MATHEMATICS COLLEGE

Find the zeros of f(x) =x^3+2x^2-3x

Answers

Answer 1

Answer:

Answer: 0, 1, -3

Step-by-step explanation:

f(x) = x^3 +2x^2 -3x

f(x) = x(x^2 +2x-3)

f(x) = x(x-1)(x+3)

to find zeroes, plug into f(x) and solve for each value of x. doing this gives 0, 1, and -3

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The manager of a computer retails store is concerned that his suppliers have been giving him laptop computers with lower than average quality. His research shows that replacement times for the model laptop of concern are normally distributed with a mean of 4.2 years and a standard deviation of 0.6 years. He then randomly selects records on 38 laptops sold in the past and finds that the mean replacement time is 3.9 years. Assuming that the laptop replacement times have a mean of 4.2 years and a standard deviation of 0.6 years, find the probability that 38 randomly selected laptops will have a mean replacement time of 3.9 years or less.

Answers

Answer: 0.0010

Step-by-step explanation:

Given the following :

Population Mean(m) = 4.2 years

Sample mean (s) = 3.9

Standard deviation (sd) = 0.6

Number of samples (n) = 38

Calculate the test statistic (z) :

(sample mean - population mean) / (sd / √n)

Z = (3.9 - 4.2) / (0.6 / √38)

Z = (- 0.3) / (0.6 / 6.1644140)

Z = -0.3 / 0.0973328

Z = - 3.0822086

Z = - 3.08

From the z table :

P(Z ≤ - 3.08) = 0.0010

Final answer:

The probability that the mean replacement time of a random sample of 38 laptops is 3.9 years or less, assuming the true mean replacement time is 4.2 years, is approximately 0.0038 or 0.38%

Explanation:

To solve this problem, we will use the formula for the Z score of a sample mean:
Z = (x - µ) / (σ/ √n)

In this case, the mean µ is 4.2 years, the standard deviation σ is 0.6 years, the sample mean x is 3.9 years, and the sample size n is 38.

Substituting the given values into the formula, we get:
Z = (3.9 - 4.2) / (0.6/√38) = -2.67.

We can then look up this Z score in the Z score table or use statistical software to find the corresponding probability. The probability associated with Z = -2.67 is approximately 0.0038. This means there's about a 0.38% chance that the mean replacement time of a random sample of 38 laptops will be 3.9 years or less, assuming the true mean replacement time is 4.2 years.

Learn more about Probability here:

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Consider the following hypothesis test. The following data are from matched samples taken from two populations. Population Element 1 2 1 21 20 2 28 26 3 18 18 4 20 20 5 26 24 Compute the difference value for each element. If your answer is "0", enter 0.a. Compose the difference valuefor each element.b. Compute d.c.Compute the standard deviation sd.d.Conduct a hypothesis test using α=.05.What is yourconclusion?

Answers

Answer:

a) d: 1 2 0 0 2

b) $\bar d= (\sum_(i=1)^n d_i)/(n)= (5)/(5)=1$

c) $s_d =(\sum_(i=1)^n (d_i -\bar d)^2)/(n-1) =1$

d) $p_v =P(t_((4))>2.236) =0.0445$

So the p value is lower than the significance level provided, so then we can conclude that we reject the null hypothesis that the difference mean between population 1 and 2 is less than 0. And we can say that the mean difference is higher than 0 at 5% of significance.

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.

Let put some notation

x=Population 1 , y = population 2

x: 21 28 18 20 26

y: 20 26 18 20 24

a. Compose the difference valuefor each element

Let d =x-y, so the values for the difference are:

d: 1 2 0 0 2

b. Compute d.

We need the mean for the difference. If we use the following formula we got:

$\bar d= (\sum_(i=1)^n d_i)/(n)= (5)/(5)=1$

c.Compute the standard deviation sd.

For the standard deviation we can use the following formula:

$s_d =(\sum_(i=1)^n (d_i -\bar d)^2)/(n-1) =1$

d.Conduct a hypothesis test using α=.05.What is yourconclusion?

If we assume that the system of hypothesis for this case are:

Null hypothesis: $\mu_x- \mu_y \leq 0$

Alternative hypothesis: $\mu_x -\mu_y > 0$

We can calculate the statistic given by :

$t=(\bar d -0)/((s_d)/(√(n)))=(1 -0)/((1)/(√(5)))=2.236$

The next step is calculate the degrees of freedom given by:

$df=n-1=5-1=4$

Now we can calculate the p value, since we have a right tailed test the p value is given by:

$p_v =P(t_((4))>2.236) =0.0445$

Final answer:

To complete this hypothesis test, calculate the difference value for each element, compute the sample mean difference (d), calculate the standard deviation (sd), and then conduct the hypothesis test using a significance level of α = 0.05.

Explanation:

To compute the difference value for each element, subtract the second value from the first value for each pair. For example, the difference for the first pair is 1-2 = -1. Repeat this for each pair of elements in the data set.

To compute the sample mean difference (d), add up all the difference values and divide by the total number of pairs. In this case, d = (-1 + 1 + 2 + 0 + 2)/5 = 0.8.

To compute the standard deviation (sd), first calculate the squared difference value for each pair and sum them up. Then divide the sum by (n-1), where n is the total number of pairs. Finally, take the square root of the result. In this case, sd = sqrt(((1-0.8)^2 + (2-0.8)^2 + (0-0.8)^2 + (2-0.8)^2)/4) = 1.32.

For the hypothesis test, we compare the sample mean difference (d) to the population mean difference (μd) under the null hypothesis. The null hypothesis states that there is no difference between the two populations. If the difference between d and μd is statistically significant at a significance level of α = 0.05, we reject the null hypothesis and conclude that there is a significant difference between the two populations. Otherwise, we fail to reject the null hypothesis.

Without information about the population mean difference (μd), we cannot perform the hypothesis test or draw a conclusion.

Learn more about Hypothesis Testing here:

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Let be a continuous random variable that follows a normal distribution with a mean of and a standard deviation of . Find the value of so that the area under the normal curve to the right of is . Round your answer to two decimal places.

Answers

Complete Question

Let x be a continuous random variable that follows a normal distribution with a mean of 550 and a standard deviation of 75.

Find the value of x so that the area under the normal curve to the left of x is .0250.

Find the value of x so that the area under the normal curve to the right ot x is .9345.

Answer:

$x = 403$

$x = 436.75$

Step-by-step explanation:

From the question we are told that

The mean is $\mu = 550$

The standard deviation is $\sigma = 75$

Generally the value of x so that the area under the normal curve to the left of x is 0.0250 is mathematically represented as

$P( X < x) = P( (x - \mu )/( \sigma) < (x - 550 )/(75 ) ) = 0.0250$

$(X -\mu)/(\sigma ) = Z (The \ standardized \ value\ of \ X )$

$P( X < x) = P( Z < z ) = 0.0250$

Generally the critical value of 0.0250 to the left is

$z = -1.96$

=> $(x- 550 )/(75) = -1.96$

=> $x = [-1.96 * 75 ]+ 550$

=> $x = 403$

Generally the value of x so that the area under the normal curve to the right of x is 0.9345 is mathematically represented as

$P( X < x) = P( (x - \mu )/( \sigma) < (x - 550 )/(75 ) ) = 0.9345$

$(X -\mu)/(\sigma ) = Z (The \ standardized \ value\ of \ X )$

$P( X < x) = P( Z < z ) = 0.9345$

Generally the critical value of 0.9345 to the right is

$z = -1.51$

=> $(x- 550 )/(75) = -1.51$

=> $x = [-1.51 * 75 ]+ 550$

=> $x = 436.75$

Answer the following true or false. Justify your answer.(a) If A is a subset of B, and x∈B, then x∈A.
(b) The set {(x,y) ∈ R2 | x > 0 and x < 0} is empty.
(c) If A and B are square matrices, then AB is also square.
(d) A and B are subsets of a set S, then A∩B and A∪B are also subsets of S.
(e) For a matrix A, we define A^2 = AA.

Answers

Answer:

a) False

b) True

c) True

d) True

e) True

Step-by-step explanation:

a) Consider the sets $B=\{1,2,3,4,5,6\}$ and $A=\{4,5,6\}$ . Observe that A is a subset of B and $1\in B$ but $1\notin A$

b) Any number different of 0 can be positive and negative simultaneusly. Then doesn't exist $x\in\mathbb{R}$ such that $x<0 and x>0$ . Then the set $\{(x,y) \in \mathbb{R}^2 | x > 0\; \text{and}\; x < 0}$ is empty.

c) If the multiplication AB is defined and A and B are square matrices with A of size nxn, then B is the size nxn and the matrix AB is the size nxn.

d) Let A and B subsets of a set S. Since each element of A and B are in S then each element of $A\cup B$ is in S. Also, if $x\in A\cap B$ , the $x\in A\subset S$ and $x\in B\subset S$ then $x\in S$ . This shows that $A\cap B \subset S$ .