Answer:
-0.36666666666...
Step-by-step explanation:
not sure how to explain, that's a repeating number.
Which numbers are irrational? Check all that apply.
1.9/15
2. sqrt of 18
3.sqrt of 9
4.sqrt of 169
5.sqrt of 78
6.pi
Answer: B, E, and F
Step-by-step explanation: :)
Orange M&M’s: The M&M’s web site says that 20% of milk chocolate M&M’s are orange. Let’s assume this is true and set up a simulation to mimic buying 200 small bags of milk chocolate M&M’s. Each bag contains 55 candies. We made this dotplot of the results. Normal sampling distribution of proportion of orange candies Now suppose that we buy a small bag of M&M’s. We find that 25.5% (14 of the 55) of the M&M’s are orange. What can we conclude? Group of answer choices This result is not surprising because we expect to see many samples with 14 or more orange candies. This result is surprising because we expect the orange candies to make up no more than 20% of the candies in a packet. This result is surprising because it is unlikely that we will select a random sample with 25.5% or more orange candies if 20% of milk chocolate M&M’s are orange.
The initial expectation is that 20% of the M&Ms in a bag would be orange. However, the occurrence of 25.5% orange M&Ms in a bag can be considered surprising but doesn't negate the initial percentage due to statistical sampling variability.
Explanation:This question is dealing with the concept of statistical sampling and proportions. Given that the M&M’s website states that 20% of milk chocolate M&M’s are orange, we would expect, on average, that 20% or 11 (0.20 * 55) of the M&M’s in a small bag of 55 M&M’s to be orange. However, in this instance, we observe that 25.5% (or 14 of the 55) of the M&M’s are orange.
This result might be surprising because it is more than the average expectation of 20%. But it's important to remember the concept of sampling variability. Even though we expect 20%, the actual results can vary from this expectation due to random chance.
In conclusion, this result can be considered surprising because we have got a higher percentage of orange candies than expected, but it does not necessarily prove that the initial assertion of 20% orange M&M’s is incorrect due to the inherent variability in statistical sampling.
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The observed result of 25.5% orange candies per bag is surprising given the stated 20% proportion. However, due to sampling variability, it falls within one standard deviation from the mean, indicating it's a possible result in a sample.
Explanation:The conclusion based on your provided information would be that this result is surprising because it is unlikely to select a sample with 25.5% or more orange candies if only 20% of milk chocolate M&M's are orange. This observation is compared to the expected proportion of 20%, i.e., the stated percentage of orange M&M's as per the website information.
To calculate how surprising this outcome is, we can use the concept of standard deviation. Based on the binominal distribution, the standard deviation (σ) is calculated as √npq, where n is the number of trials (in this case, the number of candies per bag, 55), p is the probability of success (the expected proportion of orange candies, 0.20), and q is the probability of failure (1-p).
Applying these values, σ = √(55)(0.20)(0.80), we get σ ≈ 4.7. Therefore, observing 14 orange candies is about one standard deviation above the expected value (11 candies, or 20% of 55). Although that might seem like a significant deviation, in a normal distribution about 68% of the data falls within one standard deviation of the mean. So while it is surprising, it is still a possible variation to have 25.5% or more orange candies in a single bag due to sampling variability.
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A company offers collision insurance and liability insurance. The claim amount under collision insurance is normally distributed with mean 10, 000 and standard deviation 2, 000; the claim amount under liability insurance is normally distributed with mean 9, 000 and standard deviation 2, 000. Assuming independence, find the probability that the claim amount under liability insurance exceeds the claim amount under collision insurance.
Answer:
36.32% probability that the claim amount under liability insurance exceeds the claim amount under collision insurance.
Step-by-step explanation:
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.
Two variables:
X1 and X2, probability X1 exceeds X2
[tex]\mu(X_{1} - X_{2}) = \mu_{X_{1}} - \mu_{X_{2}}[/tex]
[tex]\sigma^{2}(X_{1} - X_{2}) = \sigma_{X_{1}}^{2} + \sigma_{X_{2}}^{2}[/tex]
In this problem, we have:
X1: Liability insurance:
So [tex]\mu_{X_{1}} = 9000, \sigma_{X_{1}} = 2000[/tex]
X2: Collusion insurance:
So [tex]\mu_{X_{2}} = 10000, \sigma_{X_{2}} = 2000[/tex]
Then
[tex]\mu(X_{1} - X_{2}) = \mu_{X_{1}} - \mu_{X_{2}} = 9000 - 10000 = -1000[/tex]
[tex]\sigma^{2}(X_{1} - X_{2}) = 2000^{2} + 2000^{2} = 8000000[/tex]
[tex]\sigma = \sqrt{8000000} = 2828.43[/tex]
Assuming independence, find the probability that the claim amount under liability insurance exceeds the claim amount under collision insurance.
This is [tex]P(X_{1} - X_{2} > 0)[/tex], which is 1 subtracted by the pvalue of Z when X = 0. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0 - (-1000)}{2828.43}[/tex]
[tex]Z = 0.35[/tex]
[tex]Z = 0.35[/tex] has a pvalue of 0.6368
1 - 0.6368 = 0.3632
36.32% probability that the claim amount under liability insurance exceeds the claim amount under collision insurance.
Somebody please help me with this i have no clue how to do any of this
Ms. Vargas recorded the hourly wage of each employee at her business.
Which of the following statements are true?
Chose all correct answers.
A. The mean of the data describes how the hourly wages vary.
B. The mean absolute deviation of the data describes how the hourly wages vary. C. The interquartile range of the data summarizes the hourly wages.
D. The median of the data summarizes the hourly wages.
E.The maximum value of the data summarizes the hourly wages.
Answer:
I think it is A and B. Not one hundred percent sure but,50 percent sure.
Step-by-step explanation:
Answer:
B.) and D.)
Step-by-step explanation:
What is the measure of side x in the isosceles triangle that is shown below
Answer:
x = 29 cm
Step-by-step explanation:
An isosceles triangle has two equal sides that have equal angles at the bases
Since the 80 degree angles are at the bases of the 29 cm side, the length of x must be 29 cm
Answer:
29 cm
Step-by-step explanation:
Because both angles are 80 degrees. If your angles are the same so are your sides.
Find the Jacobian ∂(x, y, z) ∂(u, v, w) for the indicated change of variables. If x = f(u, v, w), y = g(u, v, w), and z = h(u, v, w), then the Jacobian of x, y, and z with respect to u, v, and w is ∂(x, y, z) ∂(u, v, w) = ∂x ∂u ∂x ∂v ∂x ∂w ∂y ∂u ∂y ∂v ∂y ∂w ∂z ∂u ∂z ∂v ∂z ∂w . x = 1 6 (u + v), y = 1 6 (u − v), z = 6uvw
Answer:
The Jacobian ∂(x, y, z) ∂(u, v, w) for the indicated change of variables
= -3072uv
Step-by-step explanation:
Step :-(i)
Given x = 1 6 (u + v) …(i)
Differentiating equation (i) partially with respective to 'u'
[tex]\frac{∂x}{∂u} = 16(1)+16(0)=16[/tex]
Differentiating equation (i) partially with respective to 'v'
[tex]\frac{∂x}{∂v} = 16(0)+16(1)=16[/tex]
Differentiating equation (i) partially with respective to 'w'
[tex]\frac{∂x}{∂w} = 0[/tex]
Given y = 1 6 (u − v) …(ii)
Differentiating equation (ii) partially with respective to 'u'
[tex]\frac{∂y}{∂u} = 16(1) - 16(0)=16[/tex]
Differentiating equation (ii) partially with respective to 'v'
[tex]\frac{∂y}{∂v} = 16(0) - 16(1)= - 16[/tex]
Differentiating equation (ii) partially with respective to 'w'
[tex]\frac{∂y}{∂w} = 0[/tex]
Given z = 6uvw ..(iii)
Differentiating equation (iii) partially with respective to 'u'
[tex]\frac{∂z}{∂u} = 6vw[/tex]
Differentiating equation (iii) partially with respective to 'v'
[tex]\frac{∂z}{∂v} =6 u (1)w=6uw[/tex]
Differentiating equation (iii) partially with respective to 'w'
[tex]\frac{∂z}{∂w} =6 uv(1)=6uv[/tex]
Step :-(ii)
The Jacobian ∂(x, y, z)/ ∂(u, v, w) =
[tex]\left|\begin{array}{ccc}16&16&0\\16&-16&0\\6vw&6uw&6uv\end{array}\right|[/tex]
Determinant 16(-16×6uv-0)-16(16×6uv)+0(0) = - 1536uv-1536uv
= -3072uv
Final answer:-
The Jacobian ∂(x, y, z)/ ∂(u, v, w) = -3072uv
James is 5 feet 4 inches tall, how tall is James in inches
Answer:
64 in.
Step-by-step explanation:
One foot is 12 inches, so you would have to multiply 12 and 5 to get 60. You then add the 60 to the 4 to get 64 inches. Hope this helped!
what is the value for x 3x - 2 = 16
Answer:
18
Step-by-step explanation:
I meant 6x3=18 -2=16 sorry for confusion
To find the value of x, you need to isolate/get the variable "x" by itself in the equation:
3x - 2 = 16 Add 2 on both sides
3x - 2 + 2 = 16 + 2
3x = 18 Divide 3 on both sides to get "x" by itself
[tex]\frac{3x}{3} =\frac{18}{3}[/tex]
x = 6
PROOF
3x - 2 = 16 Substitute/plug in 6 into "x" since x = 6
3(6) - 2 = 16
18 - 2 = 16
16 = 16
Recipes for the same type of cookies can vary in terms of ingredients and baking times. From a collection of chocolate chip cookie recipes, a baker randomly selected 5 recipes. From a collection of oatmeal raisin cookie recipes, the baker randomly selected 4 recipes. The mean baking times, in minutes, for each sample were recorded as x¯C and x¯O, respectively. What is the correct unit of measure for the standard deviation of the sampling distribution of x¯C−x¯O?
Answer:
answer should be minutes.
Step-by-step explanation:
in the problem, it is stated that the mean is in "baking times." this can lead one to assume that the standard deviation between the two xC-xO can be the variation in minutes that the two cookies types have.
1/3=n+3/4 what dies n equal??
Which solid does the net form?
c. traingular prism
Answer:
Triangular Prism
I'm guessing this means what 3D shape it produces, which would lead my conclusion to it being a triangular prism.
Davis performed an experiment in which he spun a 4-color spinner 20 times. He recorded his results in the frequency table.
Davis predicts that if he spins the spinner another 20 times, the frequency for the red and green segments will still be equal. Is Davis correct?
Yes. Another experiment will give the same numbers as the first experiment.
Yes. The experimental probabilities remain the same for all experiments.
No. The frequencies will be multiplied by 20.
No. The frequencies and probabilities for red and green segments might change in the next experiment.
Answer:
No. The frequencies and probabilities for red and green segments might change in the next experiment.
Step-by-step explanation:
I just did it and it said my answer was right
Answer:
Step-by-step explanation:
No. The frequencies and probabilities for red and green segments might change in the next experiment.
i just did it and got it right so i hope you do too
What 4c=12
4c/4=12.4
Answer:
c =3
Step-by-step explanation:
4c = 12
Divide each side by 4
4c/4 =12/4
c = 3
pls help!!!!!!!!! if possible will mark brainliest!
Answer:
i think you may have to use sin or cos
Step-by-step explanation:
Answer:
9m (You were correct!!!!)
Step-by-step explanation:
Use Pythagorean Theorem:
[tex]a^{2} +b^{2} = c^{2}[/tex]
⬆️ ⬆️ ⬆️
legs hypotenuse
x² + 12² = 15²
x² + 144 = 225
x² = 81
x = 9
Which events have a probability of 25 percent? Select three options.
Answer:
The answer is A, C, E
A: choosing a green jelly bean out of a bag that contains 2 green jelly beans, 1 red jelly bean, and 5 yellow jelly beans
C: spinning a number less than 2 on a spinner that has four equal sections numbered from 1 to 4
D: choosing a spade out of a standard deck of cards that contains 13 hearts, 13 clubs, 13 diamonds, and 13 spades
Therefore the answer is A,C,E
Edg : )
Joey read in his biology book that fish activity increases with water temperature, and he decided to investigate this issue by conducting an experiment. On nine successive days, he measured fish activity and water temperature in his aquarium. He found the equation of the LSRL to be y t ˆ = + 152.1 37.65 , where yˆ is the predicted amount of activity. If one of Joey’s days had a water temperature of 69.0 degrees Fahrenheit and the fish activity on that day was 265, what is the residual activity for this particular temperature?
The residual value is the difference between the actual and predicted values. Hence, the residual activity at a temperature of 69°F will be −2484.95
The equation of the Least Square Regression Line is given as :
y = 152.1 + 37.65tThe predicted amount of activity, y at a temperature of 69°F will be :
y = 152.1 + 37.65(69)
y = 2749.95
The actual amount of activity at a temperature of 69°F is 265
The residual activity at 69°F temperature can be calculated thus :
Residual = Actual - Predicted
Residual = 265 - 2749.95 = −2484.95
Therefore, the residual activity at 69°F is −2484.95
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What is the area of this tile ?
Use the empirical rule to solve the problem. The annual precipitation for one city is normally distributed with a mean of 288 inches and a standard deviation of 3.7 inches. Fill in the blanks. In 95.44% of the years, the precipitation in this city is between ___ and ___ inches. Round your answers to the nearest tenth as needed.
Answer:
[tex]z=-1.99<\frac{a-288}{3.7}[/tex]
[tex]z=1.99<\frac{a-288}{3.7}[/tex]
And if we solve for a we got
[tex]a=288 -1.99*3.7=214.4[/tex]
[tex]a=288 +1.99*3.7=295.4[/tex]
And the limits for this case are: (214.4; 295.4)
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 annual precipitation of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(288,3.7)[/tex]
Where [tex]\mu=288[/tex] and [tex]\sigma=3.7[/tex]
The confidence level is 95.44 and the signficance is [tex] 1-0.9544=0.0456[/tex] and the value of [tex]\alpha/2 =0.0228[/tex]. And the critical value for this case is [tex]z = \pm 1.99[/tex]
Using this condition we can find the limits
[tex]z=-1.99<\frac{a-288}{3.7}[/tex]
[tex]z=1.99<\frac{a-288}{3.7}[/tex]
And if we solve for a we got
[tex]a=288 -1.99*3.7=214.4[/tex]
[tex]a=288 +1.99*3.7=295.4[/tex]
And the limits for this case are: (214.4; 295.4)
Cars on Campus. Statistics students at a community college wonder whether the cars belonging to students are, on average, older than the cars belonging to faculty. They select a random sample of 18 cars in the student parking lot and find the average age to be 8.4 years with a standard deviation of 7.3 years. A random sample of 15 cars in the faculty parking lot have an average age of 4.2 years with a standard deviation of 3.8 years.
1. The null hypothesis is H0:????????=???????? H 0 : μ s = μ f . What is the alternate hypothesis? A. H????:????????≠???????? H A : μ s ≠ μ f B. H????:????????<???????? H A : μ s < μ f C. H????:????????>???????? H A : μ s > μ f
2. Calculate the test statistic. =
3. Calculate the p-value for this hypothesis test. p value =
4. Suppose that students at a nearby university decide to replicate this test. Using the information from the community college, they calculate an effect size of 0.7. Next, they obtain samples from the university student and faculty lots and, using their new sample data, conduct the same hypothesis test. They calculate a p-value of -0.0213 and an effect size of 0.487. Do their results confirm or conflict with the results at the community college? A. It contradicts the community college results because the effect size is much smaller. B. It confirms the community college results because the effect size is nearly the same. C. It contradicts the community college results because the p-value is much bigger D. It can neither confirm or contradict the community college results because we don't know the sample sizes the university students used. E. It confirms the community college results because the p-value is much smaller.
Answer:
Let the 1st group be of students and 2nd group be of faculty.
n1 = 18
\bar{x}_{1} = 8.4
s1 = 7.3
n2 = 15
\bar{x}_{2} = 4.2
s2 = 3.8
1.
H0: =
HA: >
2.
Equality of variance test:
F = s12/s22
= 53.29/14.44
= 3.69
df-num = n1-1 = 17
df-den = n2-1 = 14
p-value = 0..008 < 0.05 i.e.H0 can be rejected and hence we can say that both populations do not have the same variances.
So, test-statistic will be calculated as follows:
[ Find the solution of this part in the attachment]
3.
df = 26
p-value = 0.0217
4.
E. It confirms the community college results because the p-value is much smaller.
The equation, y=−16x2+32x+24 , represents the height, in feet, of a firework x seconds after it is launched. What is the maximum height of the firework? Enter your answer in the box. feet
Answer:
40
Step-by-step explanation:
got 100 on quiz
The maximum height of the firework is 40 feet.
What is maximum height?
'The maximum height of the object is the highest vertical position along its trajectory.'
According to the given problem,
y = -16x²+32x+24
For maximum height,
The slope of y with respect to x has to be 0.
⇒ [tex]\frac{dy}{dx}[/tex] = 0
Now, [tex]\frac{d(-16x^{2} +32x+24)}{dx}[/tex] = -32x +32
⇒-32x + 32 = 0
⇒ 32 = 32x
⇒1 = x or x = 1 second
Therefore after 1 second, the firework reaches maximum height.
Now, we put the value of x = 1 second in the equation: y = -16x²+32x+24
⇒ -16(1)² + 32(1) +24
= -16+32+24
= 40 feet
Hence, we can conclude that after 1 second, the firework reaches a maximum height of 40 feet.
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The mayor of a town has proposed a plan for the annexation of a new bridge. A political study took a sample of 900 voters in the town and found that 75% of the residents favored annexation. Using the data, a political strategist wants to test the claim that the percentage of residents who favor annexation is above 72%. Determine the P-value of the test statistic. Round your answer to four decimal places.
Answer:
P-value of the test statistic: P=0.0266.
Step-by-step explanation:
Hypothesis test on a proportion.
The claim is that the percentage of residents who favor annexation is above 72% This is going to be stated in the alternative hypothesis.
Then, the null and alternative hypothesis are:
[tex]H_0: \pi=0.72\\\\H_a:\pi>0.72[/tex]
The sample size is n=900 and the sample proportion is p=0.75.
The standard devaition is calculated as:
[tex]\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.72*0.28}{900}}=\sqrt{0.000224}=0.015[/tex]
Then, the z-statistic is:
[tex]z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.75-0.72-0.5/900}{0.015}=\dfrac{0.029}{0.015} = 1.9333[/tex]
For this right tailed test, the P-value is:
[tex]P-value=P(z>0.1933)=0.0266[/tex]
Private universities claim to offer a higher quality education compared to public universities as justification for their more expensive tuition rates. The implication is that better education in the classroom leads to better employees after graduation. As a HR specialist in charge of selecting the best possible future employees from a pool of applicants, you propose giving preference to applicants with degrees from private universities. Hoping to control for the influence of academic performance on on-the-job performance, you randomly select 25 pairs of employees based on their cumulative GPA, one with a degree from a private university and another with a degree from a public university, and record their last performance evaluation score in an Excel file. The data appears as follows:
GPA Private University Alumni Public University Alumni
4.0 85 89
3.9 82 78
3.8 75 77
3.7 91 72
(a) How should this study design be described?
O True experiment
O Posttest only quasi experiment
O Non-experimental study
O Quasi experiment
Answer:quasi experiment
Step-by-step explanation:
Because it is basically for comparison usually by measuring outcomes if a program by participants and non participants
Find the volume
A.I0 cubic units
B.20 cubic units
C.22 cubic units
D.10 cubic units
Answer:
its b
Step-by-step explanation:
FOR THE LENGTH YOU SEE HOW LONG IT IS THEN FINALLY YOU SEE THE WIDTH ITS ONLY ONE AND THEN THE HIGHT HOW TALL IT IS AND THEN YOU MULTIPLY TO GET Your answer
i need helppp please :(
Step-by-step explanation:
2 ( 8 - 3c) = 4c - 34
16 - 6c = 4c - 34
6c + 4c = 16 + 34
10c = 50
c = 50/ 10
therefore c = 5
Hope it will help :)
What property is 1/5+1/8=1/8+1/5?
Answer: This is the Communitve Property because the equation has the same thing on both sides, but the numbers "flip flopped" or switched around.
the quoteint of 2 and a number x,times 3
Answer:
just look it up on m..a.t.h. w.a.y.!!
Step-by-step explanation:
it works for me
In ΔBCD, the measure of ∠D=90°, CB = 25, BD = 7, and DC = 24. What is the value of the sine of ∠C to the nearest hundredth?
Answer: The value of the sine of ∠C is 0.28.
Step-by-step explanation:
Let's draw a triangle again (see the picture below).
Remember SOHCAHTOA? It's:
-sin = opposite/hypotenuse
-cos = adjacent/hypotenuse
-tan = opposite/adjacent.
We need to use sin in here.
The sin of C is [tex]\frac{7}{25}[/tex], which in a calculator, gives [tex]0.28[/tex]
A national magazine took a poll to determine company's opinions about outsourcing. Outsourcing is defined as the assignment of critical, but non-core, business functions to outside specialists. The magazine found that 26 percent of business executives felt that the benefits of outsourcing were less than what they expected. A sample of 100 business executives was polled. In this sample 19% of the business executives felt the benefits of outsourcing were less than what they expected. You want to determine whether the proportion of business executives in the sample differs from the national magazine poll. Which statistical test should you use?
Answer:
Hypothesis test on a proportion.
There is no enough evidence to support the claim that the proportion of business executives in the sample differs from the national magazine poll.
Step-by-step explanation:
We should test the hypothesis of the proportion, with a z-statistic.
The claim is that the proportion of business executives in the sample differs from the national magazine poll.
Then, the null and alternative hypothesis are:
[tex]H_0: \pi=0.26\\\\H_a:\pi<0.26[/tex]
The significance level is assumed to be 0.05.
The sample, of size n=100, has a proportion p=0.19.
The standard error for the proportion is:
[tex]\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.26*0.74}{100}}=\sqrt{0.001924}=0.044[/tex]
Then, the z-statistic can be calculated as:
[tex]z=\dfrac{p-\pi+0.5/n}{\sigma_p}=\dfrac{0.19-0.26+0.5/100}{0.044}=\dfrac{-0.065}{0.044}=-1.477[/tex]
The P-value for this one-tailed test is:
[tex]P-value=P(z<-1.477)=0.07[/tex]
As the P-value is bigger than the significance level, the effect is not significant. The null hypothesis failed to be rejected.
There is no enough evidence to support the claim that the proportion of business executives in the sample differs from the national magazine poll.
Angles opposite to one another at the intersection of two lines are called ____ angles.
Answer:
A Vertical Angle
Step-by-step explanation:
Answer:
vertical angles
Step-by-step explanation:
What is the measure of angle DEG on circle O? Please help! 50 points!
Answer:
50 degrees
Step-by-step explanation: