If you deposit $5,000 into an account that pays 4% simple interest, what will the balance be in 6 years?

Answer:

(A) The mean for the differences is 2.0.

(B) The test statistic is 1.617.

(C) At 90% confidence the null hypothesis should not be rejected.

Step-by-step explanation:

We are given that a random sample of eight cars from each manufacturer is selected, and eight drivers are selected to drive each automobile for a specified distance.

The following data (in miles per gallon) show the results of the test;

Driver         Manufacturer A               Manufacturer B

   1                      32                                       28

  2                      27                                       22

  3                      26                                       27

  4                      26                                       24

  5                      25                                       24

  6                      29                                       25

  7                       31                                       28

  8                      25                                       27

Let [tex]\mu_1[/tex] = mean MPG for the fuel efficiency of Manufacturer A brand

[tex]\mu_2[/tex] = mean MPG for the fuel efficiency of Manufacturer B brand

SO, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu_1-\mu_2=0[/tex]  or  [tex]\mu_1= \mu_2[/tex]    {means that there is a not any significant difference in the mean MPG (miles per gallon) when testing for the fuel efficiency of these two brands of automobiles}

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu_1-\mu_2\neq 0[/tex]  or  [tex]\mu_1\neq \mu_2[/tex]   {means that there is a significant difference in the mean MPG (miles per gallon) for the fuel efficiency of these two brands of automobiles}

The test statistics that will be used here is Two-sample t test statistics as we don't know about the population standard deviations;

                      T.S.  = [tex]\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex]  ~ [tex]t__n__1+_n__2-2[/tex]

where, [tex]\bar X_1[/tex] = sample mean MPG for manufacturer A = [tex]\frac{\sum X_A}{n_A}[/tex] = 27.625

[tex]\bar X_2[/tex] = sample mean MPG for manufacturer B =[tex]\frac{\sum X_B}{n_B}[/tex] = 25.625

[tex]s_1[/tex] = sample standard deviation for manufacturer A = [tex]\sqrt{\frac{\sum (X_A-\bar X_A)^{2} }{n_A-1} }[/tex] = 2.72

[tex]s_2[/tex] = sample standard deviation manufacturer B = [tex]\sqrt{\frac{\sum (X_B-\bar X_B)^{2} }{n_B-1} }[/tex] = 2.20

[tex]n_1[/tex] = sample of cars selected from manufacturer A = 8

[tex]n_2[/tex] = sample of cars selected from manufacturer B = 8

Also, [tex]s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }[/tex]   =  [tex]\sqrt{\frac{(8-1)\times 2.72^{2}+(8-1)\times 2.20^{2} }{8+8-2} }[/tex]  = 2.474

(A) The mean for the differences is = 27.625 - 25.625 = 2

(B) The test statistics  =  [tex]\frac{(27.625-25.625)-(0)}{2.474 \times \sqrt{\frac{1}{8}+\frac{1}{8} } }[/tex]  ~  [tex]t_1_4[/tex]

                                     =  1.617

(C) Now at 10% significance level, the t table gives critical values between -1.761 and 1.761 at 14 degree of freedom for two-tailed test. Since our test statistics lies within the range of critical values of t, 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 there is a not any significant difference in the mean MPG (miles per gallon) when testing for the fuel efficiency of these two brands of automobiles.

The mean for the differences is 2.5 MPG. The test statistic is 2.256. At 90% confidence, the null hypothesis should be rejected.

In order to determine if there is a significant difference in the mean MPG between the two manufacturers, we need to calculate the mean for the differences and the test statistic. The mean for the differences is calculated by subtracting the Manufacturer B MPG from the Manufacturer A MPG for each driver and then taking the average of those differences. In this case, the mean for the differences is 2.5 MPG. The test statistic is calculated by dividing the mean for the differences by the standard deviation of the differences and then multiplying by the square root of the sample size. In this case, the test statistic is 2.256.

At 90% confidence, we can compare the test statistic to the critical value for a two-tailed test. The critical value for a 90% confidence level is 1.645. Since the test statistic (2.256) is greater than the critical value (1.645), we reject the null hypothesis. Therefore, the answer to question C is: the null hypothesis should be rejected.

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Answer:

For this case we want to determine  if pennies are really fair when flipped, meaning equally likely to land head up or tails, so then the correct system of hypothesis are:

Null hypothesis: [tex]p=0.5[/tex]

Alternative hypothesis: [tex]p \neq 0.5[/tex]

Step-by-step explanation:

Previous concepts

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".  

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".  

Solution to the problem

For this case we want to determine  if pennies are really fair when flipped, meaning equally likely to land head up or tails, so then the correct system of hypothesis are:

Null hypothesis: [tex]p=0.5[/tex]

Alternative hypothesis: [tex]p \neq 0.5[/tex]

Final answer:

The appropriate null hypothesis for the experiment is that pennies are fair when flipped, with the alternative hypothesis being that they are not.

Explanation:

The appropriate null hypothesis for this experiment is that the true probability of a penny landing heads up when flipped is 0.5, meaning that pennies are fair when flipped. The alternative hypothesis, denoted as the alternative to the null hypothesis, would be that the true probability of a penny landing heads up when flipped is not 0.5.

To summarize:

Null hypothesis (H0): p = 0.5

Alternative hypothesis (Ha): p ≠ 0.5

Answer:

1 Gallon is equal to 4 quarts. To convert gallons to quarts, multiply the gallon value by 4.

Step-by-step explanation:

Answer:

divide the quart value by 4 and than multiply it by o.25

Answer:

P(X>4)= 0.624

Step-by-step explanation:

n = 10

p= 0.5 ,q= 1 - p = 0.5

Two fifth of 10 = 2/5 x 10 =4

It means that we have to find probability P(X>4).

P(X>4)= 1 -P(X=0)-P(X=1)-P(X=2)-P(X=3)-P(X=4)

We know that

P(X>4)= 1 -P(X=0)-P(X=1)-P(X=2)-P(X=3)-P(X=4)

P(X>4)= 1 -0.0009 - 0.0097 - 0.043 - 0.117-0.205

The probability that more than 12 of the entry forms will include an order is 0.624.

It is defined as the ratio of the number of favorable outcomes to the total number of outcomes, in other words, the probability is the number that shows the happening of the event.

We have:

The Magazine Mass Marketing Company has received 16 entries in its latest sweepstakes.

The probability that more than 12 of the entry forms will include an order can be found as:

P(X > 4) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P( X = 3) - P(X = 4)

P(X > 4) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3) - P(X = 4)

P(X > 4) = 1 - 0.0009 - 0.0097 - 0.043 - 0.117 - 0.205

P(X > 4) =  0.624

Thus, the probability that more than 12 of the entry forms will include order is 0.624.

Learn more about the probability here:

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Answer:

Step-by-step explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ = 6.8 psi

For the alternative hypothesis,

µ ≠ 6.8

This is a 2 tailed test

Since the population standard deviation is given, z score would be determined from the normal distribution table. The formula is

z = (x - µ)/(σ/√n)

Where

x = sample mean

µ = population mean

σ = population standard deviation

n = number of samples

From the information given,

µ = 6.8

x = 6.9

σ = 0.7

n = 140

z = (6.9 - 6.8)/(0.7/√140) = 0.17

Therefore, the value of the test statistic us 0.17

Answer:

20 yards

Step-by-step explanation:

The radius of a circle is half of the diameter

r=d/2

We know the diameter is 40 yards, so we can substitute that in

r=40/2

r=20

So, the radius is 20 yards

Answer:

From the question 10 to 13:Hint: check Lectures 26 and 27 in the Book, the answer to the questions are, question 10 (option B), question 11 (option b), question 12 (option c), question 13 (option b)

Step-by-step explanation:

From the given question, we simulate this scenario by answering this questions.

A random experiment was carried out a Person A tossed five coins and recorded the number of "heads"., the two dice rolled and  recorded the average the two numbers was carried out by the Person B.

Then,

When a person A tossed five coins and recorded the number of heads, person B rolled two dice and recorded the smaller out of the two dice.

Question 10 : The person A is likely to get the number 5, this is so because he has a better chance to get it. the answer here is option (b) from the lectures 26 and 27 in the book.

Question 11: The person B will have higher variations in their outcomes that is higher standard deviation, because the person B is having the smaller number out of the two dice. the correct answer here is option (b) from the the lectures 26 and 27 in the book.

Question 12: The probability that a person B get the numbers 5 or 6 is 0.03. the correct answer to the question is option (C) from the the lectures 26 and 27 in the book.

Question 13: The person's that will have a higher probability number 3 or larger is Person B. correct answer to the question is option (b) from the the lectures 26 and 27 in the book.

Step-by-step explanation:

To find out the points which are 4 units apart, plot the points on the XY plot.

Another condition is that either the X or Y co-ordinates should be the same point and the other co-ordinate should vary by 4 units to satisfy the given condition in the question.

Comparing all the given points, B(3,-6) and D(3,-2) are 4 units apart.

Such that their X co=ordinate is same and Y differs by 4 units.

Therefore, Option D, (B and  D) is correct

Answer:

D

Step-by-step explanation:

Answer:

Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] [tex]\geq[/tex] $13.04  

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < $13.04

Step-by-step explanation:

We are given that the mean cost of a flank steak, broccoli, and rice bought at the grocery store is $13.04.

A sample of 100 neighborhood restaurants showed a mean price of $12.65 and a standard deviation of $2 for a comparable restaurant meal.

Let [tex]\mu[/tex] = mean cost of a restaurant meal

So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] [tex]\geq[/tex] $13.04  

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < $13.04

Here, null hypothesis states that the mean cost of a restaurant meal is more than or equal to fixing a comparable meal at home.

On the other hand, alternate hypothesis states that the mean cost of a restaurant meal is less than fixing a comparable meal at home.

The test statistics that we can use for conducting this hypothesis would be t test statistics as we don't know about population standard deviation;

                            T.S. = [tex]\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }[/tex]  ~ [tex]t_n_-_1[/tex]

where, [tex]\bar X[/tex] = sample mean price of a restaurant meal $12.65

             s = sample standard deviation = $2

             n = sample of neighborhood restaurants = 100

The null hypothesis (H0) is that the cost of a restaurant meal is equal to or more than dining in. The alternative hypothesis (Ha) is that the restaurant meal is cheaper. If the test statistic lies in the critical region, we would then reject the null hypothesis in favor of the alternative.

To determine whether the mean cost of a restaurant meal is cheaper than fixing a similar meal at home, we have to consider hypothesis testing. In this case, the null hypothesis (H0) would be that the restaurant meal is equally or more expensive than dining in, and the alternative hypothesis (Ha) would be that the meal at the restaurant is cheaper.

H0: μ ≥ $13.04

Ha: μ < $13.04

Where μ stands for the mean cost of a restaurant meal. This hypotheses are based on a sample of 100 neighborhood restaurants with a mean price of $12.65 and a standard deviation of $2. If we get a test statistic that lies in the critical region, we could then reject the null hypothesis in favor of the alternative, thereby concluding that it is cheaper to eat at a restaurant.

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Answer:

a)From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

And the standard error for the mean would be:

[tex]\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14[/tex]

b) We want this probability:

[tex] P(\bar X >120) [/tex]

And we can use the z score formula given by:

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

And replacing we got:

[tex] z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143[/tex]

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

[tex] P( z>2.143) = 1-P(Z<2.143) = 1-0.984 = 0.016[/tex]

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 central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Part a

From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

And the standard error for the mean would be:

[tex]\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14[/tex]

Part b

We want this probability:

[tex] P(\bar X >120) [/tex]

And we can use the z score formula given by:

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

And replacing we got:

[tex] z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143[/tex]

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

[tex] P( z>2.143) = 1-P(Z<2.143) = 1-0.984 = 0.016[/tex]

Answer:

100 mL

Step-by-step explanation:

If x is the volume of 35% solution, then:

0.35x + 0.20(50) = 0.30(x + 50)

0.35x + 10 = 0.30x + 15

0.05x = 5

x = 100

You need 100 mL of 35% solution.

Answer:

Step-by-step explanation:

(6)2 + 5(6) -8

12 + 30 -8

42-8

Answer=34

Answer:

58

Step-by-step explanation:

x^2 + 5x – 8

Let  x = 6

6^2 +5(6) -8

36+30-8

58

Answer:

a

Step-by-step explanation:

first multiply 9*6.5*.5 for the triangle

Then multiply 9*4.5

Then add up to get 69.75

Answer:

a is your answer

Step-by-step explanation:

Answer:

[tex]n=(\frac{2.58(1.2)}{0.25})^2 =153.36 \approx 154[/tex]

So the answer for this case would be n=154 rounded up to the nearest integer

Step-by-step explanation:

Assuming the following question: It is desired to estimate the mean GPA of each undergraduate class at a large university. How large a sample is necessary to estimate the GPA within 0.25 at the 99% confidence level? The population standard deviation is 1.2. If needed, round your final answer up to the next whole number.

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

[tex]\bar X[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The margin of error is given by this formula:

[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]    (a)

And on this case we have that ME =0.25 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex]   (b)

The critical value for 99% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.005;0;1)", and we got [tex]z_{\alpha/2}=2.58[/tex], replacing into formula (b) we got:

[tex]n=(\frac{2.58(1.2)}{0.25})^2 =153.36 \approx 154[/tex]

So the answer for this case would be n=154 rounded up to the nearest integer

Answer:

A point in geometry is a location. It has no size i.e. no width, no length and no depth.

Step-by-step explanation:

If you need more definitions, I suggest you go to mathisfun which is an amazing website! Hope this helped :)

Answer:

Wag

Step-by-step explanation:

C72 или Wag ладно помогите

Answer:

The game developer tested by 24 participants and these 24 participants were divided evenly between the treatments and the required total number of possible replications combinations is 4.

Step-by-step explanation:

The Concepts and reason

The ANOVA can be used to analyze the data obtained from experimental or observational studies. A factor is a variable that the experimenter has taken for investigation.

A treatment is a level of a factor, and its experimental units are the objects of interest in the experiment. The variation between treatments groups captures the effect of the treatment and the variation within treatment groups represents random error not explained by the experimental treatments.

The replication means an independent repeat of each factor combination.

Fundamentals

The formula for replications is: Number of participants/number of cells

The total number of participants is: 24

The number of cells is: 6

The total number of participants in the game is 24 and the number of possible combination cells is 6

so,

Number of replications: Number of participants/number of cells

which is 24/6= 4

Therefore the number of replications is 4.

In this scenario, each treatment combination of music type and interaction method would be replicated 4 times, as there are 24 participants divided evenly among six treatment groups.

The developer's experiment introduces two variables: the type of music and the method of interaction (mouse or touchpad). First, the music has 3 levels: soft, hard rock, and none. The interaction method has 2 levels: mouse and touchpad. The participants are divided evenly among the resulting combinations of these levels (3 music types x 2 interaction types = 6 groups). Given that there are 24 participants, each group would have 24 divided by 6, or 4 participants each. Thus, there would be 4 replications for each of the treatment combinations.

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Final answer:

The question touches on elements of sociology, dealing with social influence, group behavior, and persuasion as evidenced through scenarios where individuals are impacted by their group affiliations. It illustrates real-life applications of sociological principles such as persuasion in deciding on a dinner venue or peer pressure in case of vandalism.

Explanation:

The subject matter here touches upon aspects of sociology, specifically related to social influence, group behaviour, and persuasion. These examples illustrate how individuals are often influenced by the groups they belong to and how persuasive tactics can be employed by individuals within a social context. The scenarios highlight how group membership and social pressures can impact decisions, such as where to eat dinner or participating in group activities like vandalism. The statement by John Donne encapsulates the concept that no person is isolated; instead, each individual is part of a broader society, emphasizing the importance of social groups in shaping behavior and attitudes.

In the given examples, we encounter various situations where group dynamics play a role - from peer pressure to persuade Jake to hang out against his parents' wishes, to a group's collective decision about whether to dine at a particular restaurant. These are everyday applications of sociological principles, demonstrating the study of group life, group behavior, and group processes, which are key concepts in understanding how individuals behave within and are influenced by their social environment.

The researchers conducted an experiment to test the effectiveness of a placebo in reducing pain. A statistical test called the paired t-test can be used to analyze the data and determine if the mean BOLD measurements are higher in the first session than in the second. The results of this test will provide evidence to confirm or refute the effectiveness of the placebo.

The researchers conducted an experiment to test the effectiveness of a placebo in reducing pain. They measured the blood oxygen level-dependent (BOLD) signal during two consecutive sessions with electric shocks applied to the volunteers' arms. In the second session, a cream that was actually a placebo was applied to the volunteers. The mean and standard deviation of the differences in BOLD measurements were calculated to determine if there was evidence to confirm that the placebo was effective.

To test if the mean BOLD measurements were higher in the first session than in the second, a statistical test can be used. The paired t-test is suitable for this scenario, as it compares the means of two related samples. The t-test calculates a t-value which can be compared to a critical value to determine if there is evidence of a significant difference. In this case, with a significance level of α = 0.05, if the t-value is greater than the critical value, it would indicate evidence that the mean BOLD measurements are higher in the first session than the second.

If the calculated t-value is greater than the critical value, there is evidence to confirm that the placebo is effective in reducing pain. Conversely, if the calculated t-value is not greater than the critical value, there is not enough evidence to confirm that the placebo is effective. It is important to note that statistical significance does not necessarily imply practical significance, so further investigation may be required to understand the magnitude of the effect.

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Answer:

E(Y) = $0.5

Var(Y)  = 14.25

you should pay the same amount  $0.5

Step-by-step explanation:

E(Y) =  = Σ(YP)

P = probability of each outcomes.

Var(Y) = Σ[tex]Y^{2}[/tex]p − (μ x μ)

E(Y) = (2 x 0.25) +(6 x 0.25) + (0.5 x (-3)) = $0.5

Var(Y) = ([tex]2^{2}\\[/tex]x 0.25) + ([tex]6^{2}[/tex] x 0.25) +([tex]-3^{2}[/tex] x 0.5) - ([tex]0.5^{2}[/tex])

       = 14.5 - 0.25

Var(Y)  = 14.25

for the difference between the payoff and cost of playing to have mean 0, you should pay the same amount  $0.5

The mean of your winnings in this coin tossing game is $0.50 and the variance is $3.75. To break even on average, the cost of playing the game should be $0.50.

To calculate the mean and variance of your winnings in this coin tossing game, we first need to compute the expected value and probabilities of each outcome.

Both coins coming up heads (HH) or tails (TT) each have a probability of 0.25 (1/2 * 1/2), while the coins not matching (HT or TH) has a probability 0.5.

The expected winnings, E(Y), can be calculated using the formula E(Y) = (sum of (value x probability)).

So, E(Y) = $2(0.25) + $6(0.25) + (-$3)(0.50) = $0.50 + $1.50 - $1.50 = $0.50.

The variance can be calculated using the formula Var(Y) = E(Y^2) - [E(Y)]^2.

First, find E(Y^2) = ($2^2x0.25) + ($6^2x0.25) + ((-$3)^2x0.50) = $4. Then, calculate [E(Y)]^2 = ($0.50)^2 = $0.25. Substituting in the formula gives Var(Y) = $4 - $0.25 = $3.75.

If your net winnings are to have mean 0, the cost of playing the game should be equivalent to the expected winnings, which is $0.50.

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Step-by-step explanation:

By trigonometric conversion,

Cos -90° = 0

Cos -60° = 0.5

Cos -30° = 0.866

Cos 0° = 1

From the above values, Cos -90° to Cos 0° = 0 to 1

As theta increases from [tex]\frac{-theta}{2}[/tex] to 0, the value of cos theta will increase from 0 to 1 (Option 4)

Answer:

The type II error might have been committed because of small sample size or small significance level.

Step-by-step explanation:

A type II error is a statistical word used within the circumstance of hypothesis testing that defines the error that take place when one is unsuccessful to discard a null hypothesis that is truly false. It is symbolized by β i.e.  

β = Probability of accepting H₀ when H₀ is false.

In this case we need to test the hypothesis whether the new advertising campaign increases the sales or not.

The hypothesis can be defined as:

H₀: The new advertising campaign does not increases the sales.

Hₐ: The new advertising campaign increases the sales.

The confidence level wanted here is 99%.

The type II error will be made if we conclude that the new advertising campaign does not increases the sales when in fact the sales are increased after the advertising campaign.

The type II error could have been made because of the following reasons:

Thus, the type II error might have been committed because of small sample size or small significance level.

Answer:

[tex]V = \frac{2L\sqrt{3}}{3} \pi}[/tex]

[tex]A = 4\pi + \sqrt{3L^{2} + 16}[/tex]

Step-by-step explanation:

Figure of cone is missing. See attachment

Given

Radius, R = 2m

Let L = KL=LM=KM

Required:

Volume, V and Surface Area, A

Calculating Volume

Volume is calculated using the following formula

[tex]V = \frac{1}{3} \pi R^{2} H[/tex]

Where R is the radius of the cone and H is the height

First, we need to determine the height of the cone

The height is represented by length OL

It is given that KL=LM=KM in triangle KLM

This means that this triangle is an equilateral triangle

where OM = OK = [tex]\frac{1}{2} KL[/tex]

OK = [tex]\frac{1}{2}L[/tex]

Applying pythagoras theorem in triangle LOM,

|LM|² =  |OL|² + |OM|²

By substitution

L² = H² + ( [tex]\frac{1}{2}L[/tex])²

H² = L² -  [tex]\frac{1}{4}L[/tex]²

H² = L² (1 -  [tex]\frac{1}{4}[/tex])

H² = L² [tex]\frac{3}{4}[/tex]

H² = [tex]\frac{3L^{2} }{4}[/tex]

Take square root of bot sides

[tex]H = \sqrt{\frac{3L^{2} }{4}}[/tex]

[tex]H = \frac{L\sqrt{3}}{2}[/tex]

Recall that [tex]V = \frac{1}{3} \pi R^{2} H[/tex]

[tex]V = \frac{1}{3} \pi 2^{2} * \frac{L\sqrt{3}}{2}[/tex]

[tex]V = \frac{1}{3} \pi * 4} * \frac{L\sqrt{3}}{2}[/tex]

[tex]V = \frac{1}{3} \pi} * {2L\sqrt{3}}[/tex]

[tex]V = \frac{2L\sqrt{3}}{3} \pi}[/tex]

in terms of [tex]\pi[/tex] an d L where L = KL = LM = KM

Calculating Surface Area

Surface Area is calculated using the following formula

[tex]H = \frac{L\sqrt{3}}{4}[/tex]

[tex]A=\pi r(r+\sqrt{h^{2} +r^{2} } )[/tex]

[tex]A=\pi * 2(2+\sqrt{((\frac{L\sqrt{3}}{2})^{2} +2^{2} } )[/tex])

[tex]A=\pi * 2(2+\sqrt{{\frac{3L^{2}}{4} } + 4 }[/tex] )

[tex]A=\pi * 2(2+\sqrt{{\frac{3L^{2}+16}{4} } })[/tex]

[tex]A=2\pi(2+\sqrt{{\frac{3L^{2}+16}{4} } })[/tex]

[tex]A = 2\pi (2 + \frac{\sqrt{3L^{2} + 16}}{\sqrt{4}} )[/tex]

[tex]A = 2\pi (2 + \frac{\sqrt{3L^{2} + 16}}{2} )[/tex]

[tex]A = 2\pi (2 + {\frac{1}{2} \sqrt{3L^{2} + 16})[/tex]

[tex]A = 4\pi + \sqrt{3L^{2} + 16}[/tex]

Answer: 23/111

Step-by-step explanation:

Answer:

23/111

Step-by-step explanation:

Answer:

185

Step-by-step explanation:

sorry. had to do something but im bsck :)

To establish control limits for Choudhury's bowling ball factory, you need to calculate the mean weight of the balls over the observed period. The control limits are then defined as two standard deviations (0.12 pounds) above and below this mean

The subject of this question is related to quality control and statistical measures, which is a part of Mathematics. Specifically, we need to create a control chart and find the control limits based on the average weights of bowling balls produced at Choudhury's factory.

First, we need to calculate the average weight of all the bowling balls produced over the 24 day period. Then, we calculate the control limits which are two standard deviations away from this mean. The standard deviation has been given as 0.12 pounds.

Upon calculation, you'll obtain an average weight, µ. Your upper control limit (UCL) would be µ + 2*(Standard Deviation) and your lower control limit (LCL) would be µ - 2*(Standard Deviation).

Therefore, assuming your calculated average weight of all balls is µ pounds, your UCL would be µ + 2*(0.12) pounds and your LCL would be µ - 2*(0.12) pounds.

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Answer:

96.08% probability that the average length of a randomly selected bundle of steel rods is less than 178.3-cm.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

[tex]\mu = 177.5, \sigma = 1.2, n = 7, s = \frac{1.2}{\sqrt{7}} = 0.4536[/tex]

Find the probability that the average length of a randomly selected bundle of steel rods is less than 178.3-cm.

This is the pvalue of Z when X = 178.3. So

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

By the Central Limit Theorem

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

[tex]Z = \frac{178.3 - 177.5}{0.4536}[/tex]

[tex]Z = 1.76[/tex]

[tex]Z = 1.76[/tex] has a pvalue of 0.9608

96.08% probability that the average length of a randomly selected bundle of steel rods is less than 178.3-cm.

Answer: the probability that the average length of a randomly selected bundle of steel rods is less than 178.3-cm is 0.96

Step-by-step explanation:

Since the lengths of the steel rods are normally distributed, then according to the central limit theorem,

z = (x - µ)/(σ/√n)

Where

x = sample mean lengths of the steel rods

µ = population mean length of the steel rods

σ = standard deviation

n = number of samples

From the information given,

µ = 177.5 cm

x = 178.3 cm

σ = 1.2 cm

n = 7

the probability that the average length of a randomly selected bundle of steel rods is less than 178.3-cm is expressed as

P(x < 178.3)

Therefore,

z = (178.3 - 177.5)/(1.2/√7) = 1.76

Looking at the normal distribution table, the probability corresponding to the z score is 0.96

Therefore,

P(x < 178.3) = 0.96

The ASA and AAS are methods of proving triangle congruence. ASA requires two angles and the included side in one triangle to match the respective parts in the other triangle. AAS requires two angles and a non-included side in one triangle to match the respective parts in the other triangle.

The question refers to two of the many methods to prove that triangles are congruent: Angle-Side-Angle (ASA) and Angle-Angle-Side (AAS). In triangle congruence, congruent means that two triangles have the same size and shape.

For the ASA congruence, two angles and the included side in one triangle must be congruent to the corresponding two angles and the included side in another triangle. For example, if we have two triangles, Triangle ABC and Triangle DEF, if angle A is congruent to angle D, angle B is congruent to angle E, and side AB is congruent to side DE, then the two triangles are congruent by ASA.

For AAS congruence, two angles and a non-included side in one triangle must be congruent to the corresponding two angles and the non-included side in another triangle. For instance, in Triangle ABC and Triangle DEF, if angle A is congruent to angle D, angle B is congruent to angle E, and side BC is congruent to side EF, then the two triangles are congruent by AAS.

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Answer:

The answer is 30 sq in i had that question, good luck :)