Answer:
500
Step-by-step explanation:
Lets make the unknown number the letter a
0.9% can also be shown as 0.9/100
so 0.9 of a = 4.5
x(0.9/100) = 4.5
multiply 0.9 by x
(0.9×a)/100 = 4.5
multiply both sides by 100
0.9×a = 450
divide both sides by 0.9
a = 500
there you have it
To determine the number that 4.5 is 0.9% of, we convert 0.9% to a decimal by dividing by 100 and solve the resulting equation. The answer is that 4.5 is 0.9% of 500.
Explanation:To find the number that 4.5 represents 0.9% of, we should first convert the percent to a decimal. According to B.4, a percent is converted to a decimal by dividing the percent value by 100. In our case, 0.9% becomes 0.009.
Once we have the decimal equivalent of the percent, we can set up the equation where 4.5 is 0.9% (0.009 in decimal form) of some number x:
4.5 = 0.009x
To solve for x, divide both sides of the equation by 0.009:
x = 4.5 / 0.009
x = 500
Therefore, 4.5 is 0.9% of 500.
hree TAs are grading a final exam. There are a total of 60 exams to grade. (a) How many ways are there to distribute the exams among the TAs if all that matters is how many exams go to each TA? (b) Now suppose it matters which students' exams go to which TAs. How many ways are there to distribute the exams? (c) Suppose again that we are counting the ways to distribute exams to TAs and it matters which students' exams go to which TAs. The TAs grade at different rates, so the first TA will grade 25 exams, the second TA will grade 20 exams and the third TA will grade 15 exams. How many ways are there to distribute the exams?
Final answer:
The solutions to the three parts of the question use different combinatorial methods: for part (a), the stars and bars method is used; for part (b), permutations are appropriate; and for part (c), combinations with fixed capacities are needed. Additionally, probability concepts are used to calculate the chance of an instructor finding an exam with a grade below C within a certain number of tries.
Explanation:
The student's question revolves around combinatorics, which is a field of mathematics that deals with counting, both as an art and as a science. Let's break down the responses to parts (a), (b), and (c) of the question provided by the student:
Part (a): We need to determine the number of ways to distribute 60 exams among three TAs regardless of which specific exams they receive. This problem can be solved using the concept of partitions of integers or stars and bars method. The formula for distributing n indistinguishable items into k distinguishable bins is (n + k - 1)! / [n!(k - 1)!]. Here, n=60 exams, and k=3 TAs.
Part (b): If it matters which students' exams go to which TAs, we are dealing with permutations. The total ways to distribute the exams in this case is 60!, because each exam is distinct and can be assigned to each TA.
Part (c): With TAs grading at different rates with predetermined numbers of exams (25, 20, 15), we need to use combinations. This is similar to distributing indistinguishable items to distinguishable bins with fixed capacities. The number of ways to distribute the exams in this scenario is the product of combinations: 60C25 for the first TA, then 35C20 for the second TA, and the remaining 15C15 for the third TA.
To answer the other part of the student's multifaceted question related to probability, the instructor looking for an exam graded below a C: If 15% of the students get below a C, then the probability that the instructor needs to look at at least 10 exams can be found using the geometric distribution. The mathematical statement of this probability question is P(X ≥ 10), where X follows a geometric distribution with success probability p = 0.15.
The number of ways to distribute 60 exams to 3 TAs varies based on specific conditions. If only the count of exams per TA matters, there are 1891 ways. If specific exams matter, there are approximately 4.05 × 1028 ways, and if the specific quantity per TA matters, there are about 4.28 × 1016 ways.
Distribution of Exams Among TAs
Let's break down the problem into three parts:
(a) Distribution Based on Number of Exams Each TA Grades
→ This problem can be approached using the stars and bars combinatorial method. We need to distribute 60 → → indistinguishable exams to 3 TAs.
→ The formula for this is:
C(n + r - 1, r - 1) where n = 60 exams and r = 3 TAs.
C(60 + 3 - 1, 3 - 1) = C(62, 2)
→ Calculating this combination:
C(62, 2) = 62! / (2!(60!))
62! / (2! × 60!) = (62 × 61) / (2 × 1)
= 1891
Thus, there are 1891 ways to distribute the exams such that only the number of exams per TA matters.
(b) Distribution Where Specific Exams Matter
Now, we are interested in which specific exams go to which TA.
→ This is a permutations problem with repetition. Each of the 60 exams can go to any of the 3 TAs.
3⁶⁰
→ Calculating this value:
3⁶⁰ ≈ 4.0528564 × 10²⁸
Therefore, there are approximately 4.05 × 10²⁸ ways to distribute the specific exams to the TAs.
(c) Distribution with Specific Numbers and Specific Exams
Here, we need to distribute the exams where each TA has a predetermined number of exams (25, 20, and 15).
→ This scenario uses the multinomial coefficient:
C(60, 25, 20, 15)
→ This is calculated as:
60! / (25! 20! 15!)
→ Finding the exact value:
60! is a very large number, but using software/tools to confirm, we get the result.
Thus, there are 60! / (25! 20! 15!) ≈ 4.28 × 10¹⁶ ways to distribute the exams under these conditions.
Does the function model exponential growth or decay?
$(t) = 5. (3/7)*
Answer:
decay
Step-by-step explanation:
Answer:
khan
Step-by-step explanation:
A study found that 1% of Social Security recipients are too young to vote. If 800 social security recipients are randomly selected find the Mean, Variance and the Standard deviation of social security recipients who are too young to vote. Present your answer in two decimal places and in order: mean, Variance, Standard deviation.
Answer:
8, 7.92, 2.81
Step-by-step explanation:
For each Social Security recipient, there are only two possible outcomes. Either they are too young to vote, or they are not. The probability of a Social Security recipient is independent of any other Social Security recipient. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The variance of the binomial distribution is:
[tex]V(X) = np(1-p)[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
In this problem, we have that:
[tex]n = 800, p = 0.01[/tex]
So
Mean:
[tex]E(X) = np = 800*0.01 = 8[/tex]
The variance of the binomial distribution is:
[tex]V(X) = np(1-p) = 800*0.01*0.99 = 7.92[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{800*0.01*0.99} = 2.81[/tex]
Formatted answer: 8, 7.92, 2.81
how many 1/4 kilograms are there in 15 1/2 kg
Answer:
20
Step-by-step explanation:
as u have to divide the answr and convert it
Answer:
62 is the answer because
15x4=60 + 1/2x4 = 62
PLEASE HURRY!!! Dwight has 3 baseball cards, and Ellis has 9 baseball cards. If Dwight and Ellis put their baseball cards together and then divide them up equally, how many will each one of them have?
3
6
9
12
Answer:
6
Step-by-step explanation:
3+9=12,
So since there are 2 people,
12 divided by 2 is 6.
Your answer would be 6.
9+3=12
12÷2=6
Dwight and Ellis both have 6 cards equally.
Hope this helps!!!
Which set of sample characteristics is most likely to produce a significant value for the independent-measures t statistic and a large effect size?
a. A small mean difference and small sample variances
b. A large mean difference and small sample variances
c. A small mean difference and large sample variances
d. A large mean difference and large sample variances
Answer:
Correct option: (b).
Step-by-step explanation:
Effect size (η) is a statistical measure that determines the strength of the association (numerically) between two variables. For example, if we have data on the weight of male and female candidates and we realize that, on average, males are heavier than females, the difference between the weight of male and the weight of female candidates is known as the effect size.
The larger the effect size, the larger the weight difference between male and female will be.
Statistic effect size helps us in analyzing if the difference is factual or if it is affected by a change of factors.
In hypothesis testing, effect size, power, sample size, and critical significance level are related to each other.
The effect size formula for a hypothesis test of mean difference is:
[tex]\eta =\frac{\bar x_{1}-\bar x_{2}}{\sqrt{s^{2}}}[/tex]
The denominator s² is combined sample variance.
[tex]s^{2}=\frac{n_{1}s_{1}^{2}+n_{2}s_{2}^{2}}{n_{1}+n_{2}-2}[/tex]
The effect size is affected by two components:
Sample mean differenceSample variance.As the sample mean difference is directly proportional to the effect size, on increasing the sample mean difference value the effect size will also increase.
Ans the sample variance is inversely proportional to the the effect size, on decreasing the sample variance value the effect size will increase and vice-versa.
Thus, the correct option is (b).
) The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponential distribution with mean 100 cfs (cubic feet per second). a. Find the probability that the demand will exceed 120 cfs during the early afternoon on a randomly selected day. (Round your answer to four decimal places.) b. What water-pumping capacity, in cubic feet per second, should the station maintain during early afternoons so that the probability that demand will exceed capacity on a randomly selected day is only 0.09? (Round your answer to two decimal places.)
Answer:
a) 0.3012 = 30.12% probability that the demand will exceed 120 cfs during the early afternoon on a randomly selected day.
b) 240.79 cfs.
Step-by-step explanation:
Exponential distribution:
The exponential probability distribution, with mean m, is described by the following equation:
[tex]f(x) = \mu e^{-\mu x}[/tex]
In which [tex]\mu = \frac{1}{m}[/tex] is the decay parameter.
The probability that x is lower or equal to a is given by:
[tex]P(X \leq x) = \int\limits^a_0 {f(x)} \, dx[/tex]
Which has the following solution:
[tex]P(X \leq x) = 1 - e^{-\mu x}[/tex]
The probability of finding a value higher than x is:
[tex]P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}[/tex]
In this problem, we have that:
[tex]m = 100, \mu = \frac{1}{100} = 0.01[/tex]
a. Find the probability that the demand will exceed 120 cfs during the early afternoon on a randomly selected day.
This is [tex]P(X > 120)[/tex]
[tex]P(X > 120) = e^{-0.01*120} = 0.3012[/tex]
0.3012 = 30.12% probability that the demand will exceed 120 cfs during the early afternoon on a randomly selected day.
b. What water-pumping capacity, in cubic feet per second, should the station maintain during early afternoons so that the probability that demand will exceed capacity on a randomly selected day is only 0.09?
We want x for which
[tex]P(X > x) = 0.09[/tex]
So
[tex]e^{-0.01x} = 0.09[/tex]
[tex]\ln{e^{-0.01x}} = \ln{0.09}[/tex]
[tex]-0.01x = \ln{0.09}[/tex]
[tex]0.01x = -\ln{0.09}[/tex]
[tex]x = -\frac{\ln{0.09}}{0.01}[/tex]
[tex]x = 240.79[/tex]
So 240.79 cfs.
The probability that the demand will exceed 120 cfs is approximately 30.12%. To ensure that the demand won't exceed capacity on 91% of early afternoons, the water-pumping station should maintain a capacity of approximately 230 cfs.
Explanation:a. Finding the Probability That Demand Will Exceed 120 cfsThe mean (λ) of the exponential distribution equals the rate (1/λ), which in this case is 100 cfs. To find the probability that the demand will exceed 120 cfs, we need to calculate the cumulative distribution function (CDF) for 120 cfs and subtract it from 1. The formula for the CDF is F(x) = 1 - e^(-λx). Replacing x with 120 and λ with 1/100, we get: F(120) = 1 - e^(-120/100) = 1 - e^-1.2. The value of e^-1.2 is approximately 0.3012. Thus, F(120) = 1 - 0.3012 = 0.6988. Therefore, the probability that the demand will exceed 120 cfs is 0.3012 or 30.12%, rounded to four decimal places.
b. Finding the Water-Pumping Capacity Needed to Limit the Probability of Exceeding Demand to 0.09We want to find the volume of water (x) such that the probability that the demand will exceed x is 0.09. To do this, we set F(x) = 1 - 0.09 (or 0.91), and use the CDF formula: F(x) = 1 - e^(-λx). Solving the equation 0.91 = 1 - e^(-x/100) for x yields x = -100ln(1 - 0.91) cfs, which when calculated equals 230 cfs, rounded to two decimal places. Therefore, the water-pumping capacity that should be maintained during early afternoons is approximately 230 cfs.
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Find the perimeter of the figure to the nearest hundredth.
Please consider helping!
Any help is appreciated!
Hello!
Your answer should be 26.85.
We can use the formula of pi d or 2 pi r for the circumference.
You would get 5 pi.
But since its not a full circle you would divide it in half and get 2.5 pi.
2.5 pi is equal to 7.85.
Now we can use the equation 7.85 + 7 + 7 + 5
That would equal our answer... 26.85!
Hope this helps!
What is the area of a rectangle with the numbers as 6 and 14
Answer:
I believe its 84
Step-by-step explanation:
6 multiplied by 14
Weights of Elephants A sample of 7 adult elephants had an average weight of 11,647 pounds. The standard deviation for the sample was 24 pounds. Find the 95 % confidence interval of the population mean for the weights of adult elephants. Assume the variable is normally distributed. Round intermediate answers to at least three decimal places. Round your final answers to the nearest whole number.
The 95% confidence interval for the population mean weight of adult elephants is approximately (11628, 11666) pounds.
Explanation:To find the 95% confidence interval of the population mean for the weights of adult elephants, we can use the formula:
CI = x ± Z * (σ/√n)
where CI is the confidence interval, x is the sample mean, Z is the z-score corresponding to the desired confidence level (in this case, 95%), σ is the population standard deviation, and n is the sample size.
Using the given values, x = 11,647 lb, σ = 24 lb, and n = 7, we can calculate the confidence interval:
CI = 11647 ± 1.96 * (24/√7)
CI ≈ 11647 ± 19.12
Therefore, the 95% confidence interval for the population mean weight of adult elephants is approximately (11628, 11666) pounds.
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Assume that 0 < x < pi/2 and 0 < y < pi/2. Find the exact value of cos(x-y) if cos(x)=3/5 and cos(y)=4/5
a. 25/24
b. -25/24
c. 24/25
d. -24/25
Answer: The answer is C
I ready probability concepts quiz
Total
Last July, 160 babies were born in a hospital in Maine;
Š of the babies were girls. Seventy babies weighed 8
pounds or more. Fifty boys weighed 8 pounds or more.
b
REFERATER
O a = 64, b = 14, c = 76, d = 20, e = 90
o a = 14, b = 64, c = 90 d = 20, e = 76
14, b = 76, c = 64, d = 90, e = 20
a = 14, b = 64, c = 76, d = 20, e = 90
11
Answer:
D
Step-by-step explanation:
If you work backwards and add everything up it works.
Answer:
D
Step-by-step explanation:
find a coterminal angle to 20 degrees answer choices r 320 760 690 and 740
Answer:
740°
Step-by-step explanation:
2 *360° + 20° = 720° + 20° = 740°
Which number is bigger? 0.183 or 0.18
Answer:
.183 give me brainliest
Step-by-step explanation:
The National Collegiate Athletic Association (NCAA) measures the Graduation Success Rate (GSR), which is the percentage of eligible athletes who graduate within six years of entering college. According to the NCAA, the GSR for all scholarship athletes in a particular division is 57%. The GSR for all students in this division is 62%. Suppose the NCAA report was based on a sample of 500 student-athletes, of which 285 graduated within six years. Is this sufficient information to conclude that the GSR for all scholarship athletes in this division differs from 62%? Carry out the test using a Type I error rate of 0.05.
Answer:
Yes. There is enough evidence to support the claim that the GSR for all scholarship athletes in this division differs from 62%.
Step-by-step explanation:
We have to perform a hypothesis test on a proportion.
The claim is that the GSR for all scholarship athletes in this division differs from 62%. Then, the null and alternative hypothesis are:
[tex]H_0: \pi=0.62\\\\H_a:\pi<0.62[/tex]
The significance level, named here as Type I error rate, is 0.05.
The sample size is n=500.
The sample proportion is:
[tex]p=X/n=285/500=0.57[/tex]
The standard deviation of the proportion is:
[tex]\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.62(1-0.62)}{500}}=\sqrt{0.0004712}=0.022[/tex]
The z-statistic is then:
[tex]z=\dfrac{p-\pi+0.5/n}{\sigma_p}=\dfrac{0.57-0.62+0.5/500}{0.022}=\dfrac{-0.049}{0.022} = -2.227[/tex]
The P-value for this left tailed test is:
[tex]P-value=P(z<-2.227)=0.013[/tex]
The P-value is smaller than the significance level, so the effect is significant. The null hypothesis is rejected.
There is enough evidence to support the claim that the GSR for all scholarship athletes in this division differs from 62%.
Answer:
[tex]z=\frac{0.57 -0.62}{\sqrt{\frac{0.62(1-0.62)}{500}}}=-2.303[/tex]
[tex]p_v =2*P(z<-2.303)=0.0213[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of student athletes who graduate within 6 years is significantly different from 0.62 or 62%
Step-by-step explanation:
Data given and notation
n=500 represent the random sample taken
X=285 represent the student athletes who graduate within 6 years
[tex]\hat p=\frac{285}{500}=0.57[/tex] estimated proportion of student athletes who graduate within 6 years
[tex]p_o=0.62[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
Confidence=95% or 0.95
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that true proportion differs from 0.62.:
Null hypothesis:[tex]p=0.62[/tex]
Alternative hypothesis:[tex]p \neq 0.62[/tex]
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.57 -0.62}{\sqrt{\frac{0.62(1-0.62)}{500}}}=-2.303[/tex]
Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a bilateral test the p value would be:
[tex]p_v =2*P(z<-2.303)=0.0213[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of student athletes who graduate within 6 years is significantly different from 0.62 or 62%
2.06. In a study to estimate the proportion of residents in a certain city and its suburbs who favor the construction of a nuclear power plant, it is found that 74 of 100 urban residents favor the construction while only 70 of 125 suburban residents are in favor. Is there a significant difference between the proportions of urban and suburban residents who favor the construction of the nuclear plant at 5% significance level
Answer:
[tex]z=\frac{0.74-0.56}{\sqrt{0.64(1-0.64)(\frac{1}{100}+\frac{1}{125})}}=2.795[/tex]
[tex]p_v =2*P(Z>2.795)= 0.005[/tex]
So if we compare the p value and using any significance level for example [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the proportions are different at 5% of significance.
Step-by-step explanation:
Data given and notation
[tex]X_{1}=74[/tex] represent the number of residents in a certain city and its suburbs who favor the construction of a nuclear power plant
[tex]X_{2}=70[/tex] represent the number of people suburban residents are in favor
[tex]n_{1}=100[/tex] sample 1 selected
[tex]n_{2}=125[/tex] sample 2 selected
[tex]p_{1}=\frac{74}{100}=0.74[/tex] represent the proportion of residents in a certain city and its suburbs who favor the construction of a nuclear power plant
[tex]p_{2}=\frac{70}{125}=0.56[/tex] represent the proportion of suburban residents are in favor
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the value for the test (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to check if the proportions are different, the system of hypothesis would be:
Null hypothesis:[tex]p_{1} = p_{2}[/tex]
Alternative hypothesis:[tex]p_{1} \neq p_{2}[/tex]
We need to apply a z test to compare proportions, and the statistic is given by:
[tex]z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}[/tex] (1)
Where [tex]\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{74+70}{100+125}=0.64[/tex]
Calculate the statistic
Replacing in formula (1) the values obtained we got this:
[tex]z=\frac{0.74-0.56}{\sqrt{0.64(1-0.64)(\frac{1}{100}+\frac{1}{125})}}=2.795[/tex]
Statistical decision
The significance level provided is [tex]\alpha=0.05[/tex] ,and we can calculate the p value for this test.
Since is a two tailed test the p value would be:
[tex]p_v =2*P(Z>2.795)= 0.005[/tex]
So if we compare the p value and using any significance level for example [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the proportions are different at 5% of significance.
Which functions have a maximum and are transformed to the left and down of the parent function, f(x) = x2? Check all that apply. p(x) = 14(x + 7)2 + 1
q(x) = –5(x + 10)2 – 1
s(x) = –(x – 1)2 + 0.5
g(x) = 2x2 + 10x – 35
t(x) = –2x2 – 4x – 3
Answer:q(x)=-5(x+10)2-1
t(x)=-2x2-4x-3
Step-by-step explanation:
The functions that have a maximum and are transformed to the left and down of the parent function, f(x) = x2 include:
q(x) = –5(x + 10)2 – 1t(x) = –2x2 – 4x – 3What is a function?It should be noted that a function simply means a rule the shows the relationship between the variables. The variables are the dependent and the independent variables.
In order to determine whether the function will have a minimum or a maximum depending on the coefficient of the x² term. When the x² coefficient is positive, the function has a minimum and when it is negative, the function has a maximum.
In this case, the above functions have a maximum and are transformed to the left and down of the parent function, f(x) = x2.
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The graph below show Layla’s Pay as a hairdresser. How much does Layla charge per haircut?
A. $5
B. $20
C. $1
D. $10
Answer:
i assume 20 dollars per haircut
Step-by-step explanation:
i do not know what the bottom numbers stand for because there is no label
Answer:
The Answer is B. $20 per haircut.
Step-by-step explanation:
work out 5 power 8 x 5 to power -2 divide by 5 to power 4
Answer: 25
Step-by-step explanation:
A hang glider is soaring over a 100-acre area that consists of thick forest and open fields. In the diagram below, the forested area is shaded in green and the open field is the white space. Upon landing, the hang glider realizes she has dropped her keys.
Answer:
the answer will be 0.8
Step-by-step explanation:
hard to explain
Answer:
C. 0.8
Step-by-step explanation:
When we slice a three-dimensional object, we expose new faces that are two dimensional. The two-dimensional face is called ______________.
Answer:
Cross section
Step-by-step explanation:
Cross section refers to the new two dimensional face exposed when we slice through a three dimensional objects.
It can also be the surface or shape exposed by making a straight cut through something, especially at right angles to an axis.
Cross section is the plane surface(two-dimensional objects) formed by cutting across a solid shape (three-dimensional shape) especially perpendicular to its longest axis.
The distance a car travels can be found using the formula d=r t where d is distance, r is the rate of speed, and t is time. How many miles does the car travel, if it drives at an average speed of 70 miles per hour for 1/2 of an hour?
Answer:
35 miles
Step-by-step explanation:
70miles per hour x1/2 hour =35distance the car travels
A citrus grower anticipates a profit of $100,000 this year if the nightly temperatures remain mild. Unfortunately, the weather forecast indicates a 10% chance that the temperature will drop below freezing during the next week. Such freezing weather will destroy 40% of the crop and reduce the profit to $60,000. However, the grower can protect the citrus fruit against the possible freezing at a cost of $5000. Should the grower spend the $5000 and thereby reduce the profit to $95,000? Hint: Compute E(X), where X is the profit the grower will get if he does nothing to protect the fruit.
Answer:
No, the citrus grower shouldn't spend the $5000 and thereby reduce the profit to $95,000 as the expected profit from doing nothing to protect the citrus plants ($96,000) is more than the profit that'll be available if $5,000 is spent on protection.
Step-by-step explanation:
First of, we compute the probability distribution of X.
X represents the profit if the citrus grower does nothing to protect the citrus fruits.
If the citrus grower does nothing, there are two possibilities as to what will happen.
1) The temperatures can drop below freezing point at a chance of 10% and the profit plummets to $60,000
2) The temperature can remain mild at a chance of 90% (100%-10%) and the profit stays at $100,000.
The probability distribution will then be
X ||||||||||||||| P(X)
60,000 ||| 0.10
100,000 | 0.90
The expected value of any probability distribution is given as
E(X) = Σ xᵢpᵢ
xᵢ = each variable
pᵢ = probability of each variable
E(X) = (60,000×0.10) + (100,000×0.90)
= 6,000 + 90,000 = $96,000
The expected amount of profits from doing nothing to protect the citrus fruits = $96,000
The expected amount of profits expected from spending $5,000 to protect the citrus fruits = $95,000
$96,000 > $95,000
Hence, the citrus grower is better off doing nothing to protect the citrus fruits.
Hope this Helps!!!
Question 1 Show all your work. Indicate clearly the methods you use, because you will be scored on the correctness of your methods as well as on the accuracy and completeness of your results and explanations. A recent survey collected information on television viewing habits from a random sample of 1,000 people in the United States. Of those sampled, 37 percent indicated that their favorite sport to watch on television was American football.
(a) Construct and interpret a 95 percent confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football.
(b) Based on your answer to part (a), is it reasonable to believe that 33 percent is the actual percent of people in the United States whose favorite sport to watch on television is American football? Justify your answer.
Answer:
(a) The 95% confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football is (0.34, 0.40).
(b) Not reasonable.
Step-by-step explanation:
The information provided is:
n = 1000
[tex]\hat p[/tex] = 0.37
(a)
The (1 - α)% confidence interval for the population proportion p is:
[tex]CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
Here,
[tex]\hat p[/tex] = sample proportion
n = sample size
[tex]z_{\alpha/2}[/tex] = critical value of z.
Compute the critical value of z for 95% confidence interval as follows:
[tex]z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96[/tex]
*Use a z-table for the value.
Compute the 95% confidence interval for the population proportion p as follows:
[tex]CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
[tex]=0.37\pm 1.96\times\sqrt{\frac{0.37(1-0.37)}{1000}}[/tex]
[tex]=0.37\pm 0.03\\=(0.34, 0.40)[/tex]
Thus, the 95% confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football is (0.34, 0.40).
(b)
Now we need to determine whether it is reasonable to believe that the actual percent of people in the United States whose favorite sport to watch on television is American football is 33%.
The hypothesis can be defined as:
H₀: The percentage of people in the United States whose favorite sport to watch on television is American football is 33%, i.e. p = 0.33.
Hₐ: The percentage of people in the United States whose favorite sport to watch on television is American football is different from 33%, i.e. p ≠ 0.33
The hypothesis can be tested based on a confidence interval.
The decision rule:
If the (1 - α)% confidence interval includes the null value of the test then the null hypothesis will not be rejected. And if the (1 - α)% confidence interval includes the null value of the test then the null hypothesis will be rejected.
The 95 confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football is (0.34, 0.40).
The confidence interval does includes the null value of p, i.e. 0.33.
So, the null hypothesis will be rejected.
Hence, concluding that is is not reasonable to believe that 33% is the actual percent of people in the United States whose favorite sport to watch on television is American football.
95% confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football is (0.34, 0.40).
Concluding that is is not reasonable to believe that 33% is the actual percent of people in the United States whose favorite sport to watch on television is American football
Given that,
A recent survey collected information on television viewing habits from a random sample of 1,000 people in the United States.
Of those sampled, 37 percent indicated that their favorite sport to watch on television was American football.
We have to determine,
Construct and interpret a 95 percent confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football.
According to the question,
Sample proportion p = 37% = 0.37
Sample space n = 1000
The (1 - α)% confidence interval for the population proportion,[tex]C.I. = P \pm Z_\frac{\alpha}{2} \sqrt{\dfrac{p(1-p)}{n} }[/tex]
To compute the critical value of z for 95% confidence interval as follows:
[tex]z_\frac{ \alpha}{2} = z_\frac{0.05}{2} = 1.96[/tex]
By using a z-table for the value.
Compute the 95% confidence interval for the population proportion p as follows:
[tex]C.I. = p\pm Z_\frac{\alpha}{2} \sqrt{\dfrac{p(1-p)}{n} }\\\\C.I. = 0.37\pm 1.96 \sqrt{\dfrac{0.43(1-0.34)}{1000} }\\\\C.I.= 0.03 \pm 0.09\\\\C.I. = (0.34, \ 0.40)[/tex]
Hence, 95% confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football is (0.34, 0.40).
The hypothesis can be defined as:H₀: The percentage of people in the United States whose favorite sport to watch on television is American football is 33%, i.e. p = 0.33.
Hₐ: The percentage of people in the United States whose favorite sport to watch on television is American football is different from 33%, i.e. p ≠ 0.33
The hypothesis can be tested based on a confidence interval.
The (1 - α)% confidence interval includes the null value of the test then the null hypothesis will not be rejected.
And if the (1 - α)% confidence interval includes the null value of the test then the null hypothesis will be rejected.
The 95 confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football is (0.34, 0.40).
The confidence interval does includes the null value of p, i.e. 0.33.
So, the null hypothesis will be rejected.
Hence, Concluding that is is not reasonable to believe that 33% is the actual percent of people in the United States whose favorite sport to watch on television is American football
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Find the 107th term of the sequence -9, -5, -1, 3, 7
Answer:
4×107 = 428
428-(5×4) =
428-20= 408
In ΔSTU, the measure of ∠U=90°, TS = 73, SU = 55, and UT = 48. What ratio represents the sine of ∠S?
sin S = [tex]$\frac{48}{73}[/tex] is the ratio found.
Step-by-step explanation:
It is given that m∠U = 90°
TS is the hypotenuse = 73 units
UT is the adjacent side of the right angle = 48 units
SU is the base of the triangle = 55 units
Now we have to find the ratio as,
sin S = [tex]$\frac{opp}{hyp}[/tex]
sin S = [tex]$\frac{UT}{TS}[/tex]
Plugin the values, we will get,
sin S = [tex]$\frac{48}{73}[/tex]
So the ratio was found.
Answer:
73/48
Step-by-step explanation:
he put it backwards
if the perimeter of a square is 236 inches, how long is each side?
Answer:
59 inches
Step-by-step explanation:
a square has 4 equal sides
if the sum of all sides(perimeter) is 236 you can do the inverse of that and divide by 4
so your answer should be 59
Answer:
59 inches
Step-by-step explanation:
All of the 4 sides in a square are equal, and the perimeter is all the sides added together.
s+s+s+s=236
4s=236
Divide both sides by 4
s=59
Each side is 59 inches
Find each value and measure. Assume that segments that appear to be tangent are tangent.
Given:
Given that AC and BD are chords of the circle.
The two chords intersect at the point E which makes an angle 93°
The measure of arc BC is 161°
We need to determine the measure of arc AD.
Measure of arc AD:
The measure of arc AD can be determined using the property that "if two chords intersect in the interior of the circle, then the measure of each angle is half the sum of the arcs intercepted by the angles and its vertical angle".
Thus, applying the above theorem, we have;
[tex]m \angle E=\frac{1}{2}(m \widehat{BC}+m \widehat{AD})[/tex]
Substituting the values, we have;
[tex]93^{\circ}=\frac{1}{2}(161^{\circ}+m \widehat{AD})[/tex]
[tex]186^{\circ}=161^{\circ}+m \widehat{AD}[/tex]
[tex]25^{\circ}=m \widehat{AD}[/tex]
Thus, the measure of arc AD is 25°
Arc angle AD is 25 degrees
What are secant:Secant are lines that intersect a circle at two points.
Secant AC intersect secant BD at angle 93 degree.
Using secant rule , in circle theorem,
93° = 1 /2 (arcBC + arcAD)Therefore,
93° = 1 / 2(AD + 161)
93 = AD / 2 + 161 / 2
93 = AD + 161/ 2
cross multiply
186 = AD + 161
AD = 186 - 161
arc AD = 25°
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Bob is the owner of a home improvement store. He has hired you to check his machine’s calibration prior to starting production on a large order. To check this, you set the machine to create 1.5 inch bolts and manufacture a random sample of 200 bolts. That sample of bolts has an average length of 1.521 inches with a standard deviation of 0.204 inches. Does this sample provide convincing evidence that the machine is working properly or should it be shut down for repairs?
Parameter:
Null hypothesis: μ = 1.5 (the machines work as needed)
Alternative hypothesis: μ ≠ 1.5 (The machines don't work properly)
Since we don't know the population deviation, we will apply a t-test to compare the actual mean to the reference value
Conditions:
Simple random sample: The problem states the sample was chosen at random.
Independence: You can assume there are more than 10(200) = 2000 screws.
Normality: (200 ≥ 30) the sample is large enough for sampling distribution to assume Normality
Calculations:
Since the conditions are met we will carry out a T-test using a calculator for μ≠μ0
μ = population mean = 1.5
σ= standard dev = 0.204
xbar = sampe mean = 1.521
n = sample size= 200
After adding all of this data into the calculator in the T-test program we get a p-value of 0.147
Conclusion:
We will assume a 0.05 sig level for our conclusion.
Since 0.147 > 0.05 we will fail to reject the null hypothesis meaning that we have enough evidence to show that the machines work as needed.
Given the average length and standard deviation of the manufactured bolts, the machine might require recalibration since the lengths produced are slightly larger than desired, and there's a wide spread in lengths. Application of the empirical rule can provide further insight about the need for machine repair. A larger sample size might give a more accurate assessment.
Explanation:Given that the machine is set to manufacture bolts of 1.5 inches and a random sample of 200 bolts showed an average length of 1.521 inches with a standard deviation of 0.204 inches, it seems the machine may not be calibrated correctly. The average length is slightly larger than the desired length, a factor that may be important depending on the tolerances required for these bolts. The standard deviation is also relatively high, implying that there is a wide spread in the lengths of bolts being produced.
One way to determine if the machine needs to be repaired or not is to apply the empirical rule, also known as the 68-95-99.7 rule, which says approximately 68% of data falls within one standard deviation from the mean, 95% falls within two standard deviations, and 99.7% falls within three standard deviations in a normal distribution. In this case, it means 95% of the bolts should fall between 1.113 inches (1.521 - 2*0.204) and 1.929 inches (1.521 + 2*0.204). If these lengths are acceptable for the operation, the machine can continue working. If not, it might need to be shut down for repair.
Also, it's also worth noting that a larger sample size could provide a more accurate assessment of whether the machine is working correctly or not. While a sample size of 200 is decent, a larger sample size would reduce the margin of error from sample to population.
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