A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 430 gram setting. It is believed that the machine is underfilling the bags. A 21 bag sample had a mean of 429 grams with a variance of 289. Assume the population is normally distributed. A level of significance of 0.01 will be used. State the null and alternative hypotheses.

Answers

Answer 1

Answer:

Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] [tex]\geq[/tex] 430 gram   {means that the machine is not under filling the bags}

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 430 gram   {means that the machine is under filling the bags}

Step-by-step explanation:

We are given that a manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 430 gram setting.

It is believed that the machine is under filling the bags. A 21 bag sample had a mean of 429 grams with a variance of 289.

Let [tex]\mu[/tex] = mean weight bag filling capacity of machine.

SO, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] [tex]\geq[/tex] 430 gram   {means that the machine is not under filling the bags}

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 430 gram   {means that the machine is under filling the bags}

The test statistics that will be used here is One-sample t test statistics as we don't know about the 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 = 429 grams

             s = sample standard deviation = [tex]\sqrt{289}[/tex] = 17 grams

             n = sample of bags = 21

So, test statistics  =  [tex]\frac{429-430}{\frac{17}{\sqrt{21} } }[/tex]  ~ [tex]t_2_0[/tex]   

                               =  -0.269

Now at 0.01 significance level, the t table gives critical value of -2.528 at 20 degree of freedom for left-tailed test. Since our test statistics is higher than the critical value of t as -0.269 > -2.528, 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 machine is not under filling the bags.

Answer 2

Final answer:

The student's question involves setting up null and alternative hypotheses for a hypothesis test related to a bag filling machine's weight accuracy. The null hypothesis posits no difference from the 430-gram setting, while the alternative hypothesis suggests underfilling. The test uses a 0.01 significance level to determine whether to reject the null hypothesis.

Explanation:

The student is asking about hypothesis testing in the context of quality control for a bag filling machine. The null hypothesis (H0) assumes that there's no difference between the sample's mean weight and the population mean weight that the machine is set to, which is 430 grams; hence H0:
μ = 430 grams. The alternative hypothesis (Ha) suggests that the actual mean weight is less than the setting, given the concern about underfilling; thus Ha: μ < 430 grams.

For the given scenario, the bag sample's mean is 429 grams with a variance of 289 grams. Since a 0.01 level of significance will be used, if the test statistic calculated from the sample data falls into the critical region determined by this significance level, the null hypothesis would be rejected in favor of the alternative hypothesis.


Related Questions

The manager of a paint supply store wants to determine whether the mean amount of paint contained in 1-gallon cans purchased form a nationally known manufacture is actually 1 gallon. You know from the manufacturer’s specifications that the standard deviation of the amount of pant is 0.02 gallon. You select a random sample of 50 cans, and the mean amount of paint per 1-gallon cans is 0.995 gallon.
a. Is there evidence that the mean amount is different from 1.0 gallon (use α = 0.01)?
b. Compute the p-value and interpret the meaning
c. Construct a 99% confidence interval estimate of the population mean amount of paint.
d. Compare the results of (a) and (c). What conclusions do you reach?

Answers

Answer:

a) There is no significant evidence to conclude that there there is significant difference in the mean amount of paint per 1-gallon cans and 1 gallon.

b) The p-value obtained = 0.076727 > significance level (0.01), hence, we fail to reject the null hypothesis and conclude that there is no significant evidence to conclude that there there is significant difference in the mean amount of paint per 1-gallon cans and 1 gallon.

That is, the mean amount of paint per 1-gallon cans is not significantly different from 1 gallon.

c) The 99% confidence for the population mean amount of paint per 1-gallon cans is

(0.988, 0.999) in gallons.

d) The result of the 99% confidence interval does not agree with the result of the hypothesis testing performed in (a) because the right amount of paint in 1-gallon cans, 1 gallon, does not lie within this confidence interval obtained.

Step-by-step explanation:

a) This would be answered after solving part (b)

b) For hypothesis testing, the first thing to define is the null and alternative hypothesis.

The null hypothesis plays the devil's advocate and is always about the absence of significant difference between two proportions being compared. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.

While, the alternative hypothesis takes the other side of the hypothesis; that there is indeed a significant difference between two proportions being compared. It usually contains the signs ≠, < and > depending on the directions of the test.

For this question, the null hypothesis is that there is no significant difference in the mean amount of paint per 1-gallon cans and 1 gallon. That is, the mean amount of paint per 1-gallon cans should be 1 gallon.

And the alternative hypothesis is that there is significant difference in the mean amount of paint per 1-gallon cans and 1 gallon. That is, the mean amount of paint per 1-gallon cans is not 1 gallon.

Mathematically,

The null hypothesis is

H₀: μ₀ = 1 gallon

The alternative hypothesis is

Hₐ: μ₀ ≠ 1 gallon

To do this test, we will use the z-distribution because we have information on the population standard deviation.

So, we compute the z-test statistic

z = (x - μ)/σₓ

x = the sample mean = 0.995 gallons

μ₀ = what the amount of paint should be; that is 1 gallon

σₓ = standard error = (σ/√n)

σ = standard deviation = 0.02 gallon

n = sample size = 50

σₓ = (0.02/√50) = 0.0028284271 = 0.00283 gallons.

z = (0.995 - 1) ÷ 0.00283

z = -1.77

checking the tables for the p-value of this z-statistic

p-value (for z = -1.77, at 0.01 significance level, with a two tailed condition) = 0.076727

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 0.01

p-value = 0.076727

0.076727 > 0.01

Hence,

p-value > significance level

So, we fail to reject the null hypothesis and conclude that there is no significant evidence to conclude that there there is significant difference in the mean amount of paint per 1-gallon cans and 1 gallon.

That is, the mean amount of paint per 1-gallon cans is not significantly different from 1 gallon.

c) To compute the 99% confidence interval for population mean amount of paint per 1-gallon paint cans.

Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample Mean) ± (Margin of error)

Sample Mean = 0.995 gallons

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error)

Critical value at 99% confidence interval is obtained from the z-tables because we have information on the population standard deviation.

Critical value = 2.58 (as obtained from the z-tables)

Standard error = σₓ = 0.00283 (already calculated in b)

99% Confidence Interval = (Sample Mean) ± [(Critical value) × (standard Error)]

CI = 0.995 ± (2.58 × 0.00283)

CI = 0.995 ± 0.0073014

99% CI = (0.9876986, 0.9993014)

99% Confidence interval = (0.988, 0.999) in gallons.

d) The result of the 99% confidence interval does not agree with the result of the hypothesis testing performed in (a) because the right amount of paint in 1-gallon cans, 1 gallon, does not lie within this confidence interval obtained.

Hope this Helps!!!

Final answer:

By calculating a z-value and comparing it to the critical value at alpha = 0.01, evidence can be determined. The p-value, as extreme as the calculated one assuming the null hypothesis is true, can be used to interpret the findings. Additionally, a 99% confidence interval estimate can be constructed to provide a range of values that is 99% confident in containing the true population mean amount of paint.

Explanation:

a. To determine whether the mean amount of paint contained in 1-gallon cans is different from 1.0 gallon, we can perform a hypothesis test. The null hypothesis (H0) is that the mean amount is 1.0 gallon, and the alternative hypothesis (Ha) is that the mean amount is different from 1.0 gallon. We can perform a z-test using the formula Z = (sample mean - population mean) / (standard deviation / sqrt(sample size)). In this case, the sample mean is 0.995 gallon, the population mean is 1.0 gallon, the standard deviation is 0.02 gallon, and the sample size is 50. By calculating the z-value, we can compare it to the critical value at alpha = 0.01 to determine whether there is evidence to reject the null hypothesis.

b. The p-value is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. In this case, we can calculate the p-value using the standard normal distribution table or a calculator. If the p-value is less than the significance level (alpha = 0.01), we reject the null hypothesis. The interpretation of the p-value is that there is strong evidence to suggest that the mean amount of paint is different from 1.0 gallon.

c. To construct a 99% confidence interval estimate of the population mean amount of paint, we can use the formula CI = sample mean ± (z-score * (standard deviation / sqrt(sample size))). In this case, the z-score for a 99% confidence level is approximately 2.61. Plugging in the values, we can calculate the confidence interval, which gives us a range of values that we are 99% confident contains the true population mean amount of paint.

d. By comparing the results of (a) and (c), we can draw conclusions about whether the mean amount of paint is different from 1.0 gallon. If the null hypothesis is rejected in (a) and the 99% confidence interval in (c) does not include 1.0 gallon, then we can conclude that there is evidence to suggest that the mean amount is different from 1.0 gallon.

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the specification for a plastic handle calls for a length of 6.0 inches +- .2 inches. The standard deviation of the process is estimated to be 0.05 Inches. what are the upper and lower specification limits for this product.

Answers

Answer:

a)

USL = 6.2 inches

LSL = 5.8 inches

b) Cp = 1.33

Cpk = 0.67

c)

Yes it meets all specifications

Step-by-step explanation:

The specification for a plastic handle calls for a length of 6.0 inches ± .2 inches. The standard deviation of the process is estimated to be 0.05 inches. What are the upper and lower specification limits for this product? The process is known to operate at a mean thickness of 6.1 inches. What is the Cp and Cpk for this process?   Is this process capable of producing the desired part?

Given that:

Mean (μ) = 6.1 inches, Standard deviation (σ) = 0.05 inches and the length of the plastic handle is 6.0 inches ± .2

a) Since the length of the plastic handle is 6.0 inches ± .2  = (6 - 0.2, 6 + 0.2)

The Upper specification limits (USL) = 6 inches + 0.2 inches = 6.2 inches

The lower specification limits (LSL) = 6 inches - 0.2 inches = 5.8 inches

b) The Cp is given by the formula:

[tex]Cp=\frac{(USL-LSL)}{6\sigma} =\frac{(6.2-5.8)}{6*0.05} =1.33[/tex]

The Cpk is given by the formula:

c)

The upper specification limit lies about 3 standard deviations from the centerline, and the lower specification limit is further away, so practically all units will meet specifications

[tex]Cpk=min(\frac{USL-\mu}{3\sigma},\frac{\mu -LSL}{3\sigma})=min(\frac{6.2-6.1}{3*0.05},\frac{6.1-5.8}{3*0.05})=min(0.67,2)=0.67[/tex]

Final answer:

The upper specification limit for the plastic handle is 6.2 inches, and the lower specification limit is 5.8 inches, with these limits defining the acceptable range for the handle length.

Explanation:

The specification for a plastic handle is given as a length of 6.0 inches with a tolerance of ± 0.2 inches. This means the upper specification limit (USL) and the lower specification limit (LSL) are defined by adding and subtracting the tolerance to the target length respectively. The process standard deviation is 0.05 inches, but this does not affect the USL and LSL directly; it's a measure of the process variation.

The USL and LSL are calculated as follows:

USL = Target Length + Tolerance = 6.0 inches + 0.2 inches = 6.2 inchesLSL = Target Length - Tolerance = 6.0 inches - 0.2 inches = 5.8 inches

These limits are the range within which the plastic handle lengths should fall according to the given specifications.

A store sells televisions for $360 and video cassette recorders for $270. At the beginning of the week its entire stock is worth $56,430. During the week it sells three quarters of the televisions and one third of the video cassette recorders for a total of $32,310. How many televisions and video cassette recorders did it have in its stock at the beginning of the week

Answers

Answer:

The number of Television at the beginning was 90

The number of video cassette recorders at the beginning was 89

Step-by-step explanation:

Selling Price of 1 Television =$360.

Selling Price of 1 video cassette recorders for $270.

Let the number of Television at the beginning=x

Let the number of video cassette recorders at the beginning =y

Opening Stock =$56,430.

Therefore:

360x+270y=$56,430.

It sells three quarters of the televisions and one third of the video cassette recorders for a total of $32,310.

[tex]\frac{3}{4}(360)x+\frac{1}{3}(270)y= \$32,310[/tex]

270x+90y=32310

We then solve the two equations to obtain x and y.

360x+270y=$56,430. (Multiply by 270)270x+90y=32310          (Multiply by 360)

97200x+72900y=15236100

97200x+32400y=11631600

Subtract

40500y=3604500

y=89

Substitute y=89 into 270x+90y=32310 to obtain x

270x+90(89)=32310

270x=32310-8010=24300

x=90

Therefore:

The number of Television at the beginning was 90

The number of video cassette recorders at the beginning was 89

Answer:

90 televisions and 89 video cassette recorders

Step-by-step explanation:

The unit cost of television = $360

The unit cost of video cassette recorder = $270

Let "T" represent the number of televisions and "R" represent the number of recorders, so that we can make representations using equations from the statements.

At the beginning of the week, Total Stock is worth $56,430, where

Total Stock = Total cost of televisions + Total cost of recorders

Total Cost = Unit Cost × Number of items

$56,430 = 360T + 270R  This is the first equation

Next, During the week, Number of Sales = [tex]\frac{3}{4}[/tex] T + [tex]\frac{1}{3}[/tex] R

Total Sales Price = 360 ([tex]\frac{3}{4}[/tex] T) + 270 ([tex]\frac{1}{3}[/tex] R)

$32,310 = 270T + 90R     This is the second equation

Solving both equations simultaneously, let us use elimination method which involves equating one of the two terms in both equations. Let us multiply the second equation by 3. This doesn't affect the equation, since we are doing it to all the terms in it.

56,430 = 360T + 270R

32,310 = 270T + 90R            × 3

So, we have;

56,430 = 360T + 270R

96,930 = 810T + 270R

Subtracting both equations, we have;

96,930 - 56,430 = 810T - 360T

40,500 = 450T

T = [tex]\frac{40,500}{450}[/tex] = 90

Since we now have the number of televisions, we can get the number of recorders by putting 90 in any (say, the second) equation.

32,310 = 270 (90) + 90R

32,310 = 24,300 + 90R

32,310 - 24,300 = 90R

8010 = 90R

R = [tex]\frac{8010}{90}[/tex] = 89

At the beginning of the week, the store had 90 televisions and 89 video cassette recorders

To transfer into a particular technical department, a company requires an employee to pass a screening test. A maximum of 3 attempts are allowed at 6-month intervals between trials. From past records it is found that 40% pass on the first trial; of those that fail the first trial and take the test a second time, 60% pass; and of those that fail on the second trial and take the test a third time, 20% pass. For an employee wishing to transfer:
(A) What is the probability of passing the test on the first or second try?
(B) What is the probability of failing on the first 2 trials and passing on the third?
(C) What is the probability of failing on all 3 attempts?

Answers

Answer:

a) 0.760

b) 0.048

c) 0.192

Step-by-step explanation:

The step by step solution is attached as an image.

A) The probability of passing the test on the first or second try is 0.760.

That is he pass in the first trial or second trial.

(B) The probability of failing on the first 2 trials and passing on the third is 0.048.

That is the employee fail the first the trial and pass the third trial.

(C) The probability of failing on all 3 attempts is 0.192.

That is the employee fail all the three trial.

Final answer:

The probability of passing on the first or second try is 76%, the probability of failing the first 2 trials and passing on the third is 4.8%, and the probability of failing all 3 attempts is 19.2%.

Explanation:

This problem relates to the field of probability. Let's break it down.

For part A, the probability of passing on the first or second try is the sum of the probability of passing on the first try and the product of the probability of failing on the first try and passing on the second. This is calculated as 0.4 + (0.6*0.6) = 0.76 or 76%.

For part B, the probability of failing the first 2 trials and passing on the third is calculated by multiplying the probability of failing the first trial, failing the second, and passing the third: (0.6*0.4*0.2) = 0.048 or 4.8%.

For part C, the probability of failing all 3 attempts is equal to the product of the probability of failing each attempt: (0.6*0.4*0.8) = 0.192 or 19.2%.

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x g(x)
−2 1/4
-1 1/2
0 1
1 2
3 8

Consider that f(x) = x + 2, while the table represents y = g(x). Which statement is true when comparing the rate of change for the functions?
A) The rate of increase for the functions is the same.
B) f(x) has a greater rate of increase than function g(x).
C) g(x) has a greater rate of increase than function f(x).
D) g(x) has a greater rate of decrease than function f(x).

Answers

Answer:

A

Step-by-step explanation:

Cause I know

Answer: C!!

Step-by-step explanation:

USA Test Prep told me! :)

Jay was reaching into her purse and accidentally spilled her coin purse. 10 pennies fell on the floor. Jay noticed that only 2 of the pennies landed on heads. What is the theoretical probability of this happening?

Answers

Answer:

The theoretical probability of landing on 2 heads, when 10 coins are tossed is 0.0439 or 4.39%.

Step-by-step explanation:

Number of coins that fell on the floor = 10

Number of coins that landed on heads = 2

We have to find the theoretical probability of getting 2 coins landing of heads when 10 coins are tossed.

Notice that there are only 2 possible outcomes: Either that coin will land on head or it won't. Landing of each coin is independent of the others coins. Probability of each coin landing on head is constant i.e. 0.5 or 1/2. Number of trials, i.e. the number of times the experiment will be done is fixed, which is 10.

All the 4 conditions for an experiment to be considered a Binomial Experiment are satisfied. So we will use Binomial Probability to solve this problem.

Probability of success = Probability of coin landing on head = 0.5

Number of trials = n = 10

Number of success = r = 2

The formula for Binomial Probability is:

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

Using the values, we get:

[tex]P(X=2)=^{10}C_{2}(0.5)^2(0.5)^8=0.0439[/tex]

Thus, the theoretical probability of landing on 2 heads, when 10 coins are tossed is 0.0439 or 4.39%.

A study by a federal agency concludes that polygraph tests given to truthful persons have probability 0.2 of suggesting that the person is deceptive. A firm asks 12 job applicants about thefts from previous employers, using a polygraph to assess their truthfulness. Suppose that all 12 answer truthfully. Let X be the number of applicants who are classified deceptive. a) Describe the probability mass function of X. b) What is the probability that the polygraph says at least 1 is deceptive? c) What is the mean number among 12 truthful persons who will be classified as deceptive? What is the standard deviation of this number? d) What is the probability that the number classified deceptive is less than the mean?

Answers

Answer:

b) 0.9313

c) mean = 2.4

standard deviation= 1.3856

d) 0.5583

Step-by-step explanation:

Given:

p = 0.2

n = 12

a) X= number of applicants classified as deceptive.

Probability mass function of X will be:

[tex] P(X=x) = \left(\begin{array}{c}12\\x\end{array}\left) (0.2)^x(1-0.2)^1^2^-^x,x=0, 1, 2, .....,12 [/tex]

b) Probability that the polygraph says at least 1 is deceptive:

[tex] P(X≥1) = 1 -P(X=0) = 1 -\left(\begin{array}{c}12\\0\end{array}\left) (0.2)^0(1-0.2)^1^2^-^0[/tex]

= 1 - 0.0687

= 0.9313

c) The mean number among 12 truthful persons who will be classified as deceptive:

E(X) = n•p

= 12 * 0.2

= 24

Standard deviation:

[tex]s.d = \sqrt{12*0.2*(1-0.2)}[/tex]

= 1.3856

d) Probability that the number classified deceptive is less than the mean:

[tex] P(X<2.4) = P(X≤2) = E^2_x_=_0 \left(\begin{array}{c}12\\0\end{array}\left) (0.2)^0(1-0.2)^1^2^-^0 [/tex]

= 0.5583



Add together 8.03 m 1.26 m 0.5 m 4.09 m 3.5 m

Answers

Answer:

17.38m

Step-by-step explanation:

Answer:

17.38m

Step-by-step explanation:

W is less than or equal to 9 and greater than -7

Answers

Answer:

(-7, 9]

Step-by-step explanation:

Assuming you want an answer in interval notation:

W is less than or equal to 9 >> [tex]w \leq 9[/tex]

W is greater than- 7 >> [tex]-7 < w[/tex]

Combine the two >> [tex]-7 < w \leq 9[/tex]

So in interval notation, (-7, 9]

Which diagram represents a cylinder with a base area equal to 50Pi square meters?

1. A cylinder with height of 50 meters and volume = 250 pi meters cubed.

2. A cylinder with height 25 meters and volume = 250 pi meters cubed.

3. A cylinder with height 5 meters and volume = 250 pi meters cubed.

Answers

Answer: C

Step-by-step explanation:

Answer:The answer Is The Third One On The Right.

Step-by-step explanation:

The weekly amount of money spent on maintenance and repairs by a company was observed, over a long period of time, to be approximately normally distributed with mean $490 and standard deviation $10. How much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.1?

Answers

Answer:

[tex]z=1.28<\frac{a-490}{10}[/tex]

And if we solve for a we got

[tex]a=490 +1.28*10=502.8[/tex]

So the value of height that separates the bottom 90% of data from the top 10% is 502.8.  

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 weeknly amount of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(490,10)[/tex]  

Where [tex]\mu=490[/tex] and [tex]\sigma=10[/tex]

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

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

For this part we want to find a value a, such that we satisfy this condition:

[tex]P(X>a)=0.1[/tex]   (a)

[tex]P(X<a)=0.9[/tex]   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.9[/tex]  

[tex]P(z<\frac{a-\mu}{\sigma})=0.9[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=1.28<\frac{a-490}{10}[/tex]

And if we solve for a we got

[tex]a=490 +1.28*10=502.8[/tex]

So the value of height that separates the bottom 90% of data from the top 10% is 502.8.  

Final answer:

The company should budget approximately $502.80 for weekly repairs and maintenance to ensure that the probability of exceeding this amount is only 0.1.

Explanation:

We want to find how much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.1.

Since the weekly amount of money spent is normally distributed with a mean of $490 and a standard deviation of $10, we can find the amount by looking up the z-score that corresponds to the 90th percentile

(since 100% - 10% = 90%) in a standard normal distribution table or using a calculator.

Let the z-score for the 90th percentile be denoted as z.

Looking up the standard normal distribution table or using a calculator, we find that z ≈ 1.28 for 0.9 cumulative probability.

We then use the z-score formula:

z = (X - mean) / standard deviation

Plugging in our z-score and the parameters, we can solve for X:

1.28 = (X - 490) / 10

X - 490 = 12.8

X = $502.80

Therefore, the company should budget approximately $502.80 for weekly repairs and maintenance to ensure that the probability of exceeding this amount is only 0.1.

Two dice: Find the probability of rolling an odd number on the first die and an even number on the second die.

Answers

Answer:

1/4; 25%

Step-by-step explanation:

Both events happen with probability 1/2: there are 3 even numbers and 3 odd numbers in a die.

Since the two events are also independent (the outcome of the first die doesn't affect the outcome of the second), we have to multiply those probability.

So, you roll an odd number on the first die and an even number on the second die with probability

[tex]\dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{1}{4}[/tex]

Look at the number line below. The letters f, g, h, and i all represent integers. Write two inequalities to compare f and g.

Answers

Final answer:

Without the provided number line, we cannot determine the exact relationship between f and g. Inequalities f < g or f > g represent f being less than or greater than g, respectively. To write the correct inequality, one must refer to the positions of f and g on the number line.

Explanation:

Since the number line is not provided, we cannot see the exact positions of f and g. However, we can discuss how to write inequalities to compare two integers based on their positions on a number line. If f is located to the left of g on the number line, it means that f is less than g. The inequality for this scenario would be f < g. On the other hand, if f is located to the right of g, then f is greater than g, and the corresponding inequality would be f > g.

You can use an inequality symbol to show how two metric measurements are related. If two numbers are the same, the inequality symbol would be the equal sign, representing they are equivalent. However, without the number line, we cannot determine the exact relationship between f and g, so one must look at the number line to ascertain the correct inequality to use.

A study is conducted to investigate whether customer satisfaction is greater among computer companies that offer tech support versus those that do not offer tech support. A random sample of 50 customers are selected from among those that purchased computers that offer tech support. A separate random sample of 40 customers are selected from among those that purchased computers that do not offer tech support.
The study found that the mean satisfaction rating was significantly greater among customers that purchased computers that offer tech support.
Which of the following is the best description of this study?
(A) An experiment using a completely randomized design.
(B) An experiment using a randomized block design.
(C) An experiment using a matched pairs design
(D) An observational study using a simple random sample.

Answers

Answer:

The correct answer is (E).

This is not an experiment because no treatment is being imposed upon the customers. Additionally, this study used a stratified sample because independent random samples were selected from two distinct populations of customers.

Step-by-step explanation:

The correct answer is (E) An observational study using a stratified sample.

What is stratified sample?

Stratified sampling is also known as stratified random sampling. The stratified sampling process starts with researchers dividing a diverse population into relatively homogeneous groups called strata, the plural of stratum. Then, they draw a random sample from each group (stratum) and combine them to form their complete representative sample.

Given that a data of a survey, the study found that the mean satisfaction rating was significantly greater among customers that purchased computers that offer tech support.

We need to find which is the best description of this study,

This study used a stratified sample because independent random samples were selected from two distinct populations of customers.

This is not an experiment because no treatment is being imposed upon the customers.

Hence, the best description is an observational study using a stratified sample.

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The complete question is attached.

In a one-way ANOVA, if the computed F statistic is greater than the critical F value you may Question 1 options: reject H0 since there is evidence that not all the means are different. not reject H0 since there is no evidence of a difference in the means. not reject H0 because a mistake has been made. reject H0 since there is evidence all the means differ.

Answers

Answer:

And the F statistic calculated from the mean squares if [tex]F_{calc}[/tex]. And for this case we know that [tex] F_{calc}>F_{critical}[/tex]. So then we can reject the null hypothesis that all the means are equal at a significance level given [tex]\alpha[/tex]. And the best conclusion would be:

reject H0 since there is evidence all the means differ.

Step-by-step explanation:

Previous concepts

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".  

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"  

Solution to the problem

The hypothesis for this case are:

Null hypothesis: [tex]\mu_{A}=\mu_{B}=....=\mu_{k}[/tex]

Alternative hypothesis: Not all the means are equal [tex]\mu_{i}\neq \mu_{j}, i,j=A,B,...,k[/tex]

If we assume that we have [tex]p[/tex] groups and on each group from [tex]j=1,\dots,p[/tex] we have [tex]n_j[/tex] individuals on each group we can define the following formulas of variation:  

[tex]SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2 [/tex]  

[tex]SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 [/tex]  

[tex]SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 [/tex]  

And we have this property  

[tex]SST=SS_{between}+SS_{within}[/tex]  

And the F statistic calculated from the mean squares if [tex]F_{calc}[/tex]. And for this case we know that [tex] F_{calc}>F_{critical}[/tex]. So then we can reject the null hypothesis that all the means are equal at a significance level given [tex]\alpha[/tex]. And the best conclusion would be:

reject H0 since there is evidence all the means differ.

In a one-way ANOVA, if the computed F statistic is greater than the critical F value you may eject H0 since there is evidence all the means differ.

What is F-test?

F test is a method used in statistics to determine which models best fits the population from which the sample is derived.

The formula for calculating  the F-test statistic = explained variance / unexplained variance

The null hypothesis usually states that the means are the means are the same while the alternative hypothesis states that the means are not the same.

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I don’t understand my homework... and don’t go at me I’m an slow learner I never done this... before... but NOTE I already got the first one...

Answers

Answer:

52 weeks

365 days

10 years

100 years

Step-by-step explanation:

The manufacturer of an airport baggage scanning machine claims it can handle an average of 530 bags per hour. (a-1) At α = .05 in a left-tailed test, would a sample of 16 randomly chosen hours with a mean of 510 and a standard deviation of 50 indicate that the manufacturer’s claim is overstated? Choose the appropriate hypothesis. a. H1: μ < 530. Reject H1 if tcalc > –1.753 b. H0: μ < 530. Reject H0 if tcalc > –1.753 c. H1: μ ≥ 530. Reject H1 if tcalc < –1.753 d. H0: μ ≥ 530. Reject H0 if tcalc < –1.753 a b c d (a-2) State the conclusion. a. tcalc = –1.6. There is not enough evidence to reject the manufacturer’s claim. b. tcalc = –1.6. There is significant evidence to reject the manufacturer’s claim. a b

Answers

Answer:

(a) H1: μ < 530. Reject H1 if tcalc > –1.753

(b) t calc = –1.6. There is not enough evidence to reject the manufacturer’s claim.

Step-by-step explanation:

We are given that the manufacturer of an airport baggage scanning machine claims it can handle an average of 530 bags per hour.

A sample of 16 randomly chosen hours with a mean of 510 and a standard deviation of 50 is given.

Let [tex]\mu[/tex] = average bags an airport baggage scanning machine can handle

So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \geq[/tex] 530 bags     {means that an airport baggage scanning machine can handle an average of more than or equal to 530 bags per hour}

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 530 bags     {means that an airport baggage scanning machine can handle an average of less than 530 bags per hour}

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

where, [tex]\bar X[/tex] = sample mean = 510

             s = sample standard deviation = 50

            n = sample of hours = 16

So, test statistics  =  [tex]\frac{510-530}{\frac{50}{\sqrt{16} } }[/tex]  ~ [tex]t_1_5[/tex]

                              =  -1.60

The value of t test statistics is -1.60.

Now, at 0.05 significance level the t table gives critical value of -1.753 at 15 degree of freedom for left-tailed test. Since our test statistics is more than the critical values of t as -1.60 > -1.753, so we have insufficient evidence to reject our null hypothesis as it will not in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that an airport baggage scanning machine can handle an average of more than or equal to 530 bags per hour.

I don’t know how to do this. Care to explain?

Answers

Answer:

y=6x+7

Step-by-step explanation:

−6x+y=7

Step 1: Add 6x to both sides

y=6x+7

Answer:

y=6x+7

Step-by-step explanation:

whenever you solve for a letter you have to get it by itself. That means it has to be alone on the opposite side of the equation

move the -6x to the other side of the equation

that -6 changes to a  positive 6 because, your moving it to the other side of the equation

your left with y=6x+7

pls mark me brainliest

According to a survey, 62% of murders committed last year were cleared by arrest or exceptional means. Fifty murde committed last year are randomly selected, and the number cleared by arrest or exceptional means is recorded. When technology is used, use the Tech Help button for further assistance.

a. Find the probability that exactly 41 of the murders were cleared.
b. Find the probability that between 36 and 38 of the murders, inclusive, were cleared.
c. Would it be unusual if fewer than 19 of the murders were cleared? Why or why not?

a.The probability that exactly 41 of the murders were cleared is ____. (Round to four decimal places as needed.)

Answers

Answer:

a) The probability that exactly 41 of the murders were cleared is 0.0013

b) The probability that between 36 and 38 of the murders, inclusive, were cleared is 0.0809.

c) Yes, it would be unusual

Step-by-step explanation:

Let p=62% considered as the probability of having a commited that is cleared by arres or exceptional means. We assume that choosing each of the 50 commited is independent of each choose. Then, let X be the number of cleared. In this case, X is distributed as a binomial random variable. Recall that, in this case,

[tex] P(X=k) = \binom{50}{k} p^{k}(1-p)^{50-k}[/tex] for[tex]0\leq k \leq 50[/tex], with p=0.62

a) We have that

[tex] P(X=40) = \binom{50}{40} p^{40}(1-p)^{50-40} =0.001273487  [/tex]

b) We are asked for the following

[tex]P(36\leq X \leq 38) = P(X=36)+P(X=37)+P(X=38) = 0.080888936

[/tex] (The specific calculation is omitted.

c) We will check for the following probability [tex]P(X\leq 19)[/tex]

[tex]P(X\leq 19 ) = \sum_{k=0}^{19} P(X=k) = 0.000499222 [/tex]

Given that the probability of this event is really close to 0, it would be unusual if less than 19 murders are cleared.

Final answer:

The question deals with the application of binomial probability distribution in a real life situation involving crime investigation. The probability values for a certain range or exact number of cleared murders can be calculated by using the binomial probability formula. It would be statistically unusual for fewer than 19 murders to be cleared given a 62% clearance rate.

Explanation:

This question can be approached using the binomial distribution, where the number of successes in a sequence of n independent experiments (in this case, the number of murders being cleared) follows a binomial distribution.

a. The probability that exactly 41 of the murders were cleared can be found by calculating the binomial probability. This can be done by using the formula: P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k)). In this case, n=50 (number of trials/murders), k=41 (number of successes/murders cleared), and p=0.62 (probability of success/clearing a murder). You need to substitute these values into the formula and calculate the value.

b. Finding the probability that between 36 and 38 murders were cleared involves the same process, but you need to calculate for k=36, 37, and 38, and then add the results together to get the total probability.

c. If fewer than 19 of the murders were cleared, it would be statistically unusual considering the 62% clearance rate according to the survey. The reasoning being that, given a 62% probability, the expectation would be significantly higher.

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If n - 2 = 6 -2(7 + n), then n = ?

Answers

it would equal -8.8 or -84/5

Parallelogram ABCD is dilated to form parallelogram EFGH. Side AB is proportional to side EF. What corresponding side is proportional to segment AD? Type the answer in the box below. (2 points)



Answers

EH is proportional to AD.

In this assignment, your team is managing a software development project with a total project budget of $178,500. Total work effort is 1,536 hours and the timeline for completion 24 weeks. At the end of week 12, the plan was to have completed 55% of the project scope. However, actual progress was calculated at 650 worth of hours completed on project activities and actual cost (in hours) for these activities is 780. With your team, please answer the following questions. For each response, show the pertinent formula(s) as it applies and "show your work" in addition to providing a narrative statement.
a. How much money was supposed to have been spent at the end of week 12?b. Will the project finish on time and within the given budget?

Answers

Answer:

im lost good luck

Step-by-step explanation:

D. Now calculate resting Vm given the following relative permeability ratios. Assume that there is no meaningful permeability to Ca2+. (1pt) PK : PNa : PCl = 1.0 : 0.04 : 0.45 E. Let’s say that the K+ permeability increases, for example because of expression of more K+ channels. Calculate Vm with new the permeability values and discuss how the membrane potential has changed in relation to the potassium equilibrium potential EK. (roughly 1 sentence) (1pt)+

Answers

Answer:EX = RT ln [X]o

.........zF.....[X]i

EX = (1.987 cal/deg.mol)(293 deg) ln [X]o

.........z(23,062 cal/volt.mol)................[X]i

OR

EX = (8.315 joules/deg.mol)(293 deg) ln [X]o

.........z(96,485 joules/volt.mol)................[X]i

EK+ = 0.025 ln(12/400) = -0.088 V = -88 mV

ENa+ = 0.025 ln(450/55) = 0.053 V = 53 mV

ECa+2 = 0.0126 ln(10/0.0001) = 0.145 V = 145 mV

ECl- = -0.025 ln(550/56) = -0.058 V = -58 mV

Step-by-step explanation:

Final answer:

The Goldman equation is used to calculate resting membrane potential (Vm) considering the relative permeabilities of different ions. Given the permeabilities, the resting Vm can be estimated. If the K+ permeability increases, Vm will move closer to the Potassium equilibrium potential (EK).

Explanation:

The resting membrane potential (Vm) can be calculated using the Goldman equation, which considers the relative permeabilities and concentrations of different ions. The equation is: Vm = 61.5 log ((PK[K+]out + PNa[Na+]out + PCl[Cl-]in) / (PK[K+]in + PNa[Na+]in + PCl[Cl-]out)), where Px indicates the relative permeability of each ion & the square brackets contain the ion concentrations inside (in) or outside (out) the cell.

Given the permeabilities PK : PNa : PCl = 1.0 : 0.04 : 0.45, assuming concentrations inside and outside the cell in a balanced condition with no net movement of any ion, you might estimate the resting Vm.  

If the K+ permeability increases, Vm would reportedly move closer towards the Potassium equilibrium potential (EK). This is because the membrane is becoming more permeable to K+ and less responsive to the influences of other ions.

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Enter the equation that describes the line in slope-intercept form.
slope =-4, (5,6) is on the line.

Answers

Answer:

y = -4x + 26

Step-by-step explanation:

y = mx + b    m: slope  -4   b: y intercept

pass (5 , 6)

b = y - mx = 6 - (-4) x5 = 6 + 20 = 26

equation: y = -4x + 26

Multiply: (-3/10)(-2/9)

Answers

Answer:

1/15

Step-by-step explanation:

The answer will be 1/5

MODELING REAL LIFE The total height of the Statue of Liberty and its pedestal is $153$ feet more than the height of the statue. What is the height of the statue?

A picture shows the Statue of Liberty on a pedestal. The total height of the Statue of Liberty and its pedestal is labeled “305 feet”.

Answers

Answer: 152

Step-by-step explanation:

305 - 153 = 152

Final answer:

The height of the Statue of Liberty itself, without the pedestal, is 152 feet. This is found by subtracting the pedestal height (153 feet) from the total height (305 feet).

Explanation:

The student is asked to determine the height of the Statue of Liberty excluding its pedestal. Given that the total height of the Statue of Liberty including its pedestal is labeled as 305 feet, and the total height is 153 feet more than the height of the statue alone, we can set up the following equation to solve for the height of the statue (let's call it S):

S + 153 = 305

To find the height of the Statue of Liberty without the pedestal, we subtract 153 from both sides of the equation:
S = 305 - 153
S = 152

Therefore, the height of the Statue of Liberty itself is 152 feet.

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Suppose that grade point averages of undergraduate students at one university have a bell-shaped distribution with a mean of 2.542.54 and a standard deviation of 0.420.42. Using the empirical rule, what percentage of the students have grade point averages that are between 1.281.28 and 3.83.8?

Answers

Answer:

[tex] P(1.28< X< 3.8) [/tex]

And we can use the z score formula to calculate how many deviations we are within the mean

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

And if we use this formula we got:

[tex] z = \frac{1.28-2.54}{0.42}= -3[/tex]

[tex] z = \frac{3.8-2.54}{0.42}= 3[/tex]

And using the empirical rule we know that within 3 deviation from the mean we have 99.7% of the values

Step-by-step explanation:

Previous concepts

The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".

Let X the random variable who represent the grade point averages of undergraduate students.

From the problem we have the mean and the standard deviation for the random variable X. [tex]E(X)=2.54, Sd(X)=0.42[/tex]

So we can assume [tex]\mu=2.54 , \sigma=0.42[/tex]

On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:

• The probability of obtain values within one deviation from the mean is 0.68

• The probability of obtain values within two deviation's from the mean is 0.95

• The probability of obtain values within three deviation's from the mean is 0.997

For this case we want to find this probability:

[tex] P(1.28< X< 3.8) [/tex]

And we can use the z score formula to calculate how many deviations we are within the mean

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

And if we use this formula we got:

[tex] z = \frac{1.28-2.54}{0.42}= -3[/tex]

[tex] z = \frac{3.8-2.54}{0.42}= 3[/tex]

And using the empirical rule we know that within 3 deviation from the mean we have 99.7% of the values

Final answer:

Approximately 99.7 percent of the students have grade point averages between 1.28 and 3.8 according to the empirical rule.

Explanation:

The empirical rule states that approximately 68 percent of the data lies within one standard deviation of the mean, 95 percent lies within two standard deviations, and more than 99 percent lies within three standard deviations. In this case, the mean grade point average is 2.54 and the standard deviation is 0.42. To find the percentage of students with grade point averages between 1.28 and 3.8, we need to find the z-scores for both values and calculate the area under the curve between those z-scores.

First, we find the z-score for 1.28 using the formula: z = (x - µ) / σ = (1.28 - 2.54) / 0.42 = -3.0476. Then, we find the z-score for 3.8 using the same formula: z = (3.8 - 2.54) / 0.42 = 3.0476. Using a standard normal distribution table or a calculator, we can find the area between these two z-scores, which is approximately 99.7 percent.

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Create a cylinder with a height of 14 cm and a radius of 10 cm.

Create a cylinder with a height of 14 cm and a radius of 10 cm. What can be concluded about the cylinder’s volume when the radius is halved?
The volume is One-fourth of the original.
The volume is One-third of the original.
The volume is One-half of the original.
The volume is twice the original.

Answers

Answer:

a on edge2021

Step-by-step explanation:

The volume is One-fourth of the original.

What is a cylinder?

"It is a three dimensional structure having two parallel bases joined by a curved surface, at a fixed distance."

The formula of the volume of a cylinder:

V = π × r² × h, where 'r' is the radius of the circular base and 'h' is the height of the cylinder.

In the given question,

the radius of the cylinder is 10 cm, and the height is 14 cm.

⇒ r = 10 cm, h = 14 cm

The volume of the cylinder would be,

[tex]V_1=\pi\times 10^2\times 14\\\\V_1=\frac{22}{7}\times 100 \times 14\\\\V_1=4400~~cu.~cm.[/tex]

If the radius is halved, the radius becomes 5 cm.

The new volume of a cylinder would be,

[tex]V_2=\pi \times 5^2\times 14\\\\V_2=\frac{22}{7}\times 25 \times 14\\\\V_2=1100~~cu.~cm.[/tex]

The ratio between the original and the new volume of the cylinder is:

[tex]\Rightarrow \frac{1100}{4400}=\frac{1}{4}\\\\\Rightarrow V_2=\frac{1}{4}V_2[/tex]

This means, the volume is One-fourth of the original.

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Josephine’s father is 5 times as old as Josephine. In 6 years, he will be only three times as old. How old is Josephine now?

Answers

Answer:

Josephine is 6 years old.

Step-by-step explanation:

6 * 5 = 30

12 * 3=  36

Josephine is currently 6 years old.

We are given two conditions about the ages of Josephine and her father.

Josephine’s father is 5 times as old as Josephine currently.

In 6 years, he will be only three times as old as she will be at that time.

Let's let 'J' represent Josephine's current age.

Then her father's current age would be 5J. In 6 years, Josephine will be J+6 and her father will be 5J+6.

According to the second condition, at that time her father's age will be three times Josephine's age.

So, we set up the equation 5J+6 = 3(J+6).

To find Josephine's current age, we solve the equation:

5J + 6 = 3J + 18

Subtract 3J from both sides:

2J + 6 = 18

Now subtract 6 from both sides:

2J = 12

And divide both sides by 2:

J = 6

Therefore, Josephine is currently 6 years old.

If the drawing has dimensions 7.5 cm × 5 cm, what is the size of the soccer field?

Answers

Answer: 35.5cm^2

Step-by-step explanation:

I believe we're talking about a rectangular. So it would be length * width or 7.5cm * 5cm = 35.5cm^2

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