The center of mass of the lamina is located at the point (0, (39/12) (1 / ln(4))).
To find the center of mass of the given lamina, we need to calculate the moment about the x-axis and the y-axis, and then divide them by the total mass of the lamina.
Given information:
- The boundary of the lamina consists of the semicircles y = sqrt(1 - x^2) and y = sqrt(16 - x^2), and the portions of the x-axis that join them.
- The density at any point is inversely proportional to its distance from the origin.
Find the total mass of the lamina.
Let the density function be [tex]\[ \rho(x, y) = \frac{k}{\sqrt{x^2 + y^2}} \][/tex], where k is a constant.
The total mass, M, is given by the double integral of the density function over the region of the lamina.
M = ∫∫ ρ(x, y) dA
To evaluate this integral, we need to express the lamina in polar coordinates.
The semicircles can be represented as:
0 ≤ r ≤ 1, 0 ≤ θ ≤ π
0 ≤ r ≤ 4, π ≤ θ ≤ 2π
The total mass can be calculated as:
[tex]\[ M = \int_{0}^{\pi} \int_{0}^{1} \frac{k}{r} r \, dr \, d\theta + \int_{\pi}^{2\pi} \int_{0}^{4} \frac{k}{r} r \, dr \, d\theta \]\[ M = k \left( \pi \ln(1) + 2\pi \ln(4) \right) \]\[ M = 2\pi k \ln(4) \][/tex]
Calculate the moment about the x-axis.
The moment about the x-axis, Mx, is given by:
[tex]\[ M_x = \int_{0}^{\pi} \int_{0}^{1} \frac{k}{r} r^2 \sin(\theta) \, dr \, d\theta + \int_{\pi}^{2\pi} \int_{0}^{4} \frac{k}{r} r^2 \sin(\theta) \, dr \, d\theta \]\[ M_x = k \left( \frac{\pi}{2} + \frac{32\pi}{3} \right) \]\[ M_x = \frac{39\pi}{6} k \][/tex]
Calculate the moment about the y-axis.
The moment about the y-axis, My, is given by:
My = ∫∫ x ρ(x, y) dA
In polar coordinates:
[tex]\[ M_y = \int_{0}^{\pi} \int_{0}^{1} \frac{k}{r} r^2 \cos(\theta) \, dr \, d\theta + \int_{\pi}^{2\pi} \int_{0}^{4} \frac{k}{r} r^2 \cos(\theta) \, dr \, d\theta \][/tex]
My = 0 (due to symmetry)
Find the coordinates of the center of mass.
The coordinates of the center of mass (x_cm, y_cm) are given by:
x_cm = My / M
y_cm = Mx / M
Substituting the values, we get:
x_cm = 0 / (2πk ln(4)) = 0
y_cm = (39π/6) k / (2πk ln(4)) = (39/12) (1 / ln(4))
Therefore, the center of mass of the lamina is located at the point (0, (39/12) (1 / ln(4))).
Complete question:
The boundary of a lamina consists of the semicircles y=sqrt(1 − x^2) and y= sqrt(16 − x^2) together with the portions of the x-axis that join them. Find the center of mass of the lamina if the density at any point is inversely proportional to its distance from the origin.
The center of mass of the lamina is at the origin (0, 0)
To find the center of mass of the lamina, we first need to find the mass and the moments about the x- and y-axes.
The mass M of the lamina can be calculated by integrating the density function over the lamina. Since the density at any point is inversely proportional to its distance from the origin, we can express the density[tex]\( \delta \) as \( \delta(x, y) = \frac{k}{\sqrt{x^2 + y^2}} \)[/tex], where k is a constant.
Let's denote [tex]\( \delta(x, y) \) as \( \frac{k}{\sqrt{x^2 + y^2}} \)[/tex]. Then the mass M is given by the double integral of [tex]\( \delta(x, y) \)[/tex] over the region R bounded by the semicircles and the portions of the x-axis:
[tex]\[ M = \iint_R \delta(x, y) \, dA \][/tex]
Where dA represents the differential area element.
To find the moments about the x- and y -axes, we calculate:
[tex]\[ M_x = \iint_R y \delta(x, y) \, dA \]\[ M_y = \iint_R x \delta(x, y) \, dA \][/tex]
Then, the coordinates [tex]\( (\bar{x}, \bar{y}) \)[/tex] of the center of mass are given by:
[tex]\[ \bar{x} = \frac{M_y}{M} \]\[ \bar{y} = \frac{M_x}{M} \][/tex]
Now, let's proceed to find [tex]\( M \), \( M_x \), and \( M_y \)[/tex]
First, let's express the density [tex]\( \delta(x, y) \)[/tex] in terms of k:
[tex]\[ \delta(x, y) = \frac{k}{\sqrt{x^2 + y^2}} \][/tex]
Now, we'll find the mass M by integrating [tex]\( \delta(x, y) \)[/tex] over the region R :
[tex]\[ M = \iint_R \frac{k}{\sqrt{x^2 + y^2}} \, dA \][/tex]
Since the region R is symmetric about the x-axis, we can integrate over the upper half and double the result:
[tex]\[ M = 2 \iint_{R_1} \frac{k}{\sqrt{x^2 + y^2}} \, dA \][/tex]
Now, we'll switch to polar coordinates [tex]\( (r, \theta) \)[/tex]. In polar coordinates, the region [tex]\( R_1 \)[/tex]is described by [tex]\( 0 \leq \theta \leq \pi \) and \( 1 \leq r \leq 4 \).[/tex]
So, the integral becomes:
[tex]\[ M = 2 \int_{0}^{\pi} \int_{1}^{4} \frac{k}{r} \cdot r \, dr \, d\theta \]\[ = 2k \int_{0}^{\pi} \int_{1}^{4} 1 \, dr \, d\theta \]\[ = 2k \int_{0}^{\pi} (4 - 1) \, d\theta \]\[ = 2k \int_{0}^{\pi} 3 \, d\theta \]\[ = 6k \pi \][/tex]
For [tex]\( M_x \)[/tex], we integrate [tex]\( x \delta(x, y) \)[/tex] over the region R :
[tex]\[ M_x = \iint_R x \cdot \frac{k}{\sqrt{x^2 + y^2}} \, dA \]\[ = 2 \int_{0}^{\pi} \int_{1}^{4} r \cos(\theta) \cdot \frac{k}{r} \cdot r \, dr \, d\theta \]\[ = 2k \int_{0}^{\pi} \int_{1}^{4} \cos(\theta) \cdot r \, dr \, d\theta \]\[ = 2k \int_{0}^{\pi} \left[ \frac{1}{2} r^2 \cos(\theta) \right]_{1}^{4} \, d\theta \][/tex]
[tex]\[ = 2k \int_{0}^{\pi} \left( 8 \cos(\theta) - \frac{1}{2} \cos(\theta) \right) \, d\theta \]\[ = 2k \int_{0}^{\pi} \left( \frac{15}{2} \cos(\theta) \right) \, d\theta \]\[ = 2k \left[ \frac{15}{2} \sin(\theta) \right]_{0}^{\pi} \]\[ = 2k \cdot 0 \]\[ = 0 \][/tex]
Now, for[tex]\( M_y \)[/tex], we integrate [tex]\( y \delta(x, y) \)[/tex]over the region R :
[tex]\[ M_y = \iint_R y \cdot \frac{k}{\sqrt{x^2 + y^2}} \, dA \\\[ = 2 \int_{0}^{\pi} \int_{1}^{4} r \sin(\theta) \cdot \frac{k}{r} \cdot r \, dr \, d\theta \\\[ = 2k \int_{0}^{\pi} \int_{1}^{4} \sin(\theta) \cdot r \, dr \, d\theta \\\[ = 2k \int_{0}^{\pi} \left[ \frac{1}{2} r^2 \sin(\theta) \right]_{1}^{4} \, d\theta \\[/tex]
[tex]\[ = 2k \int_{0}^{\pi} \left( 8 \sin(\theta) - \frac{1}{2} \sin(\theta) \right) \, d\theta \\\[ = 2k \int_{0}^{\pi} \left( \frac{15}{2} \sin(\theta) \right) \, d\theta \\\[ = 2k \left[ -\frac{15}{2} \cos(\theta) \right]_{0}^{\pi} \\\[ = 2k \cdot 0 \\\[ = 0 \][/tex]
Now, we have [tex]\( M = 6k \pi \), \( M_x = 0 \), and \( M_y = 0 \).[/tex]
Finally, we can find the coordinates of the center of mass [tex]\( (\bar{x}, \bar{y}) \):[/tex]
[tex]\[ \bar{x} = \frac{M_y}{M} = \frac{0}{6k \pi} = 0 \]\[ \bar{y} = \frac{M_x}{M} = \frac{0}{6k \pi} = 0 \][/tex]
So, the center of mass of the lamina is at the origin (0, 0) .
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.
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,
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,
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.
what is the measure of angle C
Answer:
pls I can't see any diagram anywhere
The life span of a battery is the amount of time the battery will last. The distribution of life span for a certain type of battery is approximately normal with mean 2.5 hours and standard deviation 0.25 hour. Suppose one battery will be selected at random. Which of the following is closest to the probability that the selected battery will have a life span of at most 2.1 hours?A:0.055B: 0.110C: 0.445D: 0.890E: 0.945
Answer:
A:0.055
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.
In this problem, we have that:
[tex]\mu = 2.5, \sigma = 0.25[/tex]
Which of the following is closest to the probability that the selected battery will have a life span of at most 2.1 hours?
This is the pvalue of Z when X = 2.1. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{2.1 - 2.5}{0.25}[/tex]
[tex]Z = -1.6[/tex]
[tex]Z = -1.6[/tex] has a pvalue of 0.0548
So the correct answer is:
A:0.055
The circumference of the inner circle is 88 ft. The distance between the inner circle and the outer circle is 3 ft. By how many feet is the circumference of outer circle greater than the circumference of the inner circle? Use StartFraction 22 Over 7 EndFraction
for pi. (3.14)
Answer:
18.86 feet
Step-by-step explanation:
The circumference of the inner circle is 88 ft.
Circumference of a Circle[tex]=2\pi r[/tex]
Therefore:
[tex]2\pi r =88\\r=88 \div (2*\frac{22}{7})=14 ft[/tex]
Radius of the inner circle=14 feet
If the distance between the inner circle and the outer circle is 3 ft, the radius of the outer circle=14+3 =17 feet.
Therefore. circumference of the outer Circle[tex]=2\pi r[/tex]
[tex]=2*\frac{22}{7}*17=106.86 ft[/tex]
Difference in Circumference=106.86-88 =18.86 feet
The circumference of the outer circle is greater than that of the inner circle by 18.86 feet.
18.9
Step-by-step explanation:
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.
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.
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.
Hopefully that’s better :)))
Answer: Top right, a rectangle has all the properties of a square
Step-by-step explanation: A rectangle does not have all the properties of a square.
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.
He purchased $12.00 worth of lemons and $4.00 worth of glasses to make the lemonade.
He adds $0.02 worth of sugar to each glass of lemonade.
He sells each glass of lemonade for $0.25.
What is the minimum number of glasses of lemonade that Michael needs to sell to begin to make a profit?
The student must sell at least 70 glasses of lemonade at $0.25 each, considering the initial costs of $16.00 and the additional $0.02 cost per glass for sugar, to begin to make a profit.
Explanation:To calculate the minimum number of glasses of lemonade that must be sold to start making a profit, we need to consider the total costs and the revenue per glass. The student spent $12.00 on lemons and $4.00 on glasses, totaling $16.00 in costs. Additionally, each glass of lemonade has an added cost of $0.02 for sugar. The revenue from selling one glass of lemonade is $0.25.
To break even, the total revenue must equal the total costs. As costs are fixed at $16.00 and the variable cost is $0.02 per glass, we can set up the equation: (Number of glasses × $0.25) - (Number of glasses × $0.02) = $16.00. This simplifies to (Number of glasses × $0.23) = $16.00, and solving for Number of glasses gives us Number of glasses = $16.00 / $0.23, which is approximately 69.57 glasses. Since you can't sell a fraction of a glass, rounding up means the student needs to sell at least 70 glasses to begin to make a profit.
-4x + (-3) =x +3 What is X equivalent to
Answer: x= −6 /5 is the answer
Answer:
x= -1 1/5 or x= -1.2
Step-by-step explanation:
-4x+-3=x+3
-5x-3=3
-5x=6
-x=6/5
x=-6/5
x=-1 1/5
or x=-1.2
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Which fraction is equivalent to 3/4+1/6
Answer:
1/2
Step-by-step explanation:
Answer:
11/12
Step-by-step explanation:
If you convert 3/4 and 1/6,
you will get 9/12+2/12.
If you add both of them, you will get your answer.
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
3. For the school Band, Marcia decides to order t-shirts for all of the participants. It will
cost $4 per shirt for the medium size and $5 per shirt for the large size. Marcia orders
a total of 70 T-shirts and spends $320. Determine the two linear equations to
represent the information. [4A) **Hint: One equation is for amount of t-shirts and the second is
for money. YOU DO NOT NEED TO SOLVE,
Let x represent number of medium size shirts
Let y represent total large size shirts
Answer:
4x+5y = 320
x+y = 70
Step-by-step explanation:
We need one equation for the total number of shirts, and one for the total cost.
Total number = 70
So that means medium shirts + large shirts = 70
So our first equation is x+y = 70
Total cost = $320
So that means $4 times medium shirts + $5 times large shirts = 320
So our second equation is 4x+5y = 320
The demand of the computers is surely increasing in recent years. A survey shows that this phenomenon can be calculated by the function D(t) = 23.2sqrt(5 + 2.7t) Here, D represents the demand (measured by millions ) and t the time (measured by years ). It will take years that the demand of the computers reaches 132.2 millions . Round your answer to one decimal point
Answer:
10.2
Step-by-step explanation:
Apparently, you want to find the solution for ...
132.2 = 23.2√(5 +2.7t)
132.2/23.2 = √(5 +2.7t) . . . . divide by 23.2
(132.2/23.2)² = 5 +2.7t . . . . . square both sides
(132.2/23.2)² -5 = 2.7t . . . . . subtract 5
((132.2/23.2)² -5)/2.7 = t . . . . divide by the coefficient of t
10.2 ≈ t
It will take about 10.2 years for the demand for computers to reach 132.2 million.
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
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
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
https://brainly.com/question/11326971
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Steroids, which are dangerous, are sometimes used to improve athletic performance. A study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school seniors in Illinois. Results showed that 34 of the freshmen and 24 of the seniors had used anabolic steroids. Estimate at a 95% confidence level the difference between the proportion of freshmen using steroids in Illinois and the proportion of seniors using steroids in Illinois.
Answer:
At 95% confidence level, the difference between the proportion of freshmen using steroids in Illinois and the proportion of seniors using steroids in Illinois is -7.01135×10⁻³ < [tex]\hat{p}_1-\hat{p}_2[/tex] < 1.237
Step-by-step explanation:
Here we are required to construct the 95% confidence interval of the difference between two proportions
The formula for the confidence interval of the difference between two proportions is as follows;
[tex]\hat{p}_1-\hat{p}_2\pm z^{*}\sqrt{\frac{\hat{p}_1\left (1-\hat{p}_1 \right )}{n_{1}}+\frac{\hat{p}_2\left (1-\hat{p}_2 \right )}{n_{2}}}[/tex]
Where:
[tex]\hat{p}_1 = \frac{34}{1679}[/tex]
[tex]\hat{p}_2 = \frac{24}{1366}[/tex]
n₁ = 1679
n₂ = 1366
[tex]z_{\alpha /2}[/tex] at 95% confidence level = 1.96
Plugging in the values, we have;
[tex]\frac{34}{1679}- \frac{24}{1366} \pm 1.96 \times \sqrt{\frac{ \frac{34}{1679}\left (1- \frac{34}{1679}\right )}{1679}+\frac{\frac{24}{1366} \left (1-\frac{24}{1366} \right )}{1366}}[/tex]
Which gives;
-7.01135×10⁻³ < [tex]\hat{p}_1-\hat{p}_2[/tex] < 1.237.
At 95% confidence level, the difference between the proportion of freshmen using steroids in Illinois and the proportion of seniors using steroids in Illinois = -7.01135×10⁻³ < [tex]\hat{p}_1-\hat{p}_2[/tex] < 1.237.
Write the equation of the tangent line to the curve x^2/8 - y^2/4 =1 at the point (4,2) by using the following facts. The slope m of the tangent line to a hyperbola at the point (x, y) is: m=b^2x/a^2y for x^2/a^2 -y^2/b^2=1 m=a^2x/b^2y for y^2/a^2 - x^2/b^2 =1
Answer:
[tex]y=x-2[/tex]
Step-by-step explanation:
So we are given the formula for the slope of a hyperbola in this form:
[tex]\frac{x^2}{a^2}-\frac{y^2}{b^2}=1[/tex].
That formula for the slope is [tex]m=\frac{b^2x}{a^2y}[/tex]
If we compare the following two equations, we will be able to find [tex]a^2[/tex] and [tex]b^2[/tex]:
[tex]\frac{x^2}{a^2}-\frac{y^2}{b^2}=1[/tex]
[tex]\frac{x^2}{8}-\frac{y^2}{4}=1[/tex]
We see that [tex]a^2=8[/tex] while [tex]b^2=4[/tex].
So the slope at [tex](x,y)=(4,2)[/tex] is:
[tex]m=\frac{b^2x}{a^2y}=\frac{4(4)}{8(2)}=\frac{16}{16}=1[/tex].
Recall: Slope-intercept form of a linear equation is [tex]y=mx+b[/tex].
We just found [tex]m=1[/tex]. Let's plug that in.
[tex]y=1x+b[/tex]
[tex]y=x+b[/tex]
To find [tex]b[/tex], the [tex]y[/tex]-intercept, we will need to use a point on our tangent line. We know that it is going through [tex](4,2)[/tex].
Let's enter this point in to find [tex]b[/tex].
[tex]2=4+b[/tex]
Subtract 4 on both sides:
[tex]2-4=b[/tex]
Simplify:
[tex]-2=b[/tex]
The equation for the tangent line at [tex](4,2)[/tex] on the given equation is:
[tex]y=x-2[/tex]
Answer: y = x - 2
Step-by-step explanation:
First you take the derivative of each term. d/dx(x²/8) - d/dx(y²/4) = d/dx(1)
x/4 - (y/2)dy/dx = 0
Then you solve for dy/dx: dy/dx = x/2y
Plug in the values: dy/dx = 1
To find the y-intercept, plug in values for y = mx+ b. 2 = 4 + b, b = -2
The equation is y = x - 2
Consider the function on the interval (0, 2π). f(x) = x − 2 sin x (a) Find the open intervals on which the function is increasing or decreasing. (Enter your answers using interval notation.)
Answer:
Increasing:
[tex](\frac{\pi}{3},\frac{5\pi}{3})[/tex]
Decreasing
[tex](0,\frac{\pi}{3})[/tex] U [tex](\frac{5\pi}{3},2\pi)[/tex]
Step-by-step explanation:
Increasing and Decreasing Intervals
To find if a function is increasing in a point x=a, we evaluate the first derivative in x=a and if:
f'(a) >0, the function is increasing f'(a) <0, the function is decreasing f'(a) =0, the function has a critical pointFor continuous functions, we can safely assume between a given critical point and the next one, the function keeps its behavior, i.e. it's increasing or decreasing in the interval formed by both points.
So, we find the critital points of
[tex]f(x)=x-2sinx[/tex]
Taking the derivative
[tex]f'(x)=1-2cosx[/tex]
Equating to 0
[tex]1-2cosx=0[/tex]
Solving
[tex]\displaystyle cosx=\frac{1}{2}[/tex]
There are two solutions in the interval [tex](0,2\pi)[/tex]
[tex]\displaystyle x=\frac{\pi}{3},\ x=\frac{5\pi}{3}[/tex]
Now we compute the second derivative
[tex]f''(x)=2sinx[/tex]
Evaluating for both critical points
[tex]\displaystyle f''(\frac{\pi}{3})=2sin\frac{\pi}{3}=\sqrt{3}[/tex]
Since it's positive, the point is a minimum
[tex]\displaystyle f''(\frac{5\pi}{3})=2sin\frac{5\pi}{3}=-\sqrt{3}[/tex]
Since it's negative, the point is a maximum
In the interval
[tex](0,\frac{\pi}{3})[/tex]
the function is decreasing
In the interval
[tex](\frac{\pi}{3},\frac{5\pi}{3})[/tex]
the function is increasing
In the interval
[tex](\frac{5\pi}{3},2\pi)[/tex]
the function is decreasing
First, you need to find the derivative of the function f(x) = x - 2sin(x).
Using the rules of calculus, the derivative of x is 1 and the derivative of -2sin(x) is -2cos(x). Therefore, the derivative of the function, denoted as f'(x), is given by:
f'(x) = 1 - 2cos(x).
To determine where the function is increasing or decreasing, we need to find the critical points, the x-values where the derivative of the function is equal to zero or undefined.
Setting the derivative equal to zero and solving:
1 - 2cos(x) = 0.
2cos(x) = 1.
cos(x) = 1/2.
The solutions to this equation on the interval 0 to 2π are x = π/3 and x = 5π/3.
With these critical points, we have divided the entire interval into three subintervals:
(0, π/3), (π/3, 5π/3), and (5π/3, 2π).
We now determine the sign of the derivative on each of these intervals. We pick a "test point" from each interval and substitute it into the derivative.
From (0, π/3), we pick x = π/6, and find that f'(π/6) is positive.
From (π/3, 5π/3), we pick x = π, and find that f'(π) is negative.
From (5π/3, 2π), we pick x = 3π/2, and find that f'(3π/2) is positive.
By the First Derivative Test, a positive derivative indicates that the function is increasing on that interval and a negative derivative indicates that the function is decreasing on that interval.
Therefore, the function increases on the intervals (0, π/3) and (5π/3, 2π),
and decreases on the interval (π/3, 5π/3).
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According to industry sources, online banking is expected to take off in the near future. The projected number of households (in millions) using this service is given in the following table. (Here, x = 0 corresponds to the beginning of 1997.)
Year, x 0 1 2 3 4 5
Households, y 4.5 7.5 10.0 13.0 15.6 18.0
(a) Find an equation of the least-squares line for these data. (Give numbers to three decimal places.)
y(x) =
(b) Use your result of part (a) to estimate the number of households using online banking at the beginning of 2007, assuming the projection is accurate.
Answer:
(a) The least-square regression line is: [tex]y=4.662+2.709x[/tex].
(b) The number of households using online banking at the beginning of 2007 is 31.8.
Step-by-step explanation:
The general form of a least square regression line is:
[tex]y=\alpha +\beta x[/tex]
Here,
y = dependent variable
x = independent variable
α = intercept
β = slope
(a)
The formula to compute intercept and slope is:
[tex]\begin{aligned} \alpha &= \frac{\sum{Y} \cdot \sum{X^2} - \sum{X} \cdot \sum{XY} }{n \cdot \sum{X^2} - \left(\sum{X}\right)^2} \\\beta &= \frac{ n \cdot \sum{XY} - \sum{X} \cdot \sum{Y}}{n \cdot \sum{X^2} - \left(\sum{X}\right)^2} \end{aligned}[/tex]
The values of ∑X, ∑Y, ∑XY and ∑X² are computed in the table below.
Compute the value of intercept and slope as follows:
[tex]\begin{aligned} \alpha &= \frac{\sum{Y} \cdot \sum{X^2} - \sum{X} \cdot \sum{XY} }{n \cdot \sum{X^2} - \left(\sum{X}\right)^2} = \frac{ 68.6 \cdot 55 - 15 \cdot 218.9}{ 6 \cdot 55 - 15^2} \approx 4.662 \\ \\\beta &= \frac{ n \cdot \sum{XY} - \sum{X} \cdot \sum{Y}}{n \cdot \sum{X^2} - \left(\sum{X}\right)^2} = \frac{ 6 \cdot 218.9 - 15 \cdot 68.6 }{ 6 \cdot 55 - \left( 15 \right)^2} \approx 2.709\end{aligned}[/tex]
The least-square regression line is:
[tex]y=4.662+2.709x[/tex]
(b)
For the year 2007 the value of x is 10.
Compute the value of y for x = 10 as follows:
[tex]y=4.662+2.709x[/tex]
[tex]=4.662+(2.709\times10)\\=4.662+27.09\\=31.752\\\approx 31.8[/tex]
Thus, the number of households using online banking at the beginning of 2007 is 31.8.
A radioactive substance decays according to the following function, where yo is the initial amount present, and y is the amount present at time t (in days).
y= y_o e^0.072t.
Find the half-life of this substance. Do not round any intermediate computations, and round your answer to the nearest tenth. days
Answer:
[tex]t_{1/2} \approx 9.6\,days[/tex]
Step-by-step explanation:
The time constant of the radioactive constant is:
[tex]\tau = \frac{1}{0.072}\,days[/tex]
[tex]\tau = 13.889\,days[/tex]
The half-life of the substance is:
[tex]t_{1/2} = \tau \cdot \ln 2[/tex]
[tex]t_{1/2} = (13.889\,days)\cdot \ln 2[/tex]
[tex]t_{1/2} \approx 9.6\,days[/tex]
The mean of a normally distributed group of weekly incomes of a large group of executives is $1,000 and the standard deviation is $100. What is the z-score (value of z) for an income of $1,100
Answer:
The z-score (value of z) for an income of $1,100 is 1.
Step-by-step explanation:
We are given that the mean of a normally distributed group of weekly incomes of a large group of executives is $1,000 and the standard deviation is $100.
Let X = group of weekly incomes of a large group of executives
So, X ~ N([tex]\mu=1,000 ,\sigma^{2} = 100^{2}[/tex])
The z-score probability distribution for a normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean income = $1,000
[tex]\sigma[/tex] = standard deviation = $100
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.
Now, we are given an income of $1,100 for which we have to find the z-score (value of z);
So, z-score is given by = [tex]\frac{X-\mu}{\sigma}[/tex] = [tex]\frac{1,100-1,000}{100}[/tex] = 1
Hence, the z-score (value of z) for an income of $1,100 is 1.
The z-score for an income of $1,100 is 1.0.
The z-score is a measure of how many standard deviations a particular value, in this case, income, is from the mean of a normally distributed dataset. The formula to calculate the z-score is:
z = (x - μ) / σ
Where:
x is the value being evaluated.μ is the mean of the distribution.σ is the standard deviation of the distribution.Given the mean (μ) is $1,000 and the standard deviation (σ) is $100, we can substitute these values into the formula to find the z-score of an income of $1,100.
z = ($1,100 - $1,000) / $100
z = $100 / $100
z = 1
Therefore, the z-score for an income of $1,100 is 1.0. This means the income of $1,100 is one standard deviation above the mean.
what is one thousand five hundred divided by five equal
Answer: 300
We know that 5×3=15. Since we are trying to get 1500 we Multiply 300×5 and get 1500.
How do we know?
As you can see if you cross out 2 of the 0 in 300 you have 3×5. When you multiply it you get 15.Also if you cross out 1 of the 0 in 300 you have 30×5. When you multiply that you get 150.As you see when you still multiply either of those problems you have 15 in it. So we should know 300×5=1500.
Note: We can also multiply 500×3 and get 1500. You still get the same answer but just switched numbers.
Answer:
[tex]300[/tex]
Step-by-step explanation:
[tex] \frac{1500}{5} = 300[/tex]
It's very easy to find if you use a calculator.
To know whether that the answer is correct or wrong you can do like this.
[tex]300 \times 5 = 1500[/tex]
hope this helps
thanks.
Use the following to answer question 39: The average score of 100 students taking a statistics final was 70 with a standard deviation of 7. Assuming a normal distribution, what is the probability that students scored less than 60
Answer:
The probability of students scored less than 60 = .0768
Step-by-step explanation:
Given -
Mean score [tex](\nu )[/tex] = 70
standard deviation [tex](\sigma )[/tex] = 7
Let X be the score of students
the probability that students scored less than 60 =
[tex]P(X< 60)[/tex] = [tex]P(\frac{X - \nu }{\sigma}< \frac{60 - 70}{7})[/tex]
= [tex]P(z < \frac{60 - 70}{7})[/tex] put[ Z= [tex]\frac{X - \nu }{\sigma}[/tex]]
= [tex]P(z < -1.428)[/tex] using z table
= .0768
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]