Answer:
The standard error of the estimate is 413.4498
Step-by-step explanation:
According to the given data of ,in order to calculate the standard error of the estimate we would have to use the following formula:
Standard Error =sqrt(SSE/(n-p-1))
Standard Error =sqrt(4102577/(27-2-1))
Standard Error = 413.4498
The standard error of the estimate is 413.4498
A rhombus has a base of 8 centimeters and a height of 4.2 centimeters. What is its area? Do not round your answer.
A = cm2
Answer:
[tex]33.6\ cm^{2}[/tex]
Step-by-step explanation:
Given that
The Base of a rhombus = 8 centimetres
Height of a rhombus = 4.2 centimetres
Based on the above information, the area of a rhombus is
[tex]Area\ of\ rhombus = Base \times height[/tex]
[tex]= 8 \times 4.2[/tex]
[tex]= 33.6\ cm^{2}[/tex]
By multiplying the base of a rhombus with the height of a rhombus we can get the area of the rhombus and the same is applied & shown in the computation part i.e to be shown above
Determine which of (a)-(d) form a solution to the given system for any choice of the free parameter. (HINT: All parameters of a solution must cancel completely when substituted into each equation.) 3x1 + 8x2 − 14x3 = 9 x1 + 3x2 − 4x3 = 1
Answer:
Please see attachment
Question:
The options are;
(a) (9 - 2s1, 3 + 3s1, s1)
solution
not a solution
(b) (-4 - 5s1, s1, -(3 + s1)/2)
solution
not a solution
(c) (11 + 10s1, -3 - 2s1, s1)
solution
not a solution
(d) ((6 - 4s1)/3, s1, -(7 - s1)/4)
solution
not a solution
Answer:
The options that form a solution of the given system are;
(b) and (c)
Step-by-step explanation:
Here we have
3·x₁ + 8·x₂ − 14·x₃ = 9
x₁ + 3·x₂ − 4·x₃ = 1
(a) (9 - 2·s₁, 3 + 3·s₁,s₁)
3·(9 - 2·s₁) + 8·(3 + 3·s₁) − 14·s₁ = 4s₁+51
Not a solution
(b) (-4 - 5s₁, s₁, -(3 + s1)/2)
3·(-4 - 5s₁) + 8·(s₁) − 14·-(3 + s1)/2 = 9
(-4 - 5s₁) + 3·(s₁) − 4·-(3 + s1)/2 = 2
Solution
(c) (11 + 10s₁, -3 - 2s₁, s₁ )
3·(11 + 10s₁) + 8·(-3 - 2s₁) − 14·s₁ = 9
(11 + 10s₁) + 3·(-3 - 2s₁) − 4·s₁ = 2
Solution
(d) ((6 - 4s1)/3, s1, -(7 - s1)/4)
3·(6 - 4s1)/3+ 8·s1− 14·-(7 - s1)/4 = 0.5s₁ +30.5
Not a solution
Your flight has been delayed: At Denver International Airport, 85% of recent flights have arrived on time. A sample of 14 flights is studied. Round the probabilities to at least four decimal places.(a) Find the probability that all 12 of the flights were on time.(b) Find the probability that exactly 10 of the flights were on time.(c) Find the probability that 10 or more of the flights were on time.(d) Would it be unusual for 11 or more of the flights to be on time?
This problem can be approached as a binomial distribution. The probability of a particular number of flights on time is calculated using the binomial probability formula. Determining 'unusual' can be subjective but normally a probability less than 0.05 is considered unusual.
Explanation:This problem is a binomial probability problem because we have a binary circumstance (flight is either on time or it isn't) and a fixed number of trials (14 flights). The binomial probability formula is P(X=k) = C(n, k) * (p^k) * ((1 - p)^(n - k)) where n is the number of trials, k is the number of successful trials, p is the probability of success on a single trial, and C(n, k) represents the number of combinations of n items taken k at a time.
(a) For all 12 flights on time, it seems there's a typo; there are 14 flights in the sample. We can't calculate for 12 out of 14 flights without the rest of the information.
(b) For exactly 10 flights, we use n=14, k=10, p=0.85: P(X=10) = C(14, 10) * (0.85^10) * ((1 - 0.85)^(14 - 10)).
(c) For 10 or more flights on time, it's the sum of the probabilities for 10, 11, 12, 13, and 14 flights on time.
(d) For determining whether 11 or more on-time flights is unusual, it depends on the specific context, but we could consider it unusual if the probability is less than 0.05.
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Analyzing a sample of 14 flights at Denver International Airport, the probability of 10 or more flights arriving on time is 0.3783, and the probability of 11 or more flights arriving on time is 0.2142, which is not considered unusual.
(a) All 12 of the flights were on time.
(b) Exactly 10 of the flights were on time.
(c) 10 or more of the flights were on time.
(d) Would it be unusual for 11 or more of the flights to be on time?
We can use the binomial probability formula to solve this problem. The binomial probability formula is:
P(k successes in n trials) = (n choose k) *[tex](p)^k[/tex] * [tex](q)^(^n^-^k^)[/tex]
where:
n is the number of trials
k is the number of successes
p is the probability of success
q is the probability of failure
In this case, n = 14, p = 0.85, and q = 0.15.
(a) To find the probability that all 12 of the flights were on time, we can plug k = 12 into the binomial probability formula:
P(12 successes in 14 trials) = (14 choose 12) * [tex](0.85)^1^2[/tex]*[tex](0.15)^2[/tex]
Using a calculator, we can find that this probability is approximately 0.0032.
(b) To find the probability that exactly 10 of the flights were on time, we can plug k = 10 into the binomial probability formula:
P(10 successes in 14 trials) = (14 choose 10) *[tex](0.85)^1^0[/tex] *[tex](0.15)^4[/tex]
Using a calculator, we can find that this probability is approximately 0.1022.
(c) To find the probability that 10 or more of the flights were on time, we can add up the probabilities of 10, 11, 12, 13, and 14 successes:
P(10 or more successes) = P(10 successes) + P(11 successes) + P(12 successes) + P(13 successes) + P(14 successes)
Using a calculator, we can find that this probability is approximately 0.3783.
(d) To determine whether it would be unusual for 11 or more of the flights to be on time, we can find the probability of this event and compare it to a common threshold for unusualness, such as 0.05.
P(11 or more successes) = P(11 successes) + P(12 successes) + P(13 successes) + P(14 successes)
Using a calculator, we can find that this probability is approximately 0.2142. This probability is greater than 0.05, so it would not be considered unusual for 11 or more of the flights to be on time.
Researchers are studying rates of homeowners in a certain town. They believe that the proportion of people ages 36-50 who own homes is signifificantly greater than the proportin of people age 21-35 who own homes and want to test this claim. The results of the surverys are: Homeowners Renters Total
Ages 21-35 18 38 56
Ages36-50 40 22 62
TOTAL 58 60 118
What are the null hypothesis and alternative hypothesis for this situation
Answer:
Being p1 the proportion for people of ages 36-50 and p2 the proportion for people of ages 21-35, the null and alternative hypothesis will be:
[tex]H_0: p_1-p_2=0\\\\H_a: p_1-p_2>0[/tex]
Step-by-step explanation:
A hypothesis test on the difference of proportions needs to be performed for this case.
We have two sample proportions and we want to test if the true population proportions differ from each other, usign the information given by the sample statistics.
The claim is that the proportion of people of ages 36-50 who own homes is significantly greater than the proportin of people age 21-35 who own homes.
The term "higher" will define the alternative hypothesis, that is the hypothesis that represents what is claimed. The null hypothesis always include the equal sign, and will state that both proportions do not differ.
Being p1 the proportion for people of ages 36-50 and p2 the proportion for people of ages 21-35, the null and alternative hypothesis will be:
[tex]H_0: p_1-p_2=0\\\\H_a: p_1-p_2>0[/tex]
The null hypothesis (H0) is that the proportion of homeowners ages 36-50 is equal to the proportion of homeowners ages 21-35 (H0: P1 = P2), and the alternative hypothesis (Ha) is that the proportion of homeowners ages 36-50 is greater than that of ages 21-35 (Ha: P1 > P2).
Explanation:To answer the question, the null hypothesis (H0) and the alternative hypothesis (Ha) must be formulated based on the given data about homeownership across different age groups. In this research, the null hypothesis would state that the proportion of homeowners who are ages 36-50 is equal to the proportion of homeowners who are age 21-35. Mathematically, this can be represented as H0: P1 = P2.
The alternative hypothesis is what the researchers are trying to support, which is that the proportion of homeowners who are ages 36-50 is significantly greater than the proportion of homeowners who are age 21-35, represented as Ha: P1 > P2.
It is important to note that a hypothesis test will be used to determine if there is enough statistical evidence to reject the null hypothesis in favor of the alternative hypothesis.
Factor the polynomial completely using the
X method
2 + 16x + 48
Which equivalent four-term polynomial can be created
using the X method?
O2 + 8x + 8x + 48
ox2 - 12x - 4x + 48
x2 + 12x + 4x + 48
ac
0x28x 8x +
() Intro
Answer:
its answer C
Step-by-step explanation:
x² - 12x + 4x + 48
Which term could have the greatest common factor of 5m squared n squared
Answer: I think it’s B and D on eduinuity 2021
Step-by-step explanation:
The required 5m²2n² is the greatest common factor of (b) 5m⁴n₃ and (d) 15m²n².
What is the greatest common factor?The greatest common factor (GCF) is the largest positive integer that divides two or more numbers without leaving a remainder. In other words, it is the largest number that is a factor of all the given numbers.
For example, the GCF of 12 and 18 is 6, because 6 is the largest positive integer that divides both 12 and 18 without leaving a remainder.
here,
The terms that could have the greatest common factor of 5m²n² are:
5m⁴n₃, since 5m²n² is a factor of both 5m⁴n³ and 5m²n².
15m²n², since 5m²n² is a factor of both 15m²n² and 5m²2n².
Therefore, the correct options are (b) 5m⁴n₃ and (d) 15m²n².
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The complete question is given in the attachment below,
Write the equation of a parabola whose graph
separates the blue points from the red points.
The equation of a parabola that separates the blue points from the red points can be written in the form y = ax² + bx + c, where a, b, and c are constants. The specific coefficients will determine the direction, width, and position of the parabola.
Explanation:To create an equation for a parabola that separates the blue and red points, we need more information about the specific characteristics desired for the parabola. The general form of the equation y = ax² + bx + c allows us to customize the parabola's shape. The coefficient a determines the direction of the parabola (opening upwards or downwards), while b and c influence its horizontal shift and vertical position.
For example, if we want a parabola that opens upwards with its vertex at the origin (0,0) and separates the points above from those below, the equation could be y = ax², where a is a positive constant. If a horizontal shift or translation is needed, adjustments to b and c can be made accordingly.
In summary, the equation of the parabola depends on the specific requirements for separating the blue and red points, and the general form y = ax² + bx + c provides the flexibility to tailor the parabola's characteristics to meet those needs.
three students are chosen at random find the probability that all three were born on Wednesday
Final answer:
The probability that all three students were born on Wednesday is 1/343.
Explanation:
To find the probability that all three students were born on Wednesday, we need to consider the total number of possible outcomes and the number of favorable outcomes.
There are 7 days in a week, so each student has a 1/7 chance of being born on Wednesday. Since the students are chosen at random and the choices are independent, we can multiply the probabilities together to find the probability that all three students were born on Wednesday:
P(all three born on Wednesday) = (1/7) * (1/7) * (1/7) = 1/343.
F (x) = -3x+5x^2+8 has blank roots.
Answer:
F (x) = -3x+5x^2+8 has complex roots.
Step-by-step explanation:
number of roots?
ok.
f(x) = 5xx - 3x + 8 = 5xx + 5x - 8x + 8 ( I was guessing what numbers would sum to -3)
nope.
ok try discriminant: (-3)^2 - 4*5 * 8 = 9 - 160 < 0
2 complex roots
find the area of a rhombus with a perimeter equal to 40 and a diagonal equal to 14 cm.
Answer:
about 100 cm²
Step-by-step explanation:
The side length of the rhombus is 1/4 of the perimeter so is 10 cm. The length of half of the other diagonal will be the length of the leg of a right triangle with hypotenuse 10 and leg 7 (half the given diagonal).
d= √(10² -7²) = √51
Then the area of the rhombus is the product of this and the given diagonal:
A = (14 cm)(√51 cm) ≈ 99.98 cm²
The area of the rhombus is about 100 cm².
janelle wishes to finance a car for $33,000. the bank's annual interest rate is 3.5%, and she can choose between durations of five or six years. calculate the monthly payment and total amount paid for both duration options.use the formula,p = ar (1+r)^n/(1+r)^n-1where a is the amount to finance, r is the monthly interest rate, and n is the number of months to pay. show all of your steps.
Answer:
For a duration of 5 years, Monthly Payment =$600.42
For a duration of 6 years, Monthly Payment =$508.83
Step-by-step explanation:
[tex]P=\dfrac{ar (1+r)^n}{(1+r)^n-1} \\[/tex]
where a= Amount to Finance=$33,000
Annual interest rate = 3.5%=0.035
r=Monthly Interest Rate= 0.035 ÷ 12 =[tex]\frac{7}{2400}[/tex]
n=number of months to pay
For a duration of 5 years
n=5X12=60 months
[tex]P=\dfrac{ar (1+r)^n}{(1+r)^n-1} \\\\P=\dfrac{33000 X\frac{7}{2400} (1+\frac{7}{2400} )^{60}}{(1+\frac{7}{2400})^{60}-1} \\=\dfrac{96.25 (1.1909)}{1.1909-1}\\=\dfrac{96.25 (1.1909)}{0.1909}\\=\dfrac{114.62}{0.1909}=\$600.42[/tex]
For a duration of 6 years
n=6X12=72 months
[tex]P=\dfrac{ar (1+r)^n}{(1+r)^n-1} \\\\P=\dfrac{33000 X\frac{7}{2400} (1+\frac{7}{2400} )^{72}}{(1+\frac{7}{2400})^{72}-1} \\=\dfrac{96.25 (1.2333)}{1.2333-1}\\=\dfrac{96.25 (1.2333)}{0.2333}\\=\dfrac{118.71}{0.2333}=\$508.83[/tex]
what is the measure of angle C
Answer:
pls I can't see any diagram anywhere
how long will it take to earn one dollar at the rate of $10 per hour
Answer:
6 minutes
Step-by-step explanation:
Since it's you have 10 dollars in one hour, and you want to find 1/10th of the dollar value, you'd multiply 1/10th, or 0.1 by the time as well. I converted it into minutes because we're finding a smaller unit of time than 1 hour.
60 min x 1/10 (or 60 min/10)= 6 minutes
I hope this helped!
6 minutes
If it takes 1 hour to earn $10, we convert that 1 hour to 60 minutes. Now we will divide 60 by 10, and that gives you 6. Therefore every 6 minutes you earn $1.
Hope this helps!
A sample of size n = 100 produced the sample mean of ܺത= 16. Assuming the population standard deviation σ= 3, compute a 95% confidence interval for the population mean μ. (b) Assuming the population standard deviation σ= 3, how large should a sample be to estimate the population mean μ with a margin of error not exceeding 0.5 with a 95% confidence interval?
The 95% confidence interval for the population mean based on a sample of 100 with mean 16 and population standard deviation 3 is (15.412, 16.588). For a margin of error not exceeding 0.5, a sample size of 139 is needed.
Explanation:To compute the 95% confidence interval for the population mean μ, we use the formula for a confidence interval which is ȳ ± Z*(σ/√n), where ȳ is the sample mean, σ is the population standard deviation, n is the sample size, and Z is the z-score corresponding to the desired confidence level (for a 95% confidence interval, Z = 1.96).
Plugging the given values into the formula, we get 16 ± 1.96*(3/√100), which simplifies to 16 ± 0.588. Thus, the 95% confidence interval for the population mean μ is (15.412, 16.588).
For the second part of the question, the formula used to find the sample size needed for a certain margin of error (E) at a certain confidence level is n = (Zσ/E)^2. Substituting the given values into this formula, we get n = (1.96*3/0.5)^2 which is equal to 138.384. Since we can't have a fraction of a sample, we round this up to the nearest whole number, so we would need a sample size of 139 to estimate the population mean μ with a margin of error not exceeding 0.5 with a 95% confidence interval.
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To compute a 95% confidence interval for the population mean μ, we can use the formula: Confidence Interval = Sample Mean ± (Z * σ/√n). The 95% confidence interval for the population mean μ is (15.412, 16.588). To estimate the sample size needed to keep the margin of error within 0.5 with a 95% confidence level, we can use the formula: n = (Z^2 * σ^2) / (E^2). We need a sample size of at least 24 to estimate the population mean μ with a margin of error not exceeding 0.5, with a 95% confidence level.
Explanation:To compute a 95% confidence interval for the population mean μ, we can use the formula:
Confidence Interval = Sample Mean ± (Z * σ/√n)
Where Z is the z-score corresponding to the desired confidence level, σ is the population standard deviation, and n is the sample size.
For this problem:
Confidence Interval = 16 ± (1.96 * 3/√100)
Simplifying, we get:
Confidence Interval = 16 ± 0.588
Therefore, the 95% confidence interval for the population mean μ is (15.412, 16.588).
To estimate the sample size needed to keep the margin of error within 0.5 with a 95% confidence level, we can use the formula:
n = (Z^2 * σ^2) / (E^2)
Where Z is the z-score corresponding to the desired confidence level, σ is the population standard deviation, and E is the maximum acceptable margin of error.
For this problem:
n = (1.96^2 * 3^2) / (0.5^2)
Simplifying, we get:
n = 23.532
Therefore, we need a sample size of at least 24 to estimate the population mean μ with a margin of error not exceeding 0.5, with a 95% confidence level.
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Suppose small aircraft arrive at a certain airport according to a Poisson process with rate α = 8 per hour, so that the number of arrivals during a time period of t hours is a Poisson rv with parameter μ = 8t. (Round your answers to three decimal places.) (a) What is the probability that exactly 9 small aircraft arrive during a 1-hour period?
Answer:
0.124 = 12.4% probability that exactly 9 small aircraft arrive during a 1-hour period.
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given time interval.
Rate of 8 per hour
This means that [tex]\mu = 8[/tex]
(a) What is the probability that exactly 9 small aircraft arrive during a 1-hour period?
This is P(X = 9).
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 9) = \frac{e^{-8}*8^{9}}{(9)!} = 0.124[/tex]
0.124 = 12.4% probability that exactly 9 small aircraft arrive during a 1-hour period.
The number of students in an elementary school t years after 2002 is given by s(t) = 100 ln(t + 5) students. The yearly cost to educate one student t years after 2002 can be modeled as c(t) = 1500(1.05t) dollars per student. (a) What are the input units of the function f(t) = s(t) · c(t)?
The input units of the function f(t) are students multiplied by dollars per student.
The function[tex]\( f(t) = s(t) \cdot c(t) \)[/tex] represents the total cost to educate all the students in the elementary school t years after 2002. To determine the input units of this function, we need to analyze the units of its individual components.
s(t) is the number of students at time t. Its unit is "students."
c(t) is the cost to educate one student at time t. Its unit is "dollars per student."
When you multiply s(t) by c(t), the units combine as follows:
[tex]\[ f(t) = s(t) \cdot c(t) = \text{"students"} \times \left(\frac{\text{"dollars"}}{\text{"student"}}\right) = \text{"students"} \cdot \text{"dollars per student"} \][/tex]
Therefore, the input units of the function f(t) are students multiplied by dollars per student.
Ben ran for president of his class is there a 410 students and he receive 72% of the vote how many students voted for Ben
Answer:
295
Step-by-step explanation:
multiply 410 x .72=295.2
you can not have .2 of a person, so you must round. Normally the question states what to round to, whether up or down, but generally .2 rounds down so: 295
Final answer:
295 students voted for Ben.
Explanation:
To calculate how many students voted for Ben in the class president election, we need to use the percentage of votes he received. Ben received 72% of the total votes from a class of 410 students.
First, convert the percentage to a decimal by dividing by 100:
72% = 72 ÷ 100 = 0.72
Then, multiply this decimal by the total number of students to find out how many voted for Ben:
Number of votes for Ben = 0.72 × 410
Now, we calculate the multiplication:
Number of votes for Ben = 295.2
Since we can't have a fraction of a vote, we'll round down to the nearest whole number. Thus, 295 students voted for Ben.
The weight of potato chip bags filled by a machine at a packaging plant is normally distributed with a mean of 15.0 ounces and a standard deviation of 0.1 ounces. What percentage of bags weigh more than 14.8 ounces
Answer:
97.73% of bags weigh more than 14.8 ounces.
Step-by-step explanation:
We are given that the weight of potato chip bags filled by a machine at a packaging plant is normally distributed with a mean of 15.0 ounces and a standard deviation of 0.1 ounces.
Let X = weight of potato chip bags filled by a machine
So, X ~ N([tex]\mu=15.0,\sigma^{2} =0.1^{2}[/tex])
The z-score probability distribution for normal distribution is given by;
Z = [tex]\frac{ X -\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean weight = 15.0 ounces
[tex]\sigma[/tex] = standard deviation = 0.1 ounces
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.
So, percentage of bags that weigh more than 14.8 ounces is given by = P(X > 14.8 ounces)
P(X > 14.8 ounces) = P( [tex]\frac{ X -\mu}{\sigma}[/tex] > [tex]\frac{14.8-15.0}{0.1}[/tex] ) = P(Z > -2) = P(Z < 2)
= 0.9773 or 97.73%
Now, in the z table the P(Z [tex]\leq[/tex] x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2 in the z table which has an area of 0.9773.
Hence, 97.73% of bags weigh more than 14.8 ounces.
To find the percentage of bags weighing more than 14.8 ounces, we calculate the z-score for 14.8 ounces and use the normal distribution properties. A z-score of -2 corresponds to 2.28% of bags weighing less, so about 97.72% weigh more.
Explanation:To determine what percentage of potato chip bags weigh more than 14.8 ounces when the mean weight is 15.0 ounces with a standard deviation of 0.1 ounces, we use the properties of the normal distribution. First, we calculate the z-score for 14.8 ounces:
Z = (X - μ) / σ = (14.8 - 15.0) / 0.1 = -2
The z-score tells us how many standard deviations away 14.8 ounces is from the mean. A z-score of -2 indicates that 14.8 ounces is 2 standard deviations below the mean. Using the z-score table or a calculator with normal distribution functions, we find the area to the left of z = -2. This area represents the percentage of bags weighing less than 14.8 ounces. Since we're interested in bags weighing more than 14.8 ounces, we subtract this value from 1 (or 100% if we're working with percentages).
For z = -2, the area to the left is approximately 0.0228 (or 2.28%). Therefore, the percentage of bags that weigh more than 14.8 ounces is about 100% - 2.28% = 97.72%.
Juan wants to know the cross-sectional area of a circular pipe. He measures the diameter which he finds, to the nearest millimeter, to be 5 centimeters.
To find the area of the circle, Juan uses the formula where A is the area of the circle and r is its radius. He uses 3.14 for π. What value does Juan get for the area of the circle? Make sure you include your units.
Answer:
Step-by-step explanation:
Hi there,
To get started, recall the area of a bound circle formula:
[tex]A = \pi r^{2}[/tex] where r is radius of the circle. However, Juan used an approximate value of π, 3.14. So for our purposes, the formula becomes:
[tex]A=[/tex] [tex](3.14)r^{2}[/tex]
Juan measured the circle's diameter, so we can find radius from diameter first. Radius is simply twice the length of the diameter; from one circle endpoint to the center, to the endpoint across, making a straight line:
[tex]d=2r[/tex] ⇒ [tex]r=\frac{d}{2} = \frac{5 \ cm}{2} = 2.5 \ cm[/tex]
Now plug in to obtain area:
[tex]A = (3.14)(2.5 cm)^{2} =19.625 \ cm^{2}[/tex]
The area is 19.265 centimetres squared.
Cross-sections are the area shapes when you cut through a 3D volume; if you cut through a pipe perpendicular to where it flows, you can see it is a circle! If you cut straight through a cube, it would be a square, etc.
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thanks,
Plot the function y(x)=e–0.5x sin(2x) for 100 values of x between 0 and 10. Use a 2- point-wide solid blue line for this function. Then plot the function y(x)=e–0.5x cos(2x) on the same axes. Use a 3-point-wide dashed red line for this function. Be sure to include a legend, title, axis labels, and grid on the plots
In the plot, I have graphed two functions on the same set of axes.
The first function, y(x) = e^(-0.5x) * sin(2x), is represented by a solid blue line with a 2-point line width.
The second function, y(x) = e^(-0.5x) * cos(2x), is shown with a dashed red line with a 3-point line width. Both functions are evaluated for 100 values of x ranging from 0 to 10.
The solid blue line represents the sine function, and the dashed red line represents the cosine function.
The legend, title, axis labels, and grid have been included to make the plot more informative and visually appealing.
In this plot, you can observe the oscillatory behavior of both functions as they decay exponentially with decreasing x.
The legend distinguishes between the two functions, and the grid helps in reading the values accurately.
The choice of line widths and colors enhances the visibility of the two functions, making it easier to compare their behavior over the specified range of x values.
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A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 433 gram setting. It is believed that the machine is underfilling the bags. A 15 bag sample had a mean of 423 grams with a standard deviation of 26. A level of significance of 0.1 will be used. Assume the population distribution is approximately normal. Determine the decision rule for rejecting the null hypothesis.
Answer:
[tex]t=\frac{423-433}{\frac{2.6}{\sqrt{15}}}=-14.896[/tex]
[tex]df=n-1=15-1=14[/tex]
We need to find in the t distribution with df=14 a value who accumulates 0.1 of the area in the left and we got [tex]t_{crit}= -1.345[/tex].
Since our calculated value for the statistic is is so much lower than the critical value we have enough evidence to reject the null hypothesis, and we can conclude that the true mean for this case is significantly less than 433 and then the machine is underfilling.
Step-by-step explanation:
Data given
[tex]\bar X=423[/tex] represent the sample mean
[tex]s=26[/tex] represent the sample standard deviation
[tex]n=15[/tex] sample size
[tex]\mu_o =433[/tex] represent the value that we want to test
[tex]\alpha=0.1[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test
System of hypothesis
We need to conduct a hypothesis in order to check if the true mean is less than 433 (underfilling), the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 433[/tex]
Alternative hypothesis:[tex]\mu < 433[/tex]
The statistic is:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Calculate the statistic
[tex]t=\frac{423-433}{\frac{2.6}{\sqrt{15}}}=-14.896[/tex]
Decision rule
The degrees of freedom are:
[tex]df=n-1=15-1=14[/tex]
We need to find in the t distribution with df=14 a value who accumulates 0.1 of the area in the left and we got [tex]t_{crit}= -1.345[/tex]
Since our calculated value for the statistic is is so much lower than the critical value we have enough evidence to reject the null hypothesis, and we can conclude that the true mean for this case is significantly less than 433 and then the machine is underfilling.
How many games did Lisa score less than 13 points?
(Group of answer choices)
A) 15
B) 7
C) 5
D) 12
Answer:
it's B) 7
Step-by-step explanation:
Answer:
The answer is B
Step-by-step explanation:
So if Lisa scored less than 13 points then the answer has to lie in the intervals 1-12. You had 2 from intervals 1-6 and 5 from intervals 7-12 and you get 7
Of the 219 white GSS2008 respondents in their 20s, 63 of them claim the ability to speak a language other than English. With 99% confidence, what is the upper limit of the population proportion based on these statistics
Answer:
The upper limit for population proportion is 0.3666
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 219
Number of people who have ability to speak a language other than English, x = 63
[tex]\hat{p} = \dfrac{x}{n} = \dfrac{63}{219} = 0.2877[/tex]
99% Confidence interval:
[tex]\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
[tex]z_{critical}\text{ at}~\alpha_{0.05} = 2.58[/tex]
Putting the values, we get:
[tex] 0.2877\pm 2.58(\sqrt{\dfrac{ 0.2877(1- 0.2877)}{219}})\\\\ = 0.2877\pm 0.0789\\\\=(0.2088,0.3666)[/tex]
is the required 99% confidence interval for population proportion.
Thus, the upper limit for population proportion is 0.3666
Answer:
The upper limit of the 99% confidence interval for the population proportion based on these statistics is 0.3665.
Step-by-step explanation:
We are given that of the 219 white GSS 2008 respondents in their 20's, 63 of them claim the ability to speak a language other than English.
So, the sample proportion is : [tex]\hat p[/tex] = X/n = 63/219
Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;
P.Q. = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = sample proportion = [tex]\frac{63}{219}[/tex]
n = sample of respondents = 219
p = population proportion
Here for constructing 99% confidence interval we have used One-sample z proportion statistics.
So, 99% confidence interval for the population proportion, p is ;
P(-2.5758 < N(0,1) < 2.5758) = 0.99 {As the critical value of z at
0.5% level of significance are -2.5758 & 2.5758}
P(-2.5758 < [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < 2.5758) = 0.99
P( [tex]-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < [tex]{\hat p-p}[/tex] < [tex]2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.99
P( [tex]\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < p < [tex]\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.99
99% confidence interval for p = [ [tex]\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] , [tex]\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex]]
= [ [tex]\frac{63}{219} -2.5758 \times {\sqrt{\frac{\frac{63}{219}(1-\frac{63}{219})}{219} } }[/tex] , [tex]\frac{63}{219} +2.5758 \times {\sqrt{\frac{\frac{63}{219}(1-\frac{63}{219})}{219} } }[/tex] ]
= [0.2089 , 0.3665]
Therefore, 99% confidence interval for the population proportion based on these statistics is [0.2089 , 0.3665].
Hence, the upper limit of the population proportion based on these statistics is 0.3665.
a bag contain 2 red balls and 2 black balls. what is the probability of picking out black ball out of the bag?
Answer:
2/4
Step-by-step explanation:
2+2=4 so 2/4
50%
Answer:
50% chance
Step-by-step explanation:
since half of the balls are red and half are black there will be a 50%chance of picking a black one from the bag
Which expression is equivalent to 10\sqrt(5)?
Answer:
A.
Its the simplkified version of the question and they both have the same out come which is 22.4
The equivalent expression for 10√5 is √500.
What is an equivalent expression?An equivalent expression is an expression that has the same value but does not look the same. When you simplify an expression, you're basically trying to write it in the simplest way possible.
For the given situation,
The expression is 10√5.
If a square number lies inside the square root, then we write that number outside the square root. So,
⇒ [tex]10\sqrt{5}[/tex]
⇒ [tex]\sqrt{10^{2}(5) }[/tex]
⇒ [tex]\sqrt{(100)(5)}[/tex]
⇒ [tex]\sqrt{500}[/tex]
Hence we can conclude that the equivalent expression for 10√5 is √500.
Learn more about equivalent expression here
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Assume that the Poisson distribution applies and that the mean number of hurricanes in a certain area is 6.7 per year. a. Find the probability that, in a year, there will be 4 hurricanes. b. In a 35-year period, how many years are expected to have 4 hurricanes? c. How does the result from part (b) compare to a recent period of 35 years in which 3 years had 4 hurricanes? Does the Poisson distribution work well here?
Answer:
a) 10.34% probability that, in a year, there will be 4 hurricanes.
b) 3.62 years are expected to have 4 hurricanes
c) Either 3 or 4 hurricanes(discrete number) are close to the mean of 3.62, which means that the Poisson distribution works well in this case.
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given interval.
6.7 per year.
This means that [tex]\mu = 6.7[/tex]
a. Find the probability that, in a year, there will be 4 hurricanes.
This is P(X = 4).
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 4) = \frac{e^{-6.7}*(6.7)^{4}}{(4)!} = 0.1034[/tex]
10.34% probability that, in a year, there will be 4 hurricanes.
b. In a 35-year period, how many years are expected to have 4 hurricanes?
Each year, 0.1034 probability of 10 hurricanes.
In 35 years
35*0.1034 = 3.62
3.62 years are expected to have 4 hurricanes
c. How does the result from part (b) compare to a recent period of 35 years in which 3 years had 4 hurricanes? Does the Poisson distribution work well here?
Either 3 or 4 hurricanes(discrete number) are close to the mean of 3.62, which means that the Poisson distribution works well in this case.
Final answer:
Using the Poisson distribution with a mean of 6.7 hurricanes per year, the probability of exactly 4 hurricanes occurring in one year is calculated. Multiplying this probability by 35 provides the expected number of years with 4 hurricanes in a 35-year period, which is then compared to an actual historical period to evaluate the fit of the Poisson distribution to the data.
Explanation:
The student's question pertains to the application of the Poisson distribution to determine the probability of certain events. Given that the mean number of hurricanes in a certain area is 6.7 per year, we can use the Poisson formula to calculate the probabilities:
To find the probability that there will be 4 hurricanes in a year, we use the formula:
P(x; μ) = ( × e^-μ) / x!
To determine how many years are expected to have 4 hurricanes in a 35-year period, we multiply the probability found in part a by 35.
When comparing the expected number of years with 4 hurricanes to an actual 35-year period where 3 years had 4 hurricanes, it can be seen whether the Poisson distribution provides a good fit for the actual data.
A light bulb factory produces 1,188 light bulbs every hour. Approximately 3.83% of the light bulbs are defective, and do not work. Using the binomial distribution, what is the standard deviation of the number of defective bulbs produced in an hour
Answer:
The standard deviation of the number of defective bulbs produced in an hour is 6.615
Step-by-step explanation:
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 standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
In this problem, we have that:
[tex]p = 0.0383, n = 1188[/tex]
What is the standard deviation of the number of defective bulbs produced in an hour
[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1188*0.0383*(1-0.0383)} = 6.615[/tex]
The standard deviation of the number of defective bulbs produced in an hour is 6.615
If the 18th term of a geometric sequence is 177,147 and the 24th term is 129,140,163 what is the 10th term?
Answer:
the term is 2/3(6)n-1
Step-by-step explanation:
Common ratio, r= 6
3rd term=24
Finding first term=
24=a 6^{2}
24=36a
a=24/36
a=2/3
The height of a radio tower is 400 feet, and the ground on one side of the tower slopes upward at an angle of 10degrees. (a) How long should a guy wire be if it is to connect to the top of the tower and be secured at a point on the sloped side 190 feet from the base of the tower? (b) How long should a second guy wire be if it is to connect to the middle of the tower and be secured at a point 190 feet from the base on the flat side?
Answer:
(a) 412 ft
(b) 276 ft
Step-by-step explanation:
Consider the attached diagram.
(a) The internal angle of triangle RBT at B is 90° -10° = 80°. Since we know lengths RB and BT, we can find the length RT using the law of cosines:
RT² = RB² +BT² -2·RB·BT·cos(80°) = 190² +400² -2·190·400·cos(80°)
RT² ≈ 169,705.477
RT ≈ √169,705.477 ≈ 411.95
The guy wire to the hillside should be about 412 feet long.
__
(b) The Pythagorean theorem can be used to find the shorter wire length.
LM² = LB² +MB²
LM = √(190² +200²) = √76,100
LM ≈ 275.86
The guy wire to the flat side should be about 276 feet long.
2•2•2•n•n using exponents. The product is?
Answer:
2^3 n^2 or 8n^2
Step-by-step explanation: