accept the null hypothesis if the level of significance is lesser than the p value.

Therefore, since the significance level, 0.05 > p value, 0.01635, then we would reject the null hypothesis. There is enough evidence that the mean speed of all cars is greater than the posted speed limit of 70 mph.

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A mechanical dart thrower throws darts independently each time, with probability 10% of hitting the bullseye in each attempt. Th
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Answer:

The probability of hitting the bullseye at least once in 6 attempts is 0.469.

Step-by-step explanation:

It is given that a mechanical dart thrower throws darts independently each time, with probability 10% of hitting the bullseye in each attempt.

The probability of hitting bullseye in each attempt, p = 0.10

The probability of not hitting bullseye in each attempt, q = 1-p = 1-0.10 = 0.90

Let x be the event of  hitting the bullseye.

We need to find the probability of hitting the bullseye at least once in 6 attempts.

P(x\geq 1)=1-P(x=0)       .... (1)

According to binomial expression

P(x=r)=^nC_rp^rq^{n-r}

where, n is total attempts, r is number of outcomes, p is probability of success and q is probability of failure.

The probability that the dart thrower not hits the bullseye in 6 attempts is

P(x=0)=^6C_0(0.10)^0(0.90)^{6-0}

P(x=0)=0.531441

Substitute the value of P(x=0) in (1).

P(x\geq 1)=1-0.531441

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1 year ago
Market-share-analysis company Net Applications monitors and reports on Internet browser usage. According to Net Applications, in
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Answer:

a) There is a 2.43% probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

b) There is an 80.50% probability that at least 3 of the 20 Internet browsers users use Chrome as their Internet browser.

c) The expected number of Chrome users is 4.074.

d) The variance for the number of Chrome users is 3.2441.

The standard deviation for the number of Chrome users is 1.8011.

Step-by-step explanation:

For each Internet browser user, there are only two possible outcomes. Either they use Chrome, or they do not. This means that we can solve this problem using concepts of the binomial probability distribution.

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 combinatios 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.

In this problem we have that:

Google Chrome has a 20.37% share of the browser market. This means that p = 0.2037

20 Internet users are sampled, so n = 20.

a.Compute the probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

This is P(X = 8).

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

P(X = 8) = C_{20,8}.(0.2037)^{8}.(0.7963)^{12} = 0.0243

There is a 2.43% probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

b.Compute the probability that at least 3 of the 20 Internet browsers users use Chrome as their Internet browser.

Either there are less than 3 Chrome users, or there are three or more. The sum of the probabilities of these events is decimal 1. So:

P(X < 3) + P(X \geq 3) = 1

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_{20,0}.(0.2037)^{0}.(0.7963)^{20} = 0.0105

P(X = 1) = C_{20,1}.(0.2037)^{1}.(0.7963)^{19} = 0.0538

P(X = 2) = C_{20,2}.(0.2037)^{2}.(0.7963)^{18} = 0.1307

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0105 + 0.0538 + 0.1307 = 0.1950

P(X \geq 3) = 1 - P(X < 3) = 1 - 0.1950 = 0.8050

There is an 80.50% probability that at least 3 of the 20 Internet browsers users use Chrome as their Internet browser.

c.For the sample of 20 Internet browser users, compute the expected number of Chrome users

We have that, for a binomial experiment:

E(X) = np

So

E(X) = 20*0.2037 = 4.074

The expected number of Chrome users is 4.074.

d.For the sample of 20 Internet browser users, compute the variance and standard deviation for the number of Chrome users.

We have that, for a binomial experiment, the variance is

Var(X) = np(1-p)

So

Var(X) = 20*0.2037*(0.7963) = 3.2441

The variance for the number of Chrome users is 3.2441.

The standard deviation is the square root of the variance. So

\sqrt{Var(X)} = \sqrt{3.2441} = 1.8011

The standard deviation for the number of Chrome users is 1.8011.

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