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Please help me in this! you get 30 points!
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Please help me in this! you get 30 points! ​ - 1

Answers

Answer 1
Answer:

Answer:

y=3x-2

Step-by-step explanation:

You can verify it's not D because the y-intercept is at -2.

You can verify it's not A because that would mean the x-intercept is 2 despite it appearing to be closer to one.

You can verify it's not B because that would mean the x-intercept is 1.5
Answer 2
Answer: It’s C. If I’m wrong sorry

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An educational psychologist wants to test whether a new teaching method negatively affects reading comprehension scores. She randomly selects 30 6th grade students that were taught under the new teaching method and finds that their scores on a standardized reading comprehension test have a mean equal to 118.8 with a variance equal to 37.2. Scores on the standardized test in the general population of 6th graders are distributed approximately normally with a mean equal to 119.8. Is there sufficient evidence to conclude that the new teaching method negatively affects reading comprehension scores

Answers

Answer:

We conclude that the new teaching method does not negatively affects reading comprehension scores.

Step-by-step explanation:

We are given that an educational psychologist wants to test whether a new teaching method negatively affects reading comprehension scores.

She randomly selects 30 6th grade students that were taught under the new teaching method and finds that their scores on a standardized reading comprehension test have a mean equal to 118.8 with a variance equal to 37.2.

Scores on the standardized test in the general population of 6th graders are distributed approximately normally with a mean equal to 119.8.

Let \mu = mean scores on a standardized reading comprehension test.

So, Null Hypothesis, H_0 : \mu \geq 119.8      {means that the new teaching method does not negatively affects reading comprehension scores}

Alternate Hypothesis,H_A : \mu < 119.8    {means that the new teaching method negatively affects reading comprehension scores}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

                      T.S. =  (\bar X-\mu)/((s)/(√(n) ) )  ~ t_n_-_1

where, \bar X = sample mean test score = 118.8

            s = sample standard deviation = √(37.2) = 6.1

            n = sample of 6th grade students = 30

So, test statistics  =  (118.8-119.8)/((6.1)/(√(30) ) )  ~ t_2_9

                              =  -0.898

The value of t test statistics is -0.898.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the t table gives critical value of -1.699 for left-tailed test.

Since our test statistic is more than the critical value of t as -0.898 > -1.699, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the new teaching method does not negatively affects reading comprehension scores.

Could someone help me out with this.

Answers

Answer:

45 units

Step-by-step explanation:

Since both figures are similar, the ratio of their similar sides will be equal to a constant. Hence;

8/10 = 12/EH

8EH = 12 * 10

8EH = 120

EH = 120/8

EH = 15

Perimeter of EFGH = 15 + 10 + 5 + 15 = 45 units

A survey among students at a certain university revealed that the number of hours spent studying the week before final exams was approximately normally distributed with mean 25 and standard deviation 6. What proportion of students studied between 25 and 34 hours

Answers

Answer:

P(25<X<34)=P((25-\mu)/(\sigma)<(X-\mu)/(\sigma)<(34-\mu)/(\sigma))=P((25-25)/(6)<Z<(34-25)/(6))=P(0<z<1.5)

And we can find this probability with this difference:

P(0<z<1.5)=P(z<1.5)-P(z<0)

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

P(0<z<1.5)=P(z<1.5)-P(z<0)=0.933-0.5=0.433

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem'

Let X the random variable that represent the hous spent studying the week before final exams of a population, and for this case we know the distribution for X is given by:

X \sim N(25,6)  

Where \mu=25 and \sigma=6

We are interested on this probability

P(25<X<34)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=(x-\mu)/(\sigma)

If we apply this formula to our probability we got this:

P(25<X<34)=P((25-\mu)/(\sigma)<(X-\mu)/(\sigma)<(34-\mu)/(\sigma))=P((25-25)/(6)<Z<(34-25)/(6))=P(0<z<1.5)

And we can find this probability with this difference:

P(0<z<1.5)=P(z<1.5)-P(z<0)

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

P(0<z<1.5)=P(z<1.5)-P(z<0)=0.933-0.5=0.433

Find the value of X, I need help solving this

Answers

Answer:

C) 8

Step-by-step explanation:

41 + 6x + 1 = 90° because this is a right angle

42 + 6x = 90 subtract 42 from both sides

6x = 48 divide both sides by 6

x = 8

Answer:

the answer should be c

Step-by-step explanation:

180-90=90

90-41=49

49-1=48

48:6=8

x=8

Determine which of the sets of vectors is linearly independent. A: The set where p 1(t) = 1, p 2(t) = t 2, p 3(t) = 1 + 5t B: The set where p 1(t) = t, p 2(t) = t 2, p 3(t) = 2t + 5t 2 C: The set where p 1(t) = 1, p 2(t) = t 2, p 3(t) = 1 + 5t + t 2 A and C C only B only A only all of them

Answers

Answer:

(A)A and C

Step-by-step explanation:

In each case, represent each of the p_i as a column vector where  each row corresponds to the constant term, coefficient of t and t^2 respectively.

A= The set where p_1(t)=1p_2(t)=t^2p_3(t)=1+5t

A=\left[\begin{array}{ccc}1&0&1\n0&0&5\n0&1&0\end{array}\right]

|A|=\left|\begin{array}{ccc}1&0&1\n0&0&5\n0&1&0\end{array}\right|=-5

B: The set where p_1(t)=t, p_2(t)=t^2, p_3(t)=2t+5t^2

B=\left[\begin{array}{ccc}0&0&0\n1&0&2\n0&1&5\end{array}\right]\n|B|=0

C: The set where p_1(t)=1, p_2(t)=t^2, p_3(t)=1+5t+t^2

C=\left[\begin{array}{ccc}1&0&1\n0&0&5\n0&1&1\end{array}\right]\n|C|=-5

Since the determinants of A and C are not 0, the set of vectors in A and C are linearly independent.



Find the standard deviation of the following data. Round your answer to one decimal place. x 0 1 2 3 4 5 P(X

Answers

Answer:

\sigma = 1.8

Step-by-step explanation:

Given

\begin{array}{ccccccc}x & {0} & {1} & {2} & {3} & {4}& {5} \ \n P(x) & {0.2} & {0.1} & {0.1} & {0.2} & {0.2}& {0.2} \ \end{array}

Required

The standard deviation

First, calculate the expected value E(x)

E(x) = \sum x * P(x)

So, we have:

E(x) = 0 * 0.2 + 1 * 0.1 + 2 * 0.1 + 3 * 0.2 + 4 * 0.2 + 5 * 0.2

E(x) = 2.7

Next, calculate E(x^2)

E(x^2) = \sum x^2 * P(x)

So, we have:

E(x^2) = 0^2 * 0.2 + 1^2 * 0.1 + 2^2 * 0.1 + 3^2 * 0.2 + 4^2 * 0.2 + 5^2 * 0.2

E(x^2) = 10.5

The standard deviation is:

\sigma = √(E(x^2) - (E(x))^2)

\sigma = √(10.5 - 2.7^2)

\sigma = √(10.5 - 7.29)

\sigma = √(3.21)

\sigma = 1.8 --- approximated
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