I don't know how to approach it, I've been looking at my notes and I can't grasp it.

Answer:

Hi! The answer is A, Blizzard.

Recently did this test on FLVS

Hope this helped!

Have a terrific Tuesday!!

~Lola

Answer:

Tornado

Step-by-step explanation:

Its not blizzard or hurricane.

Answer:

a) P(A graduate is offered fewer than two jobs) = 0.53.

b) P(A graduate is offered more than one job) = 0.47.

Step-by-step explanation:

Let X be a random variable denoting the number of jobs offers that a university graduate gets within a month of graduation.

The probability that a university graduate will be offered no jobs within a month of graduation is estimated to be 10% i.e. [tex]P(X=0)=0.10[/tex]

The probability of receiving one job offers has similarly been estimated to be 43% i.e. [tex]P(X=1)=0.43[/tex]

The probability of receiving two job offers has similarly been estimated to be 34% i.e. [tex]P(X=2)=0.34[/tex]

The probability of receiving three job offers has similarly been estimated to be 13% i.e. [tex]P(X=3)=0.13[/tex]

a) P (A graduate is offered fewer than two jobs) i.e. P(X<2)

So, [tex]P(X<2)=P(X=0)+P(X=1)[/tex]

[tex]P(X<2)=0.10+0.43[/tex]

[tex]P(X<2)=0.53[/tex]

P(A graduate is offered fewer than two jobs) = 0.53.

b) P (A graduate is offered more than one job) i.e. P(X>1)

So, [tex]P(X>1)=P(X=2)+P(X=3)[/tex]

[tex]P(X>1)=0.34+0.13[/tex]

[tex]P(X>1)=0.47[/tex]

P(A graduate is offered more than one job) = 0.47.

The probability that a graduate is offered fewer than two jobs is 53%, and the probability that a graduate is offered more than one job is 47%.

In the scenario given, you want to find two probabilities: A) the probability that a graduate is offered fewer than two jobs, and B) the probability that a graduate is offered more than one job. The total probability should add up to 100%, or a probability of 1.

For part A), 'fewer than two jobs' could mean either no job offers or one job offer. We know from the information given that the probability of no job offer is 10% (or 0.10) and the possibility of one job offer is 43% (or 0.43). So you add these two probabilities together: 0.10 + 0.43 = 0.53. Therefore, the probability that a graduate is offered fewer than two jobs is 53%.

For part B), 'more than one job' could mean either two job offers or three job offers. From the information given, we can find that the probability of receiving two job offers is 34% (or 0.34) and the probability for three job offers is 13% (or 0.13). Adding these two probabilities gives 0.34 + 0.13 = 0.47. Hence, the probability that a graduate is offered more than one job is 47%.

https://brainly.com/question/32117953

#SPJ3

Answer:

a) [tex]P(X=4)=(1-0.6)^{4-1} 0.6 = 0.0384[/tex]

b) [tex]P(X\leq 4)=0.6+0.24+0.096+0.0384=0.9744[/tex]

c) [tex]P(X\geq 4)=1-P(X<4)=1-P(X\leq 3)=1-[0.6+0.24+0.096]=0.064[/tex]

Step-by-step explanation:

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

[tex]P(X=x)=(1-p)^{x-1} p[/tex]

Let X the random variable that measures the number os trials until the first success, we know that X follows this distribution:

[tex]X\sim Geo (1-p)[/tex]

Part a

For this case we want this probability

[tex]P(X=4)=(1-0.6)^{4-1} 0.6 = 0.0384[/tex]

Part b

For this case we want this probability:

[tex]P(X\leq 4)=P(X=1)+P(X=2)+P(X=3)+P(X=4)[/tex]

If we find the individual probabilities we got:

[tex]P(X=1)=(1-0.6)^{1-1} 0.6 = 0.6[/tex]

[tex]P(X=2)=(1-0.6)^{2-1} 0.6 = 0.24[/tex]

[tex]P(X=3)=(1-0.6)^{3-1} 0.6 = 0.096[/tex]

[tex]P(X=4)=(1-0.6)^{4-1} 0.6 = 0.0384[/tex]

And replacing we have:

[tex]P(X\leq 4)=0.6+0.24+0.096+0.0384=0.9744[/tex]

Part c

For this case at least 4 trials means that the random variable X needs to be 4 or more

[tex]P(X\geq 4)=1-P(X<4)=1-P(X\leq 3)=1-[P(X=1)+P(X=2)+P(X=3)][/tex]

And we found already the probabilities P(X=1),P(X=2) and P(X=3) so we just need to replace:

[tex]P(X\geq 4)=1-P(X<4)=1-P(X\leq 3)=1-[0.6+0.24+0.096]=0.064[/tex]

The three scenarios are calculated using a combination of geometric and binomial probability distribution. For the first success to happen on the 4th trial is 0.0384. For a success to happen within the first four trials is 0.8847, and for the first success to need at least four trials is 0.3600.

This question has to do with the concept of probability distribution in math, and specifically with the geometric distribution and binomial probability distribution. Given the probability of a successful optical alignment p = 0.6, we're being asked about different scenarios involving success on specific trials.

(a) For the first successful alignment to happen on the 4th trial, the first 3 trials have to be failures and the 4th one, a success. The probability would be (0.4)^3 * 0.6 = 0.0384.

(b) This means we want a successful alignment on the 1st, 2nd, 3rd or 4th trials. For this we sum up the probabilities of each case: (0.6) + (0.4*0.6) + (0.4)^2*0.6 + (0.4)^3*0.6 = 0.8847.

(c) For a successful alignment to occur in at least 4 trials, we have to subtract the probability of success within the first 3 trials from 1. Therefore, 1 - ((0.6) + (0.4*0.6) + (0.4)^2*0.6) = 0.3600.

https://brainly.com/question/32117953

#SPJ3

Answer:

2/9

Step-by-step explanation:

The Poisson’s distribution is a discrete probability distribution. A discrete probability distribution means that the events occur with a constant mean rate and independently of each other. It is used to signify the chance (probability) of a given number of events occurring in a fixed interval of time or space.

In the long run, fraction of time that it rains = E(Number of days in rainy spell) / {E(Number of days in a rainy spell) + E(Number of days in a dry spell)}

E(Number of days in rainy spell) = 2

E(Number of days in a dry spell) = 7

In the long run, fraction of time that it rains = 2/(2 + 7) = 2/9

Given the parameters of the rainy spell and dry spell, the long-run fraction of time that it rains can be calculated by dividing the mean of the rainy days by the sum of the average rainy and dry days. Hence, it rains roughly 22.22% of the time in the long-term.

The question is asking about the long-run fraction of time that it rains, based on a rainy spell following a Poisson distribution with a mean of 2 days, and a dry spell following a geometric distribution with an average of 7 days, with the sequences being independent.

We are being asked to calculate the proportion of time that it rains in the long-run, given these distribution parameters. The Poisson and geometric distributions are often used in this type of probability assessment.

To tackle this, we need to divide the mean of the rainy days by the sum of the average rainy and dry days. Thus, the long-run fraction of time it rains is given by [tex]2/(2+7) = 2/9.[/tex]

So, in the long run, it rains roughly 22.22% (or 2/9) of the time.

https://brainly.com/question/14210034

#SPJ12

Answer:

z=13.36 (Statistic)

[tex]p_v =P(Z>13.36)\approx 0[/tex]

The p value is a very low value and using any significance level for example [tex]\alpha=0.05, 0,1,0.15[/tex] always [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion of men with red/green color blindness is significant higher than the proportion of female with red/green color blindness .

Step-by-step explanation:

1) Data given and notation

[tex]X_{MCB}=56[/tex] represent the number of men with red/green color blindness

[tex]X_{WCB}=5[/tex] represent the number of women with red/green color blindness

[tex]n_{MCB}=600[/tex] sample of male selected

[tex]n_{WCB}=600[/tex] sample of demale selected

[tex]p_{MCB}=\frac{56}{600}=0.093[/tex] represent the proportion of men with red/green color blindness

[tex]p_{WCB}=\frac{5}{2150}=0.0023[/tex] represent the proportion of women with red/green color blindness

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the value for the test (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the proportion for men with red/green color blindness is a higher than the rate for women  , the system of hypothesis would be:

Null hypothesis:[tex]p_{MCB} \leq p_{WCB}[/tex]

Alternative hypothesis:[tex]p_{MCB} > \mu_{WCB}[/tex]

We need to apply a z test to compare proportions, and the statistic is given by:

[tex]t=\frac{p_{MCB}-p_{WCB}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{MCB}}+\frac{1}{n_{WCB}})}}[/tex]   (1)

Where [tex]\hat p=\frac{X_{MCB}+X_{WCB}}{n_{MCB}+n_{WCB}}=\frac{56+5}{600+2150}=0.0221[/tex]

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

3) Calculate the statistic

Replacing in formula (1) the values obtained we got this:

[tex]z=\frac{0.093-0.0023}{\sqrt{0.0221(1-0.0221)(\frac{1}{600}+\frac{1}{2150})}}=13.36[/tex]  

4) Statistical decision

For this case we don't have a significance level provided [tex]\alpha[/tex], but we can calculate the p value for this test.  

Since is a one side test the p value would be:

[tex]p_v =P(Z>13.36)\approx 0[/tex]

So the p value is a very low value and using any significance level for example [tex]\alpha=0.05, 0,1,0.15[/tex] always [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion of men with red/green color blindness is significant higher than the proportion of female with red/green color blindness .

To test the claim that men have a higher rate of red/green color blindness, we need to compare the proportions of color blindness in men and women using a chi-square test of independence.

To test the claim that men have a higher rate of red/green color blindness, we need to compare the proportions of color blindness in men and women.

Let p_m be the proportion of men with color blindness and p_w be the proportion of women with color blindness.

Null hypothesis: p_m = p_w

Alternative hypothesis: p_m > p_w

To test this hypothesis, we can use a chi-square test of independence. We will compare the observed frequencies of color blindness in men and women to the expected frequencies under the assumption that men and women have the same rate of color blindness.

The chi-square test statistic is calculated as follows:

X^2 = (O_m - E_m)^2/E_m + (O_w - E_w)^2/E_w

where O_m and O_w are the observed frequencies of color blindness in men and women, and E_m and E_w are the expected frequencies of color blindness in men and women.

If the chi-square test statistic is large enough, we reject the null hypothesis and conclude that men have a higher rate of color blindness than women.

The distance (in feet) in which the three steps would overlap for the first time is 56 feet.

Using LCM method

We would find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.

Multiples of 24:

24, 48, 72, 96, 120, 144, 168, 192, 216, 240, 264, 288, 312, 336, 360, 384, 408, 432, 456, 480, 504, 528, 552, 576, 600, 624, 648, 672, 696, 720

Multiples of 28:

28, 56, 84, 112, 140, 168, 196, 224, 252, 280, 308, 336, 364, 392, 420, 448, 476, 504, 532, 560, 588, 616, 644, 672, 700, 728

Multiples of 32:

32, 64, 96, 128, 160, 192, 224, 256, 288, 320, 352, 384, 416, 448, 480, 512, 544, 576, 608, 640, 672, 704, 736

Therefore,

LCM(24, 28, 32) = 672

Then we convert to feet:

672/12= 56 feet.

Read more about LCM here:

https://brainly.com/question/233244

#SPJ9

Answer:

We have the matrix [tex]A=\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&8&4\end{array}\right][/tex]

To find the eigenvalues of A we need find the zeros of the polynomial characteristic [tex]p(\lambda)=det(A-\lambda I_3)[/tex]

Then

[tex]p(\lambda)=det(\left[\begin{array}{ccc}-4-\lambda&-4&-4\\0&-8-\lambda&-4\\0&8&4-\lambda\end{array}\right] )\\=(-4-\lambda)det(\left[\begin{array}{cc}-8-\lambda&-4\\8&4-\lambda\end{array}\right] )\\=(-4-\lambda)((-8-\lambda)(4-\lambda)+32)\\=-\lambda^3-8\lambda^2-16\lambda[/tex]

Now, we fin the zeros of [tex]p(\lambda)[/tex].

[tex]p(\lambda)=-\lambda^3-8\lambda^2-16\lambda=0\\\lambda(-\lambda^2-8\lambda-16)=0\\\lambda_{1}=0\; o \; \lambda_{2,3}=\frac{8\pm\sqrt{8^2-4(-1)(-16)}}{-2}=\frac{8}{-2}=-4[/tex]

Then, the eigenvalues of A are [tex]\lambda_{1}=0[/tex] of multiplicity 1 and [tex]\lambda{2}=-4[/tex] of multiplicity 2.

Let's find the eigenspaces of A. For [tex]\lambda_{1}=0[/tex]: [tex]E_0 = Null(A- 0I_3)=Null(A)[/tex].Then, we use row operations to find the echelon form of the matrix

[tex]A=\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&8&4\end{array}\right]\rightarrow\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&0&0\end{array}\right][/tex]

We use backward substitution and we obtain

1.

[tex]-8y-4z=0\\y=\frac{-1}{2}z[/tex]

2.

[tex]-4x-4y-4z=0\\-4x-4(\frac{-1}{2}z)-4z=0\\x=\frac{-1}{2}z[/tex]

Therefore,

[tex]E_0=\{(x,y,z): (x,y,z)=(-\frac{1}{2}t,-\frac{1}{2}t,t)\}=gen((-\frac{1}{2},-\frac{1}{2},1))[/tex]

For [tex]\lambda_{2}=-4[/tex]: [tex]E_{-4} = Null(A- (-4)I_3)=Null(A+4I_3)[/tex].Then, we use row operations to find the echelon form of the matrix

[tex]A+4I_3=\left[\begin{array}{ccc}0&-4&-4\\0&-4&-4\\0&8&8\end{array}\right] \rightarrow\left[\begin{array}{ccc}0&-4&-4\\0&0&0\\0&0&0\end{array}\right][/tex]

We use backward substitution and we obtain

1.

[tex]-4y-4z=0\\y=-z[/tex]

Then,

[tex]E_{-4}=\{(x,y,z): (x,y,z)=(x,z,z)\}=gen((1,0,0),(0,1,1))[/tex]

Answer:

add all of it

Step-by-step explanation:

Answer:

1.50 marks per pound

Step-by-step explanation:

Data provided in the question:

Selling price of three-wheel cars made in North Edsel = 5000 pounds

Selling price of four-wheel cars made in south Edsel = 10,000 marks

Real exchange rate between North and South Edsel

= four​ three-wheel cars for three​ four-wheel cars

i.e

⇒ 4 × 5000 pounds  = 3 × 10,000 marks

or

1 pounds = [ ( 3 × 10,000 ) ÷ ( 4 × 5,000) ]

or

1 pound = 30,000 ÷ 20,000

or

1 pound = 1.50 marks

Hence,

The nominal exchange rate = 1.50 marks per pound

Final answer:

The nominal exchange rate between North and South Edsel is 3750 pounds.

Explanation:

The real exchange rate is the ratio at which goods and services of one country can be exchanged for those of another country. In this case, the real exchange rate between North and South Edsel is four three-wheel cars for three four-wheel cars. So, for every four four-wheel cars from South Edsel, you can exchange them for three three-wheel cars from North Edsel.

Since the price of the three-wheel cars from North Edsel is 5000 pounds, and the ratio is four three-wheel cars for three four-wheel cars, the nominal exchange rate between the two countries would be:

5000 pounds * (3 four-wheel cars / 4 three-wheel cars) = 3750 pounds.

Therefore, the nominal exchange rate between the two countries is 3750 pounds.

To calculate the total cost of drilling a 500-meter well, add the fixed cost to the sum of the marginal costs for every meter drilled. The total cost comes out to 3,500,000 riyals.

The total cost of drilling a 500-meter well comprises both fixed costs and the marginal cost per meter of depth. We are given that the fixed costs amount to 1,000,000 riyals. The marginal cost function is given as C'(x) = 4000 + 10x, which means the cost per additional meter drilled increases linearly with depth.

To find the total cost of drilling a 500-meter well, we need to compute the integral (i.e., the area under the curve) of the marginal cost function from 0 to 500 and add the fixed costs. This calculation represents the sum of the increasing cost per meter for every meter drilled.

By evaluating the integral ∫ (4000 + 10x) dx from 0 to 500, we get 2,500,000 riyals. This is the total variable cost of drilling a 500m well. Adding it to the fixed cost (1,000,000 riyals), the grand total comes out to be 3,500,000 riyals.

https://brainly.com/question/32126253

#SPJ11

Answer:

[tex]1.75*10^{-27}[/tex]

Step-by-step explanation:

If, collectively, 15 people weigh more than 3500 pounds, that means each person must weigh more than 3500/15 = 233.33 pounds.

If the distribution for population weights is normal at mean = 180 and standard deviation = 25 lbs, that means the probability for 1 person to weigh higher than 233 lbs is

[tex]1 - P(x > 233, \mu = 180, \sigma = 25) = 1 - 0.984 = 0.016[/tex]

For all 15 people to have higher weigh than that then the probability is

[tex]0.016^{15} = 1.75*10^{-27}[/tex]

This is indeed very unlikely to happen

Final answer:

It would be highly unusual for 15 people in an elevator to have a combined weight of more than 3,500 lbs; the z-score of 8.27 reflects an extremely small probability, pointing to a rare event.

Explanation:

To determine how unusual it would be for 15 people in an elevator to have a combined weight of more than 3,500 lbs, we use the Central Limit Theorem. Given that each person's weight is a random variable X that is normally distributed with a mean (μ) of 180 lbs and a standard deviation (σ) of 25 lbs, the sum of the weights of 15 people will also be normally distributed with a mean (μtotal) of 15 * 180 lbs and a standard deviation (σtotal) of √15 * 25 lbs, due to the Central Limit Theorem.

The next step is to calculate these values:

Now we calculate the z-score to determine how many standard deviations away 3,500 lbs is from the mean:

Z = (X - μtotal) / σtotal

Z = (3500 - 2700) / 96.82 ≈ 8.27

Using standard normal distribution tables or a calculator, we find that the probability of a z-score of 8.27 is extremely small, indicating that it would be highly unusual for 15 people to weigh more than 3,500 lbs in total.

Answer: (6.304, 6.696)  

Step-by-step explanation:

The confidence interval for population mean is given by :-

[tex]\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}[/tex]

, where [tex]\sigma[/tex] = Population standard deviation.

n= sample size

[tex]\overline{x}[/tex] = Sample mean

z* = Critical z-value .

Let x denotes the number of  hours slept by UCF students.

Given :  [tex]\sigma=2\ hours[/tex]

n= 400

[tex]\overline{x}= 6.5\ hours[/tex]

Two-tailed critical value for 95% confidence interval = [tex]z^*=1.96[/tex]

Then, the 95%confidence interval for the true number of hours slept by UCF students will be :-

[tex]6.5\pm(1.96)\dfrac{2}{\sqrt{400}}\\\\=6.5\pm(1.96)\dfrac{2}{20}\\\\=6.5\pm0.196=(6.5-0.196,\ 6.5+0.196)=(6.304,\ 6.696)[/tex]

Hence, the 95% confidence interval for the true number of hours slept by UCF students : (6.304, 6.696)  

Answer:  The answers are given below.

Step-by-step explanation:  Given that a  3 × 3 matrix P satisfies the matrix equation P² = P.

We are to

(a) find the possibilities for the determinant of P.

(b) explain the reason behind there are no other possibilities.

(c) give an example of P, for each possible determinant.

(a) According to the given information, we have

[tex]P^2=P\\\\\Rightarrow P^2-P=0\\\\\Rightarrow P(P-I)=0\\\\\Rightarrow P=0,~~~P=I.[/tex]

So, P can be either a zero matrix of order 3 or an identity matrix of order 3.

If P = 0, then det(P) = 0     and     if P = I, then det(P) = 1.

Therefore the possible determinants of P are 0 and 1.

(b) There can be any other determinant other than 0 and 1, because if so, then the given equation P² = P will not be satisfied.

(c) If |P| = 0, then the matrix P can be can be of the form as follows:

[tex]P=\left[\begin{array}{ccc}0&0&0\\0&0&0\\0&0&0\end{array}\right] .[/tex]

If |P| = 1, the the matrix P can be of the form as follows :

[tex]P=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] .[/tex]

Thus, all the parts are answered.

Answer:

22.23

Step-by-step explanation:

Given that in a binomial experiment study, in a certain community showed that 5% of the people suffer from insomnia

i.e p = 0.05

q=0.95

n=10400

[tex]np=10400*0.05=202\\npq =494[/tex]

Var(x)=[tex]npq=494\\std dev=\sqrt{494} \\=22.23[/tex]

the standard deviation of the number of people who suffer from insomnia

=22.23

The question is about calculating the standard deviation for the number of people suffering from insomnia in a particular community. The standard deviation can be calculated using the formula for the binomial distribution σ = sqrt(n*p*q).

The question asks for determination of the standard deviation of the number of people suffering from insomnia in a community of 10,400 where 5% suffer from insomnia. This context implies a binomial distribution as there are two outcomes (those who suffer from insomnia and those who do not) and a fixed number of trials (10,400 people).

For a binomial distribution, the standard deviation is found through the formula : σ = sqrt(n*p*q), where n is the total number of trials, p is the probability of success, and q is the probability of failure (1-p). Here, n=10,400, p=0.05 and q=0.95.

Here the standard deviation would be calculated as σ = sqrt{10,400 * 0.05 * 0.95}.

https://brainly.com/question/23907081

#SPJ11

Answer:

a) [tex]\hat \mu = \bar x =\sum_{i=1}^{10} \frac{x_i}{10}=120.2[/tex]

b) [tex]\hat \tau =n\hat \mu =25000x120.2=3005000[/tex]

c) [tex]\hat p= \hat \theta =\frac{8}{10}=0.8[/tex]

d) Median =120

Step-by-step explanation:

1) Some important concepts

The mean refers to the "average that is used to derive the central tendency of data analyzed. It is determined by adding all the data points in a population and then dividing the total by the number of points".

Method of moments "involves equating sample moments with theoretical moments". For example the first sample moment about the origin is defined as [tex]M_1=\frac{1}{n} \sum_{i=1}^{n}x_i =\bar X [/tex]

The median is "the middle number in a sorted, ascending or descending, list of numbers and can be more descriptive of that data set than the average".

When we are trying to estimate the population proportion, p.

All estimation is based on the fact that the normal can be used to approximate the binomial distribution when np and nq are both at least 5. Where p is the probability of success and q the probability of failure.

2) Part a

Using the method of the moments a point of estimate for the [tex]\mu[/tex] is:

[tex]\hat \mu = \bar x =\sum_{i=1}^{10} \frac{x_i}{10}=120.2[/tex]

3) Part b

If [tex]\hat \mu[/tex] is an individual estimate for the average gas usage during January and [tex]\tau[/tex] represent "the total amount of gas used by all of these houses during January" then the estimation for the total would be given by:

[tex]\hat \tau =n\hat \mu =25000x120.2=3005000[/tex]

3) Part c

For this part we want to estimate p ="the proportion of all houses that used at least 100 therms". If X is the random variable who represent the number of houses that exceed the usage of 100, we see that 8 out of 10 values are above 100, so the random variable X would be distributed binomial

[tex]X \sim Bin(10,0.8)[/tex] where n=10 and

[tex]\hat p= \hat \theta =\frac{8}{10}=0.8[/tex]

4) Part d

In order to find the median we need to put the data in order first, like this:

86,99,103,109,118,122,125,138,149,153

Since we have 10 observations and this number is even the procedure that we need to use in order to find the median is:

a) Find the value at position[n/2]=[10/2]=[5] on the data set ordered. For this case the value at position [5] is 118

b) Find the value at position[n/2 +1]=[10/2 +1]=[6] on the data set ordered. For this case the value at position [6] is 122

c) Find the average from the values obtained on steps a) and b). for this case (118+122)/2=120

So the Median = 120

The mean of the data set is 120.2

The total gas used =  3005000

The proportion of at least therms = 0.8

The median of the set = 120

a. To get the average gas usage, we are asked to calculate the mean of the observation.

Average

[tex]\frac{ 153+103+125+149+118+109+86+122+138+99}{10} \\\\[/tex]

= 120.2

b. The question says that 25000 houses make use of natural gas, then

total gases used by these houses =

120.2 * 25000

= 3005000

c. The proportion of the houses that used above 100 therms,

The house above 100 here are, 125, 149, 118, 109, 122, 138

They are 8 in number.

8/10 = 0.8

0.8 is the proportion that used at least 100 therms.

d. We have to find the median for the set here. We arrange the details in ascending order.

86,99,103,109,118,122,125,138,149,153

Median = 118+122/2

= 240/2

= 120

Answer:

b) The annual rainfall in a city

Step-by-step explanation:

Remember, a discrete variable is one that can only take a finite number of values between any two values of a characteristic and a continuous variable is one that can take an infinite number of values between any two values of a characteristic.

a) Observe that the variable x='classes taken in one semester' can take the values 0,1,2,...,n.

Then the variable x is discrete

b) Observe that the variable x='annual rainfall in a city' can take the values 2in, 1.6in, 5.1 in, 0.1in

Then, the variable x can be take a infinite number of values between two number. So x isn't a discrete variable.

c) The variable x='attendance at a football game' can take the values 3000,5000... n. And never will be a decimal number because There cannot be a personal decimal number. Therefore, x is a discrete variable.

d) The variable x='patients treated at an emergency room in a day' can take the values 1,2,3,...,n. And never will be a decimal number because There cannot be a personal decimal number. Therefore, x is a discrete variable.

Answer:

the concentration at 10 minutes= 0.4+0.0133= 0.4133%

Step-by-step explanation:

Air containing 0.04% carbon dioxide

V, volume of room is 6000 ft3.

Q, rate of air 2000 ft3/min,

initial concentration of 0.4% carbon dioxide,

determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes?

firstly, we find the time taken for air to completely filled the room

Q = V/t

t = V/Q = 6000/2000 = 3min

so, its take 3mins for air to be completely filled in the room and for exhaust air to move out.

there is  an initial concentration of 0.4% carbon dioxide, and the air pump in is 0.04%.

therefore,

3mins = 0.04% of CO2

3*60 =180sec = 0.04%

1sec = 0.04/180 = 0.00022%/sec

so at any time the concentration of CO2 is 0.4 + 0.00022 =0.40022%/sec

What is the concentration at 10 minute

the concentration at 10minutes = the concentration for 1minute because at every minutes, the concentration moves in is moves out. = concentration for 2000ft3.

for 0.04% = 6000ft3

   ?          = 2000ft3

              = 2000* 0.04)/6000 =0.0133%

the concentration at 10 minutes= 0.4+0.0133= 0.4133%

Answer:

The probability the average mpg is less than 28 mpg is 0.8413.

Step-by-step explanation:

Given : The manufacturer of a new compact car claims the miles per gallon (mpg) for the gasoline consumption is approximately normal with

Mean [tex]\mu=26[/tex] mpg and standard deviation [tex]\sigma=12[/tex] mpg.

Number of sample n=36

To find : What is the probability the average mpg is less than 28 mpg?

Solution :

Applying z-score formula,

[tex]z=\dfrac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

The portability is given by, [tex]P(X<28)[/tex]

[tex]=P(\dfrac{x-\mu}{\frac{\sigma}{\sqrt{n}}}<\dfrac{28-26}{\frac{12}{\sqrt{36}}})[/tex]

[tex]=P(\dfrac{x-\mu}{\frac{\sigma}{\sqrt{n}}}<\dfrac{2}{\frac{12}{6}})[/tex]

[tex]=P(z<\dfrac{2}{2})[/tex]

[tex]=P(z<1)[/tex]

Using z-table,

[tex]=0.8413[/tex]

Therefore, the probability the average mpg is less than 28 mpg is 0.8413.

Answer:

he probability the average mpg is less than 28 mpg is 0.8413.

Step-by-step explanation:

Given : The manufacturer of a new compact car claims the miles per gallon (mpg) for the gasoline consumption is approximately normal with

Mean  mpg and standard deviation  mpg.

Number of sample n=36

To find : What is the probability the average mpg is less than 28 mpg?

Solution :

Applying z-score formula,

The portability is given by,  

Using z-table,

Therefore, the probability the average mpg is less than 28 mpg is 0.8413.

Step-by-step explanation:

Let [tex]\mu[/tex] denotes the average width of stripes .

As per given , we have

[tex]H_0:\mu=4\\H_a:\mu\neq4[/tex]

, since [tex]H_a[/tex] is two-tailed , so the test is a two-tailed test.

Also, population standard deviation is unknown , so we perform two-tailed t-test.

For Sample size : n= 45

Sample mean : [tex]\overline{x}=3.87[/tex]

Sample  standard deviation : s= 0.5 inches

Test statistic : [tex]t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}[/tex]

i.e.  [tex]t=\dfrac{3.87-4}{\dfrac{0.5}{\sqrt{45}}}\approx-1.74[/tex]

Significance level = [tex]\alpha=0.05[/tex]

By using t-value table,

Two-tailed critical t-value  = [tex]t_{\alpha/2,df}=t_{0.025,\ 44}=\pm2.0154[/tex]  [df = n-1]

Decision : Since the test statistic value (-1.74) lies with in the interval (-2.0154, 2.0154) , it means we are failed to reject the null hypothesis .

Conclusion: We have sufficient evidence to support the claim that the machine has slipped out of adjustment and the average width of stripes is no longer μ = 4 inches.

After calculating the z-score for the provided data, the result (-1.962) lies within the critical values for a 5% level of significance. Therefore, we cannot reject the null hypothesis, hence there's no sufficient evidence to state that the machine is out of adjustment.

To determine if the machine has slipped out of adjustment, we should conduct a hypothesis test. We can set the null hypothesis (H0) as μ = 4 (the machine is correctly adjusted), and the alternative hypothesis (Ha) as μ ≠ 4 (the machine has slipped). The sample size is large enough (>30) to use the z-score.
The z-score can be calculated using the formula: z = (x - μ)/(s/√n), where x is the sample mean, μ is the population mean, s is the standard deviation, and n is the sample size.
So, z = (3.87 - 4) / (0.5/√45) = -1.962. The critical z-value for a 5% level of significance (two-tailed test) is approximately +/- 1.96. Our calculated z-value falls within this range so we cannot reject the null hypothesis. Therefore, we don't have sufficient evidence to state that the machine has slipped out of adjustment.

https://brainly.com/question/34171008

#SPJ11

Final answer:

To reject the null hypothesis (H0) if the calculated t-value is less than -1.9842.

Explanation:

To establish the appropriate rejection region for a significance level of 0.05, one compares the test statistic (denoted as 't') with a critical value. In this case, the correct rejection region dictates that the null hypothesis (H0) should be rejected if the calculated 't' is less than -1.9842.

Simply put, if the derived 't'-value falls below -1.9842, it implies rejecting the null hypothesis. This decision leads to the conclusion that there is enough evidence to support the alternative hypothesis (Ha: μ < 60). This meticulous comparison adheres to statistical principles, ensuring a reliable interpretation of the test outcomes and providing a solid foundation for decision-making in hypothesis testing.

Answer:

a) 6.3 minutes

Step-by-step explanation:

Population mean (μ) = 7.2 minutes  

Standard deviation (σ) = 0.56 minutes

The z-score for any running time 'X' is given by:

[tex]z=\frac{X-\mu}{\sigma}[/tex]

In this scenario, the company is looking for the top 5% runners, that is, runners at and below the 5-th percentile of the normal distribution. The equivalent z-score for the 5-th percentile is 1.645.

Therefore, the minimum speed, X, a runner needs to achieve in order to be sponsored is:

[tex]-1.645=\frac{X-7.2}{0.56}\\X= 6.3\ minutes[/tex]

Answer:

We accept the alternate hypothesis. We conclude that the mean lifetime of tires is  is less than 33,000 miles.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 33,000 miles

Sample mean, [tex]\bar{x}[/tex] =  32, 450 miles

Sample size, n = 18

Alpha, α = 0.05

Sample standard deviation, s = 1200 miles

a) First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 33000\text{ miles}\\H_A: \mu < 33000\text{ miles}[/tex]

b) Level of significance:

[tex]\alpha = 0.05[/tex]

c) We use One-tailed t test to perform this hypothesis.

d) Formula:

[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex] Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{32450 - 33000}{\frac{1200}{\sqrt{18}} } = -1.9445[/tex]

Now, [tex]t_{critical} \text{ at 0.05 level of significance, 17 degree of freedom } = -1.7396[/tex]

Rejection area:

[tex]t < -1.7396[/tex]

Since,                  

[tex]t_{stat} < t_{critical}[/tex]

e) We fail to accept the null hypothesis and reject it as the calculated value of t lies in the rejection area.

f) We accept the alternate hypothesis. We conclude that the mean lifetime of tires is  is less than 33,000 miles.

Final answer:

To find the fraction of applicants with a score of 400 or above, convert the score to a z-score and look it up in a standard normal distribution table. Subtract the resulting proportion from 1 to find the fraction above 400. Approximately 77.34% of the applicants would have a score of 400 or above.

Explanation:

To find the fraction of applicants who would have a score of 400 or above, we need to find the area under the normal distribution curve to the right of 400. First, we need to convert the score of 400 to a z-score using the formula:

z = (x - μ) / σ

where z is the z-score, x is the score, μ is the mean, and σ is the standard deviation. In this case, the mean is 460 and the standard deviation is 80, so the z-score is:

z = (400 - 460) / 80 = -0.75

Once we have the z-score, we can look it up in a standard normal distribution table to find the proportion of the distribution that is below it. The table gives us a value of approximately 0.2266 for a z-score of -0.75. Since we want the fraction above 400, we can subtract this value from 1 to get:

1 - 0.2266 = 0.7734

Therefore, we would expect approximately 77.34% of the applicants to have a score of 400 or above.

Answer:  The length of the loner side is 7 in. and the length of the shorter side is 6 in.

Step-by-step explanation:  Given that a graphics designer is designing an advertising brochure for an art show. Each page of the brochure is rectangular with an area of 42 in squared and a perimeter of 26 in.

We are to find the dimensions of the brochure.

Let l and b represents the lengths of the longer side and shorter side respectively of each page of the brochure.

Then, according to the given information, we have

[tex]l\times b=42~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]

and

[tex]2(l+b)=26\\\\\Rightarrow l+b=13\\\\\Rightarrow l=13-b~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)[/tex]

Substituting the value of l from equation (ii) in equation (i), we get

[tex](13-b)b=42\\\\\Rightarrow b^2-13b+42=0\\\\\Rightarrow b^2-6b-7b+42=0\\\\\Rightarrow (b-6)(b-7)=0\\\\\Rightarrow b-6=0,~~~b-7=0\\\\\Rightarrow b=6,7.[/tex]

Since b is the length of the shorter side, so b = 6 in.

From equation (ii), we get

[tex]l=13-6=7.[/tex]

Thus, the length of the loner side is 7 in. and the length of the shorter side is 6 in.

To find the dimensions of the brochure, we use the formulas for the area and perimeter of a rectangle. Solving the system of equations produced by these formulas, we find that the length of the brochure is 7 inches and the width is 6 inches.

In order to find the dimensions of the brochure, we can use the formulas for the area and perimeter of a rectangle. Given area, A = 42 inches squared and perimeter, P = 26 inches. The formulas for the area and perimeter of the rectangle are A = length x width and P = 2(length + width).

Let's denote the length of the rectangle as 'l' and the width as 'w'. Now we know that:

l x w = 42 inches (according to area formula)

2(l + w) = 26 inches (according to perimeter formula)

This is a system of two equations which can be solved simultaneously. After solving these equations, we find that the length (longer side) is 7 inches, and the width (shorter side) is 6 inches.

https://brainly.com/question/22109593

#SPJ11

Answer:

840

Step-by-step explanation:

Answer:

Maximum: ((1,-2,5) ; 30)

Minimum: ((-1,2,-5) ; -30)

Step-by-step explanation:

We have the function f(x,y,z) = x - 2y + 5z, with the constraint g(x,y,z) = 30, with g(x,y,z) = x²+y²+z². The Lagrange multipliers Theorem states that, the points (xo,yo,zo) of the sphere where the function takes its extreme values  should satisfy this equation:

grad(f) (xo,yo,zo) = λ * grad(g) (xo,yo,zo)

for a certain real number λ. The gradient of f evaluated on a point (x,y,z) has in its coordinates the values of the partial derivates of f evaluated on (x,y,z). The partial derivates can be calculated by taking the derivate of the function by the respective variable, treating the other variables as if they were constants.

Thus, for example, fx (x,y,z) = d/dx x-2y+5z = 1, because we treat -2y and 5z as constant expressions, and the partial derivate on those terms is therefore 0. We calculate the partial derivates of both f and g

Thus, for a critical point (x,y,z) we have this restrictions:

The last equation is just the constraint given by g, that (x,y,z) should verify.

We can put every variable in function of λ, and we obtain the following equations.

Now, we replace those values with the constraint, obtaining

(1/2λ)² + (-1/λ)²+(5/2λ)² = 30

Developing the squares and taking 1/λ² as common factor, we obtain

(1/λ²) * (1/4 + 1 + 25/4) = (1/λ²) * 30/4 = 30

Hence, λ² = 1/4, or, equivalently,[tex]\lambda =^+_- \frac{1}{2} . [/tex]

If [tex]\lambda = \frac{1}{2} , [/tex] then 1/λ is 2, and therefore

and f(x,y,z) = f(1,-2,5) = 1 -2 * (-2) + 5*5 = 30

If [tex]\lambda = - \frac{1}{2} , [/tex] then 1/λ is -2, and we have

and f(x,y,z) = f(-1,2,-5) = -1 -2*2 + 5*(-5) = -30.

Since the extreme values can be reached only within those two points, we conclude that the maximun value of f in the sphere takes place on ((1,-2,5) ; 30), and the minimun value takes place on ((-1,2,-5) ; -30).

The maximum value is 30, and the minimum value is -30. These occur at the points (1, -2, 5) and (-1, 2, -5), respectively.

To find the maximum and minimum values of f(x, y, z) = x − 2y + 5z on the sphere x² + y² + z² = 30, we use Lagrange multipliers. The constraint is g(x, y, z) = x² + y² + z² - 30 = 0.

We introduce a Lagrange multiplier λ and set up the system of equations:

Solving the system:

Equating the λs:

Substituting y and z into the constraint:

Thus, x = ±1. For x = 1: y = -2, z = 5.

For x = -1: y = 2, z = -5.

Evaluating f at these points:

Hence, the maximum value is 30 and the minimum value is -30.

Answer: Margin of error = 8.32, and Confidence interval using normal distribution is narrower than confidence interval using t-distribution.

Step-by-step explanation:

Since we have given that

n = 13

Mean repair cost = $85.00

Standard deviation = $15.30

At 95% confidence interval,

z= 1.96

Since it is normally distributed.

Margin of error is given by

[tex]z\times \dfrac{\sigma}{\sqrt{n}}\\\\=1.96\times \dfrac{15.30}{\sqrt{13}}\\\\=8.32[/tex]

95% confidence interval would be

[tex]\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=85\pm 1.96\times \dfrac{15.30}{\sqrt{13}}\\\\=85\pm 8.32\\\\=(85-8.32,85+8.32)\\\\=(76.68,93.32)[/tex]

A 95​% confidence interval using the​ t-distribution was (75.8, 94.2 ).

Confidence interval using normal distribution is narrower than confidence interval using t-distribution.

Final answer:

A 95% confidence interval for the mean repair cost of microwave ovens with a known standard deviation is calculated using the z-score. The margin of error is found to be approximately $8.31, resulting in a confidence interval of ($76.69, $93.31).

Explanation:

To calculate the 95% confidence interval for the population mean when the population standard deviation is known, we can use the z-score associated with the 95% confidence level, which is 1.96. The formula for the margin of error (EBM) is EBM = z * (σ/√n), where σ is the population standard deviation, n is the sample size, and z is the z-score. Given that the sample standard deviation is $15.30, we assume it to be the population standard deviation because the question states that it is known.

With a sample mean (μ) of $85.00, a standard deviation of $15.30, and a sample size of 13, the margin of error is calculated as follows:

EBM = 1.96 * (15.30/√13) = 1.96 * 4.24 ≈ $8.31

The 95% confidence interval is therefore ($85.00 - $8.31, $85.00 + $8.31) = ($76.69, $93.31). The results using the z-distribution are similar to those obtained using the t-distribution, but usually, the t-distribution would be used when the sample size is small and the population standard deviation is unknown, which results in a wider interval due to the extra uncertainty.

The suggested mileage interval, excluding no more than 10% of the mileage on tires, is approximately 18,420 to 31,580 miles.

To determine the mileage interval that excludes no more than 10% of the mileage on tires, we can use the standard normal distribution and the properties of the normal curve. The mileage data is normally distributed with a mean [tex](\(\mu\))[/tex] of 25,000 miles and a standard deviation [tex](\(\sigma\))[/tex] of 4,000 miles.

To find the interval, we can use the Z-score formula:

[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]

To exclude no more than 10% of the mileage, we need to find the Z-score corresponding to the 5th percentile on each side of the mean, as 10% is split between the lower and upper tails of the distribution.

Using a standard normal distribution table or a calculator, the Z-score for the 5th percentile is approximately -1.645 (negative due to being in the lower tail).

Now, we can use the Z-score formula to find the values of [tex]\(X\)[/tex] (mileage) corresponding to these Z-scores:

[tex]\[ X_{\text{lower}} = \mu + Z \times \sigma \][/tex]

[tex]\[ X_{\text{upper}} = \mu - Z \times \sigma \][/tex]

Substitute the values:

[tex]\[ X_{\text{lower}} = 25,000 - (-1.645) \times 4,000 \][/tex]

[tex]\[ X_{\text{upper}} = 25,000 + (-1.645) \times 4,000 \][/tex]

Calculating these values:

[tex]\[ X_{\text{lower}} \approx 31,580 \][/tex]

[tex]\[ X_{\text{upper}} \approx 18,420 \][/tex]

Therefore, the suggested mileage interval is approximately 18,420 miles to 31,580 miles to exclude no more than 10% of the mileage on the tires.

Answer:

Let [tex]A = (a_1, ..., a_n)[/tex] and [tex]B = (b_1, ..., b_n)[/tex] bases of V. The matrix of change from A to B is the matrix n×n whose columns are vectors columns of the coordinates of vectors [tex]b_1, ..., b_n[/tex] at base A.

The, we case correspond to find the coordinates of vectors of C,

[tex]\{\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right], \left[\begin{array}{ccc}2\\0\\-1\end{array}\right], \left[\begin{array}{ccc}-3\\1\\2\end{array}\right]   \}[/tex]

at base B.

1. We need to find [tex]a,b,c\in\mathbb{R}[/tex] such that

[tex]\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right][/tex]

Then we find these values solving the linear system

[tex]\left[\begin{array}{cccc}1&-2&2&2\\-1&2&-1&-1\\0&-1&1&-1\end{array}\right][/tex]

Using rows operation we obtain the echelon form of the matrix

[tex]\left[\begin{array}{cccc}1&-2&2&2\\0&-1&1&-1\\0&0&1&1\end{array}\right][/tex]

now we use backward substitution

[tex]c=1\\-b+c=-1,\; b=2\\a-2b+2c=2,\; a=4[/tex]

Then the coordinate vector of [tex]\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right][/tex] is [tex]\left[\begin{array}{ccc}4\\2\\1\end{array}\right][/tex]

2. We need to find [tex]a,b,c\in\mathbb{R}[/tex] such that

[tex]\left[\begin{array}{ccc}2\\0\\-1\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right][/tex]

Then we find these values solving the linear system

[tex]\left[\begin{array}{cccc}1&-2&2&2\\-1&2&-1&0\\0&-1&1&-1\end{array}\right][/tex]

Using rows operation we obtain the echelon form of the matrix

[tex]\left[\begin{array}{cccc}1&-2&2&2\\0&-1&1&-1\\0&0&1&2\end{array}\right][/tex]

now we use backward substitution[tex]c=2\\-b+c=-1,\; b=3\\a-2b+2c=2,\; a=4[/tex]

Then the coordinate vector of [tex]\left[\begin{array}{ccc}2\\0\\-1\end{array}\right][/tex] is [tex]\left[\begin{array}{ccc}4\\3\\2\end{array}\right][/tex]

3. We need to find [tex]a,b,c\in\mathbb{R}[/tex] such that

[tex]\left[\begin{array}{ccc}-3\\1\\2\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right][/tex]

Then we find these values solving the linear system

[tex]\left[\begin{array}{cccc}1&-2&2&-3\\-1&2&-1&1\\0&-1&1&2\end{array}\right][/tex]

Using rows operation we obtain the echelon form of the matrix

[tex]\left[\begin{array}{cccc}1&-2&2&-3\\0&-1&1&2\\0&0&1&-2\end{array}\right][/tex]

now we use backward substitution[tex]c=-2\\-b+c=2,\; b=-4\\a-2b+2c=2,\; a=-2[/tex]

Then the coordinate vector of [tex]\left[\begin{array}{ccc}-3\\1\\2\end{array}\right][/tex] is [tex]\left[\begin{array}{ccc}-2\\-4\\-2\end{array}\right][/tex]

Then the change of basis matrix from B to C is

[tex]\left[\begin{array}{ccc}4&4&-2\\2&3&-4\\1&2&-2\end{array}\right][/tex]

To find the change of basis matrix from basis B to basis C in R3, invert basis B, multiply it by basis C, and the resulting matrix transforms coordinates from B to C: [[1 2 -2], [2 3 -4], [4 4 -7]].

Here's how to go about finding the change of basis matrix from basis B to basis C in R3:

1. Write down the vector coordinates of interest. These coordinates are given by basis B and basis C:
   basis B : {⟨2,−1,1⟩,⟨−2,2,−1⟩,⟨1,−1,0⟩}
   basis C : {⟨2,−1,−1⟩,⟨2,0,−1⟩,⟨−3,1,2⟩}

2. Find the inverse of basis B. The inverse of a matrix is such that if you multiply the original matrix by its inverse you get the identity matrix — a simple "1, 0" matrix. This step effectively reverses the transformation provided by basis B.

3. Then, calculate the product of basis B's inverse and basis C. This essentially re-projects the coordinates of basis B onto basis C.

4. The resulting matrix is your Change of Basis matrix from B to C. In our calculation, this comes out as:

Change of Basis from B to C:
   [[ 1.  2. -2.]
    [ 2.  3. -4.]
    [ 4.  4. -7.]]

This matrix will transform any vector in coordinates relative to basis B into coordinates relative to basis C. The first row indicates how much of each vector in B is needed to form the first vector in C, the second row for the second vector in C, and so on.

To learn more about basis matrix

https://brainly.com/question/31644229

#SPJ6

Answer:

The sampling distribution of the sample mean has approximately a normal distribution because of

c) the central limit theorem.

Step-by-step explanation:

Given that the average exam score of students of a large class is 70 with a standard deviation of 10.

From the above students a sample of 36 students is selected, and the mean score of these students is computed.

As per central limit theorem we have when samples are drawn at random from population, with sample size sufficiently large to represent the population then sample mean follows a normal distribution.

Here population size N = 70 and sample size n =36

we can say sample size is greater than 30 and sufficiently large to represent the population. Also we can assume that these are randomly drawn.

So the answer would be

The sampling distribution of the sample mean has approximately a normal distribution because of

c) the central limit theorem.