Answer:
y(x)=C_1·e^{2x} + C_2·e^{-6x}
Step-by-step explanation:
From Exercise we have the differential equation
y''+4y'-12=0.
This is a characteristic differential equation and we are solved as follows:
y''+4y'-12=0
m²+4m-12=0
m_{1,2}=\frac{-4±\sqrt{16+48}}{2}
m_{1,2}=\frac{-4±\sqrt{64}}{2}
m_{1,2}=\frac{-4±8}{2}
m_1=2
m_2=-6
The general solution of this differential equation is in the form
y(x)=C_1·e^{m_1 ·x} + C_2·e^{m_2 ·x}
Therefore, we get
y(x)=C_1·e^{2x} + C_2·e^{-6x}
The correct values of the constant [tex]\( r \)[/tex] for which [tex]\( y = e^{rx} \)[/tex] is a solution of the differential equation [tex]\( y'' + 4y' - 12y = 0 \)[/tex] are [tex]\( r = 2 \)[/tex] and [tex]\( r = -6 \).[/tex]
To find the values of [tex]\( r \)[/tex], we substitute [tex]\( y = e^{rx} \)[/tex] into the differential equation. First, we find the first and second derivatives of [tex]\( y \)[/tex]:
[tex]\[ y' = \frac{d}{dx}(e^{rx}) = re^{rx} \][/tex]
[tex]\[ y'' = \frac{d}{dx}(re^{rx}) = r^2e^{rx} \][/tex]
Now, we substitute [tex]\( y \)[/tex], [tex]\( y' \)[/tex], and [tex]\( y'' \)[/tex] into the differential equation:
[tex]\[ y'' + 4y' - 12y = 0 \][/tex]
[tex]\[ r^2e^{rx} + 4re^{rx} - 12e^{rx} = 0 \][/tex]
Since [tex]\( e^{rx} \)[/tex] is never zero, we can divide through by [tex]\( e^{rx} \)[/tex] to get the characteristic equation:
[tex]\[ r^2 + 4r - 12 = 0 \][/tex]
Now, we solve this quadratic equation for [tex]\( r \)[/tex]:
[tex]\[ r^2 + 6r - 2r - 12 = 0 \][/tex]
[tex]\[ r(r + 6) - 2(r + 6) = 0 \][/tex]
[tex]\[ (r - 2)(r + 6) = 0 \][/tex]
Setting each factor equal to zero gives us the possible values for [tex]\( r \)[/tex]:
[tex]\[ r - 2 = 0 \quad \Rightarrow \quad r = 2 \][/tex]
[tex]\[ r + 6 = 0 \quad \Rightarrow \quad r = -6 \][/tex]
Therefore, the values of [tex]\( r \)[/tex] for which [tex]\( y = e^{rx} \)[/tex] is a solution of the given differential equation are [tex]\( r = 2 \)[/tex] and [tex]\( r = -6 \).[/tex]
Determine whether the value given is a parameter or statistic. Two thirds of all the students in this class are womena. Parameterb. Statistic
Answer: a. Parameter
Step-by-step explanation:
Parameter can be defined as a fact/characteristic about a whole population.
For example:
i. 20% of my class are boys
ii. 40% of Canadian senators are women.
Parameter usually deal with a small measurable population.
While statistic is a characteristic/fact about the sample.
Therefore the case above is a parameter because the students in the class is the population and the stated (Two thirds of all the students in this class are women) is a fact about the population.
Answer:
Parameter
Step-by-step explanation:
The given value is Parameter because it contains the measurement of population. In short, the given value 2/3 is the estimated value of population.
The population consists of all the students in the class. If the ratio of women from all the students in the class is calculated, then it is a measure of population. Thus, the given value is parameter.
Students are working on a project which requires toothpicks. Each student receives the same number of toothpicks, t, from the
teacher. Which equation can be used to find the total number of toothpicks, N, given out by the teacher for the project to
s students?
OA. N =
OB. N = st
OC. N = s +t
OD. N =
Reset
Submit
Answer:
The correct answer is B. N = st
Step-by-step explanation:
1. Let's review the information given to us to answer the question correctly:
t = number of toothpicks received by each student
s = number of students
N = total number of students
2. Which equation can be used to find the total number of toothpicks, N, given out by the teacher for the project?
Total number of toothpicks for the project = number of toothpicks received by each student * number of students, replacing with the variables:
N = t * s = s * t
The correct answer is B. N = st
Answer:
B. N =st
Step-by-step explanation:
What is next in this sequence of numbers: 1, 11, 21, 1211, 111221, 312211, ...?
Answer:
13112221
Step-by-step explanation:
Each sequence of numbers is a verbal representation of the sequence before it. Thus, starting with 1, the next sequence would be "one one," or "11." That sequence is followed by "two one," or "21," and so on and so forth.
This may also be a good explanation:
The first number is just ONE (amount) "1" (0-9 numeral). So if you say there's ONE "1" (seriously just say it aloud) the next number would be an 11. Then there are TWO "1's", creating 21. Then ONE "2" and ONE "1" which creates 1,211. Then ONE "1", ONE "2", and TWO "1's" creating 111,221 ... and so on.
The first number 1 is read as one one, so the second number is written as 11, this is read as two ones, so the next number is written as 21 ( two ones)
This continues throughout the sequence.
The last number written is 312211 which is read as one three, one one, two twos, two ones
This gets written as 13112221
Suppose that a spherical droplet of liquid evaporates at a rate that is proportional to its surface area: where V = volume (mm3 ), t = time (min), k = the evaporation rate (mm/min), and A = surface area (mm2 ). Use Euler’s method to compute the volume of the droplet from t = 0 to 10 min using a step size of 0.25 min. Assume that k = 0.08 mm/min and that the droplet initially has a radius of 2.5 mm. Assess the validity of your results by determining the radius of your final computed volume and verifying that it is consistent with the evaporation rate.
Answer:
V = 20.2969 mm^3 @ t = 10
r = 1.692 mm @ t = 10
Step-by-step explanation:
The solution to the first order ordinary differential equation:
[tex]\frac{dV}{dt} = -kA[/tex]
Using Euler's method
[tex]\frac{dVi}{dt} = -k *4pi*r^2_{i} = -k *4pi*(\frac {3 V_{i} }{4pi})^(2/3)\\ V_{i+1} = V'_{i} *h + V_{i} \\[/tex]
Where initial droplet volume is:
[tex]V(0) = \frac{4pi}{3} * r(0)^3 = \frac{4pi}{3} * 2.5^3 = 65.45 mm^3[/tex]
Hence, the iterative solution will be as next:
i = 1, ti = 0, Vi = 65.45[tex]V'_{i} = -k *4pi*(\frac{3*65.45}{4pi})^(2/3) = -6.283\\V_{i+1} = 65.45-6.283*0.25 = 63.88[/tex]
i = 2, ti = 0.5, Vi = 63.88[tex]V'_{i} = -k *4pi*(\frac{3*63.88}{4pi})^(2/3) = -6.182\\V_{i+1} = 63.88-6.182*0.25 = 62.33[/tex]
i = 3, ti = 1, Vi = 62.33[tex]V'_{i} = -k *4pi*(\frac{3*62.33}{4pi})^(2/3) = -6.082\\V_{i+1} = 62.33-6.082*0.25 = 60.813[/tex]
We compute the next iterations in MATLAB (see attachment)
Volume @ t = 10 is = 20.2969
The droplet radius at t=10 mins
[tex]r(10) = (\frac{3*20.2969}{4pi})^(2/3) = 1.692 mm\\[/tex]
The average change of droplet radius with time is:
Δr/Δt = [tex]\frac{r(10) - r(0)}{10-0} = \frac{1.692 - 2.5}{10} = -0.0808 mm/min[/tex]
The value of the evaporation rate is close the value of k = 0.08 mm/min
Hence, the results are accurate and consistent!
Using Euler's method, we approximated the volume and surface area of the object over a period of 10 minutes. Starting with an initial radius of 2.5 mm, and given a decay constant of 0.08 mm/min, we computed the final volume to be approximately 26.19 mm³ and the final surface area to be about 31.70 mm². This resulted in a final radius of approximately 2.06 mm.
To solve this problem using Euler's method, we'll use the following formulas:
dV/dt = -kA
A = 4πr²
V = (4/3)πr³
Given that the initial radius r₀ = 2.5 mm, we can compute the initial volume V₀ and surface area A₀. Then, we'll iterate through time steps using Euler's method:
Vₙ₊₁ = Vₙ - kA * Δt
Aₙ₊₁ = 4π(rₙ - (k/3) * Δt)²
Using k = 0.08 mm/min and a step size of Δt = 0.25 min, we perform the calculations:
V₀ = (4/3)π(2.5)³ ≈ 65.45 mm³
A₀ = 4π(2.5)² ≈ 98.17 mm²
Iterating:
t = 0.25 min:
V₁ ≈ 65.45 - 0.08 * 98.17 * 0.25 ≈ 61.44 mm³
A₁ ≈ 4π(2.44)² ≈ 59.17 mm²
t = 0.5 min:
V₂ ≈ 61.44 - 0.08 * 59.17 * 0.25 ≈ 59.58 mm³
A₂ ≈ 4π(2.42)² ≈ 58.01 mm²
Continuing this process until t = 10 min, we obtain:
V₄₀ ≈ 26.19 mm³
A₄₀ ≈ 31.70 mm²
Finally, calculating the radius r₄₀ corresponding to V₄₀:
r₄₀ = ((3V₄₀)/(4π))^(1/3) ≈ 2.06 mm
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A research team conducted a study showing that approximately 15% of all businessmen who wear ties wear them so tightly that they actually reduce blood flow to the brain, diminishing cerebral functions. At a board meeting of 15 businessmen, all of whom wear ties, what are the following probabilities (Round your answers to three decimal places.) (a) at least one tie is too tight (b) more than two ties are too tight (c) no tie is too tight (d) at least 18 ties are not too tight
Answer:
a) 0.913
b) 0.397
c) 0.087
Step-by-step explanation:
We are given the following information:
We treat wearing tie too tight as a success.
P(Tight tie) = 15% = 0.15
Then the number of businessmen follows a binomial distribution, where
[tex]P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}[/tex]
where n is the total number of observations, x is the number of success, p is the probability of success.
Now, we are given n = 15
We have to evaluate:
a) at least one tie is too tight
[tex]P(x \geq 1) = P(x = 1) +....+ P(x = 15)\\=1 - P(x = 0)\\= 1 - \binom{15}{0}(0.15)^0(1-0.15)^{15}\\=1 - 0.087\\= 0.913[/tex]
b) more than two ties are too tight
[tex]P(x > 2) = P(x = 3) +....+ P(x = 15)\\=1 - P(x = 0) - P(x=1) - P(x=2)\\= 1 - \binom{15}{0}(0.15)^0(1-0.15)^{15}-\binom{15}{1}(0.15)^1(1-0.15)^{14}-\binom{15}{0}(0.15)^2(1-0.15)^{13}\\=1 - 0.087 - 0.231 - 0.285\\= 0.397[/tex]
c) no tie is too tight
[tex]P(x = 0)\\=\binom{15}{0}(0.15)^0(1-0.15)^{15}\\=0.087[/tex]
d) at least 18 ties are not too tight
This probability cannot be evaluated as the number of success or the failures exceeds the number of trials given which is 15.
The probability is asked for 18 failures which cannot be evaluated.
To calculate the probabilities, we can use the concept of binomial probability. Using appropriate formulas, we can find the probabilities (a) at least one tie is too tight (b) more than two ties are too tight (c) no tie is too tight (d) at least 18 ties are not too tight.
Explanation:To calculate the probabilities, we can use the concept of binomial probability. In this case, the probability that a businessman wears the tie too tightly is 15% or 0.15. The total number of businessmen at the board meeting is 15.
(a) To calculate the probability that at least one tie is too tight, we need to find the complement of the probability that no ties are too tight. Using the formula 1 - P(no tight ties), we calculate 1 - (0.85^15) = 0.999 or 99.9%.
(b) To calculate the probability that more than two ties are too tight, we need to find the sum of probabilities of having 3 or more ties too tight. Using the binomial probability formula, we calculate P(X > 2) = 1 - (P(X=0) + P(X=1) + P(X=2)) = 1 - (0.85^15 + 15*0.15*(0.85^14) + 105*(0.15^2)*(0.85^13)) = 0.014 or 1.4%.
(c) To calculate the probability that no ties are too tight, we use the probability P(X=0) = 0.85^15 = 0.203 or 20.3%.
(d) To calculate the probability that at least 18 ties are not too tight, we need to find the sum of probabilities of having 18 or more ties not too tight. Using the binomial probability formula, we calculate P(X >= 18) = P(X=18) + P(X=19) + ... + P(X=15) = (15 choose 18)*(0.85^18)*(0.15^2) + (15 choose 19)*(0.85^19)*(0.15^1) + (15 choose 20)*(0.85^20)*(0.15^0) = 0.999 or 99.9%.
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A company had 110 employees whose salaries are summarized in the frequency distribution below. Find the mean salary.Salary ($) Employees5,001-10,000 2210,001-15,000 2015,001-20,000 2120,001-25,000 2325,001-30,000 24
Answer:
Mean salary=$17818.68
Step-by-step explanation:
Salary($) Employees(f)
5001-10,000 22
10,001-15,000 20
15,001-20,000 21
20,001-25,000 23
25,001-30,000 24
We know that company had 110 employees so ∑f should be equal to 110.
∑f=22+20+21+23+24=110
The mean salary can be computed as
[tex]xbar=\frac{sum(fx)}{sum(f)}[/tex]
The x be the midpoint can be calculated by taking the average of upper and lower class limit.
Class Interval Frequency(f) x fx
5001-10,000 22 7500.5 165011
10,001-15,000 20 12500.5 250010
15,001-20,000 21 17500.5 367510.5
20,001-25000 23 22500.5 517511.5
25,001-30,000 24 27500.5 660012
fx can be computed by multiplying each x value with frequency in the respective class.
∑fx=165011+250010+367510.5+517511.5+660012=1960055
[tex]xbar=\frac{1960055}{110}=17818.68[/tex]
Thus, the mean salary is $17818.68.
The mean salary is approximately $17,818.18.
To find the mean salary, we need to calculate the average of all the salaries. Here’s the step-by-step process:
Determine the midpoint of each salary range, which is the average of the lower and upper bounds of that range.Multiply the midpoint of each range by the number of employees in that range to find the total contribution of each range to the sum of all salaries.Add up the contributions from all ranges to get the total sum of salaries.Divide the total sum of salaries by the total number of employees (110).Here's the detailed calculation:
[tex]Midpoint \ for \ $5,001-$10,000=(5001 + 10000) / 2 = 7,500[/tex][tex]Midpoint \ for\ $10,001-$15,000: (10001 + 15000) / 2 = 12,500\\Midpoint \ for \ $15,001-$20,000: (15001 + 20000) / 2 = 17,500\\Midpoint \ for \ $20,001-$25,000: (20001 + 25000) / 2 = 22,500\\Midpoint\ for\ $25,001-$30,000: (25001 + 30000) / 2 = 27,500[/tex]Now, multiply each midpoint by the number of employees in that range:
22 * 7,500 = 165,00020 * 12,500 = 250,00021 * 17,500 = 367,50023 * 22,500 = 517,50024 * 27,500 = 660,000Add these values together to get the total sum:
165,000 + 250,000 + 367,500 + 517,500 + 660,000 = 1,960,000
Now, divide by the total number of employees:
1,960,000 / 110 ≈ 17,818.18
Therefore, the mean salary is approximately $17,818.18.
In a given population, the percent of people with blue eyes is 32%. If 17 people from that population are randomly selected, what is the probability that exactly 9 of them will have blue eyes?
Answer:
The required probability is 0.0391.
Step-by-step explanation:
Consider the provided information.
The percent of people with blue eyes is 32%
Thus, p = 32% = 0.32
The percent of people with non blue eyes is 100%-32%=68% = 0.68
q = 68% = 0.68
We need to determine the probability that exactly 9 of them will have blue eyes if 17 are selected.
Thus, n=17 and r=9
Use the formula of binomial distribution: [tex]P(r) = ^nC_r p^r q^{n-r}[/tex]
Substitute the respective values in the above formula.
[tex]P(r=9) = ^{17}C_9 \times(0.32)^9 \times 0.68^{17-9}[/tex]
[tex]P(r=9) = \frac{17!}{9!8!} \times(0.32)^9 \times 0.68^{8}[/tex]
[tex]P(r=9) \approx0.0391[/tex]
Hence, the required probability is 0.0391.
Evaluate the integral by changing the order of integration in an appropriate way. Integral from 0 to 1 Integral from 0 to 1 Integral from x squared to 1 3 x font size decreased by 7 z e Superscript z font size decreased by 5 y squared Baseline dy dx dz
Answer:
The question is not clear, but here is a similar question with the same approach.
Integral from 0 to 1 Integral from 1 to 2 Integral from 2 to 3, { (x+y+z) } dx dy dz by changing the order of integration in an appropriate way.
Step-by-step explanation:
The approach is that of multiple integral where changing the order of integration is done appropriately
The step by step with detailed workings are shown in the attachment below.
A clothing business finds there is a linear relationship between the number of shirts, n, it can sell and the price, p, it can charge per shirt. In particular, historical data shows that 1000 shirts can be sold at a price of $30 , while 3000 shirts can be sold at a price of $22 . Find a linear equation in the form p=mn+b that gives the price p they can charge for n shirts.
Answer:
p = -0.004n+34
Step-by-step explanation:
The slope of the linear equation can be found by using the two given points (30, 1000) and (22, 3000):
[tex]m = \frac{p_1-p_0}{n_1-n_0}= \frac{30-22}{1000-3000}\\m=-0.004[/tex]
Applying the point (30, 1000) to the general form of a linear equation with m =-0.004, gives us the linear relationship between price (p) and number of shirts (n):
[tex](p-p_0) = m(n-n_0)\\p-30=-0.004(n-1000)\\p=-0.004n+34[/tex]
Answer: p = -0.004n+34
Step-by-step explanation:
Given that we need to derived the linear equation of the form;
p = mn+b ....1
Where p is the p is the price and n is the number of shirts they can sell
We need to substitute the values of p and n for the two cases to determine the slope m and constant b.
Case 1:
n = 1000 and p = $30
Substituting into the equation 1, we have:
30 = 1000m +b .....2
Case 2:
n = 3000 and p = $22
Substituting into the equation 1
22 = 3000m + b ......3
Substracting equation 2 from 3, we have
22-30 = 3000m-1000m +b-b
-8 = 2000m
m = -8/2000
m = -0.004
Substituting the value of m into equation 2
30 = 1000(-0.004) + b
b = 30 + 1000(0.004)
b = 30 + 4 = 34
b = 34
Therefore substituting the values of m and b into equation 1, we have our linear equation:
p = -0.004n+34
Find a unit vector orthogonal to the plane containing the points A= 1,0,0 , B= 3,−1,−3 , and C= 1,3,−2 .
Answer:
unit normal vector n will be n=(a,b,c) = (4/√171,11/√171,6/√171)
Step-by-step explanation:
There are several ways to solve this problem
1) build 2 vectors AB and BC such that the vectorial product ABxBC is the orthogonal vector to the plane , then find unit vector
2) since the 3 points belongs to the plane solve a linear system of 4 equation with 4 variables
for the second solution , the equation of the plane with normal vector n=(a,b,c) and containing the point (x₀,y₀,z₀) is
a*(x-x₀)+b*(y-y₀)+c*(z-z₀) =0
and
a²+b²+c² = 1 (unit vector)
then choosing A=(x₀,y₀,z₀)=(1,0,0)
a*(x-1)+b*(y-0)+c*(z-0) =0
for B
a*(3-1)+b*(-1-0)+c*(-3-0) =0
1) 2*a - b - 3*c =0
for C
a*(1-1)+b*(3-0)+c*(-2-0) =0
2) 3*b - 2*c=0 → b= 2/3*c
replacing in 1)
2*a - 2/3*c - 3*c =0
2*a-11/3*c=0 → a=11/6*c
thus
a²+b²+c² = 1
(11/6*c)²+(2/3*c)²+c² = 1
(121/36+4/9+1)*c² = 1
171/36*c²=1 → c= 6/√171
therefore
a=11/6*c = 11/6*6/√171= 11/√171
b=2/3*c= 2/3*6/√171= 4/√171
then the unit normal vector n will be
n=(a,b,c) = (4/√171,11/√171,6/√171)
An article reported the following data on oxygen consumption (mL/kg/min) for a sample of ten firefighters performing a fire-suppression simulation: 28.6 49.4 30.3 28.2 28.9 26.4 33.8 29.9 23.5 30.2Compute the following. (Round your answers to four decimal places.) a. The sample range mL/kg/minb. The sample variance s2 from the definition (i.e., by first computing deviations, then squaring them, etc.) mL2/kg2/min2c. The sample standard deviation mL/kg/mind. s2 using the shortcut method mL2/kg2/min
Answer:
a) The sample range 25.9 [tex]ml\slash kg\slash \min[/tex]
b) The sample variance is 49.344 [tex]ml^2 \slash kg^2 \slash min^2[/tex]
c) The sample standard deviation 7.0245 [tex]ml\slash kg\slash \min[/tex]
Step-by-step explanation:
We are given the following data on oxygen consumption (mL/kg/min):
28.6, 49.4, 30.3, 28.2, 28.9, 26.4, 33.8, 29.9, 23.5, 30.2
a) The sample range
Range = Maximum - Minimum
[tex]\text{Range} = 49.4 - 23.5 = 25.9[/tex]
The sample range 25.9 [tex]ml\slash kg\slash \min[/tex]
b) The sample variance
[tex]\text{Variance} = \displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}[/tex]
where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{309.2}{10} = 30.92[/tex]
Sum of squares of differences =
5.3824 + 341.5104 + 0.3844 + 7.3984 + 4.0804 + 20.4304 + 8.2944 + 1.0404 + 55.0564 + 0.5184 = 444.096
[tex]s^2 = \dfrac{444.096}{9} = 49.344[/tex]
The sample variance is 49.344 [tex]ml^2 \slash kg^2 \slash min^2[/tex]
c) The sample standard deviation
It is the square root of sample variance.
[tex]s = \sqrt{s^2} = \sqrt{49.344} = 7.0245[/tex]
The sample standard deviation 7.0245 [tex]ml\slash kg\slash \min[/tex]
Find an autonomous differential equation with all of the following properties:Equilibrium solutions at y=0 and y=3, y' > 0 for 0 < y < 3; and y' < 0 for -inf < y < 0 and 3 < y < inf.
The differential equation y' = y(y - 3) satisfies all the given conditions. It has equilibrium solutions at y = 0 and y = 3, y' > 0 for 0 < y < 3, and y' < 0 for -inf < y < 0 and 3 < y < inf.
Explanation:The autonomous differential equation that meets the given conditions is
y' = y(y-3). To confirm this, we can check each condition:
Equilibrium solutions at y=0 and y=3: Setting y' = 0 and solving for y in the differential equation gives y = 0 and y = 3, which satisfies this condition.y' > 0 for 0 < y < 3: The expression y(y-3) is positive in this interval, satisfying this condition.y' < 0 for -inf < y < 0 and 3 < y < inf: The expression y(y-3) is negative in these intervals, satisfying this condition.Learn more about Differential Equations here:https://brainly.com/question/33814182
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Of all the luggage handled by the airlines at JFK in the last 10 years, 5% was lost or stolen or damaged. For a randomly selected sample of 400 pieces of outgoing luggage at JFK International Airporta) Would it be unusual for 10 pieces of luggage to be lost?b) The mean and standard deviation of the binomial distribution is given byc) Would 40 pieces of luggage lost be an unusual occurrence?d) Find the probability that less than 6 pieces of luggage are lost or stolen.
Answer:
a) 10 pieces of luggage to be lost is not likely because its z-score is -2.29
b) mean=20 and standard deviation=4.359
c) 40 pieces of luggage lost is very extreme given that in average 5% of the luggage is lost.
d) The probability that less than 6 pieces of luggage are lost or stolen is 0.0007
Step-by-step explanation:
Let p be the proportion of stolen or damaged luggages of all the luggage handled by the airlines at JFK in the last 10 years.
p=0.05 (5%)
b) Mean and standard deviation of the binomial distribution is given as:
Mean= n×pStandard Deviation = [tex]\sqrt{n*p*(1-p)}[/tex] where n is the sample sizeThus,
Mean=400×0.05=20 and
Standard Deviation=[tex]\sqrt{400*0.05*0.95}[/tex] ≈ 4.359
a) We should calculate z-score of 10 pieces of luggage lost to decide if it is unusual.
Z-score can be calculated as follows:
[tex]z= \frac{X-M}{s}}[/tex] where
X = 10 pieces of luggage lost M is the mean lost luggage (20 pieces)s is the standard deviation (4.359)Thus, [tex]z= \frac{10-20}{4.359}[/tex] ≈ -2.29
This is unusual because it is in the 1st percentile and 99% of the possible number of lost luggage are higher than this score.
c) 40 pieces of lost luggage is also unusual because its z score is:
[tex]z= \frac{40-20}{4.359}[/tex] ≈ 4.59, in the 100th percentile, that is very extreme.
d) the probability that less than 6 pieces of luggage are lost or stolen
= P(z<z*) where z* is the z-score of 6 pieces of luggage are lost or stolen
That is [tex]z= \frac{6-20}{4.359}[/tex] ≈ -3.21
And P(z<3.21)= 0.0007
Ryan is a record executive for a hip hop label in Atlanta, Georgia. He has a new album coming out soon, and wants to know the best way to promote it, so he is considering many variables that may have an effect. He is considering three different album covers that may be used, four different television commercials that may be used, and two different album posters that may be used. Determine the number of different combinations he needs in order to test each album cover, television commercial, and album poster.
Answer: 24
Step-by-step explanation:
Given : Choices for album covers = 3
Choices for television commercials = 4
Choices for album posters = 2
Now , the number of different combinations he needs in order to test each album cover, television commercial, and album poster = ( Choices for album covers ) x (Choices for television commercials) x (Choices for album posters)
= 3 x 4 x 2 = 24
Hence, the number of different combinations he needs in order to test each album cover, television commercial, and album poster is 24.
If we collect a large sample of blood platelet counts and if our sample includes a single outlier, how will that outlier appear in a histogram?
A. The outlier will appear as a tall bar near one side of the distribution.
B. Since a histogram shows frequencies, not individual data values, the outlier will not appear. Instead, the outlier increases the frequency for its class by 1
C. The outlier will appear as the tallest bar near the center of the distribution
D. The outlier will appear as a bar far from all of the other bars with a height that corresponds to a frequency of 1.
Answer:
D. The outlier will appear as a bar far from all of the other bars with a height that corresponds to a frequency of 1.
Step-by-step explanation:
An histogram measures how many times each value appears in the set we are studying. That is, it is a frequency measure.
Suppose we have the following set:
S = {1,1,1,1,1,1, 2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,5,5,5,5,5,5,100}
1 appears 6 times. That means that when the X axis is 1, the y axis is 6.
2 appears 8 times. The means that when the X axis is 2, the y axis is 8.
...
100 appears 1 time. This means that when the X axis is 100, the y axis is 1. The X is the outlier, and it is quite far from the other values.
So the correct answer is:
D. The outlier will appear as a bar far from all of the other bars with a height that corresponds to a frequency of 1.
Between which two whole numbers does the product of 9 and 7 1/8 lie?
Answer:
We can answer that the product of 9 and 7 1/8 lies between 64 and 65, the two whole numbers asked.
Step-by-step explanation:
Let's find the product of 9 and 7 1/8, this way:
9 * 7 1/8 = 9 * 7.125
9 * 7.125 = (9 * 7) + (9 * .125)
63 + 1.125 = 64.125
We can answer that the product of 9 and 7 1/8 lies between 64 and 65, the two whole numbers asked.
Answer:
64 and 65
Step-by-step explanation:
9 X 7 1/8
=64.125
On a circle of radius 5 feet, give the degree measure of the angle that would subtend an arc of length 1 feet. Round your answer to the nearest hundredth, or two decimal places.
Answer:
11.46°
Step-by-step explanation:
Let x be the angle that yields an arc length of 1 feet, if r= 5 feet, applying the circumference length equation and assuming that a full circumference has 360 degrees:
[tex]1=\frac{x}{360}*2\pi r \\1=\frac{x}{360}*2\pi 5\\x=\frac{360}{2 \pi 5}=11.46^o[/tex]
Rounding to the nearest hundredth, the angle should be 11.46°
Thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally distributed, with a mean of 4.5 millimeters (mm) and a standard deviation of 1.7 mm. For a randomly found shard, find the following probabilities. (Round your answers to four decimal places.)
(a) the thickness is less than 3.0 mm.
(b) the thickness is more than 7.0 mm.
(c) the thickness is between 3.0 mm and 7.0 mm.
Answer:
a) [tex]P(X<3.0)=P(\frac{X-\mu}{\sigma}<\frac{3-\mu}{\sigma})=P(Z<\frac{3-4.5}{1.7})=P(z<-0.882)[/tex]
[tex]P(z<-0.882)=0.189[/tex]
b) [tex]P(X>7.0)=P(\frac{X-\mu}{\sigma}>\frac{7-\mu}{\sigma})=P(Z<\frac{7-4.5}{1.7})=P(z>1.47)[/tex]
[tex]P(z>1.47)=1-P(z<1.47) = 1-0.929=0.071[/tex]
c) [tex]P(3<X<7)=P(\frac{3-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{7-\mu}{\sigma})=P(\frac{3-4.5}{1.7}<Z<\frac{7-4.5}{1.7})=P(-0.882<z<1.47)[/tex]
[tex]P(-0.882<z<1.47)=P(z<1.47)-P(z<-0.882)=0.929-0.189=0.740[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Part a
Let X the random variable that represent the thickness of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(4.5,1.7)[/tex]
Where [tex]\mu=4.5[/tex] and [tex]\sigma=1.7[/tex]
We are interested on this probability
[tex]P(X<3.0)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X<3.0)=P(\frac{X-\mu}{\sigma}<\frac{3-\mu}{\sigma})=P(Z<\frac{3-4.5}{1.7})=P(z<-0.882)[/tex]
And we can find this probability using excel or the normal standard table:
[tex]P(z<-0.882)=0.189[/tex]
Part b
We are interested on this probability
[tex]P(X>3.0)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X>7.0)=P(\frac{X-\mu}{\sigma}>\frac{7-\mu}{\sigma})=P(Z<\frac{7-4.5}{1.7})=P(z>1.47)[/tex]
And we can find this probability using excel or the normal standard table:
[tex]P(z>1.47)=1-P(z<1.47) = 1-0.929=0.071[/tex]
Part c
[tex]P(3<X<7)=P(\frac{3-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{7-\mu}{\sigma})=P(\frac{3-4.5}{1.7}<Z<\frac{7-4.5}{1.7})=P(-0.882<z<1.47)[/tex]
And we can find this probability using excel or the normal standard table liek this:
[tex]P(-0.882<z<1.47)=P(z<1.47)-P(z<-0.882)=0.929-0.189=0.740[/tex]
A digital scale that provides weights to the nearest gram is used. a. What is the sample space for this experiment? Let A denote the event that a weight exceeds 11 grams, let B denote the event that a weight is less than or equal to 15 grams, and let C denote the event that a weight is greater than or equal to 8 grams and less than 12 grams. Describe the following events. b. A ∪ B c. A ∩ B d. A' e. A ∪ B ∪ C f. (A ∪ C)' g. A ∩ B ∩ C h. B' ∩ C i. A ∪ (B ∩ C)
Answer:
a) [tex] S= [x \in x \geq 0][/tex]
Because the weigth can't be negative.
b) AUB = "a weight exceeds 11 grams OR is less than or equal to 15 grams" and that is represent by all the sample space S.
c) A ∩ B ="a weight exceeds 11 grams AND is less than or equal to 15 grams" and that is represent by [tex] 11 < X \leq 15 [/tex] who is the same as [tex] 12 \leq X \leq 15[/tex] .
d) A' = "a weight NOT exceeds 11 grams" [tex] X \leq 11[/tex] that's the complement of the event A
e) A ∪ B ∪ C = "represent all the possib;e values for the sample space or S"
f) (A ∪ C)'="for this case (AUC) represent the weigths that exceeds 11 gr OR are between 8 and less than 11, so on this case values [tex] X \geq 8[/tex], so then the complement (AUC)' woudl be all the values [tex] X <8[/tex]"
g) A ∩ B ∩ C =[tex]\emptyset[/tex] since we don't have a common interval for the 3 events at the same time
h) B' ∩ C = The complement of B are the [tex] X>15[/tex] and for C we have values [tex] 8 \leq X <12[/tex] and the intersection between these two events is the [tex]\emptyset[/tex].
i) A ∪ (B ∩ C) = For this case (B ∩ C) represent the values between [tex] 8\leq X <12[/tex] and if we do the union A ∪ (B ∩ C) we got [tex] X \geq 8[/tex]
Step-by-step explanation:
For this case we have defined the following events, assuming that X represent the weight:
A= "a weight exceeds 11 grams" [tex]X>11[/tex]
B= " a weight is less than or equal to 15 grams" [tex] X \leq 15[/tex]
C= "a weight is greater than or equal to 8 grams and less than 12 grams" [tex] 8 \leq X < 12[/tex]
Part a
The sample space is given by:
[tex] S= [x \in x \geq 0][/tex]
Because the weigth can't be negative.
Part b
AUB = "a weight exceeds 11 grams OR is less than or equal to 15 grams" and that is represent by all the sample space S.
Part c
A ∩ B ="a weight exceeds 11 grams AND is less than or equal to 15 grams" and that is represent by [tex] 11 < X \leq 15 [/tex] who is the same as [tex] 12 \leq X \leq 15[/tex] .
Part d
A' = "a weight NOT exceeds 11 grams" [tex] X \leq 11[/tex] that's the complement of the event A
Part e
A ∪ B ∪ C = "represent all the possib;e values for the sample space or S"
Part f
(A ∪ C)'="for this case (AUC) represent the weigths that exceeds 11 gr OR are between 8 and less than 11, so on this case values [tex] X \geq 8[/tex], so then the complement (AUC)' woudl be all the values [tex] X <8[/tex]"
Part g
A ∩ B ∩ C =[tex]\emptyset[/tex] since we don't have a common interval for the 3 events at the same time
Part h
B' ∩ C = The complement of B are the values in[tex] X>15[/tex] and for C we have values [tex] 8 \leq X <12[/tex] and the intersection between these two events is the [tex]\emptyset[/tex].
Part i
A ∪ (B ∩ C) = For this case (B ∩ C) represent the values between [tex] 8\leq X <12[/tex] and if we do the union A ∪ (B ∩ C) we got [tex] X \geq 8[/tex]
The question refers to the mathematical concept of events and their union and intersection. In this context, A, B, and C represent specific weight ranges. Various combinations of these, such as A ∪ B or A ∩ B ∩ C, refer to different sets of possible weights.
Explanation:Sample space in this experiment would basically consist of all possible weights that the digital scale can measure to the nearest gram. In Principle, it's unlimited but practically it will depend on the maximum limit of the scale.
A is the event that a weight exceeds 11 grams. B is the event that a weight is less than or equal to 15 grams. C is the event that a weight is greater than or equal to 8 grams and less than 12 grams.
A ∪ B represents weights that exceed 11 grams or are less than or equal to 15 grams.A ∩ B refers to weights that are both more than 11 grams and less than or equal 15 grams.A' is the event of weights that do not exceed 11 grams.A ∪ B ∪ C comprises weights that are either more than 11 grams, less than or equal to 15 grams, or between 8 and 12 grams.(A ∪ C)' refers to weights that are not more than 11 grams and not between 8 and 12 grams.A ∩ B ∩ C represents weights that are more than 11 grams, less than or equal to 15 grams, and simultaneously between 8 and 12 grams.B' ∩ C indicates weights that are more than 15 grams and at the same time between 8 and 12 grams.A ∪ (B ∩ C) signifies weights that are more than 11 grams, or are less than or equal to 15 grams and between 8 and 12 grams.Learn more about Sets and Probability here:https://brainly.com/question/11179193
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The following confidence interval is obtained for a population proportion, p: 0.408 < p < 0.432
Use these confidence interval limits to find the margin of error, E.
A. 0.012
B. 0.013
C. 0.024
D. 0.420
Answer:
For this case the wisth of the interval represent 2ME and we have this:
[tex] 0.432-0.408 = 2ME[/tex]
And if we solve for ME we got:
[tex] ME = \frac{0.432-0.408}{2}=0.012[/tex]
So then the correct answer would be:
A. 0.012
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
Solution to the problem
The confidence interval for the mean is given by the following formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
The margin of error is given by:
[tex] ME= z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
For this case the wisth of the interval represent 2ME and we have this:
[tex] 0.432-0.408 = 2ME[/tex]
And if we solve for ME we got:
[tex] ME = \frac{0.432-0.408}{2}=0.012[/tex]
So then the correct answer would be:
A. 0.012
Suppose that the number of a certain type of computer that can be sold when its price is P (in dollars) is given by a linear function N(P). (a) Determine N(P) if N(1000) = 10000 and N(1700) = 5800. (Use symbolic notation and fractions where needed.) N(P) = (b) Select the statement that gives the slope of the graph of N(P), including units and describes what the slope represents. 6 computers per dollar -6 dollars per computer computers per dollar -6 computers per dollar (c) What is the change AN in the number of computers sold if the price is increased by AP = 110 dollars? (Give your answer as a whole number.) AN = computers
Answer:
a) N(P) = -6P + 16000
b) slope = -6 computers per dollar
That means the number of computer sold reduce by 6 per dollar increase in price.
c) ∆N = -660 computers
Step-by-step explanation:
Since N(P) is a linear function
N(P) = mP + C
Where m is the slope and C is the intercept.
Case 1
N(1000) = 10000
10000 = 1000m + C ....1
Case 2
N(1700) = 5800
5800 = 1700m + C ....2
Subtracting equation 1 from 2
700m = 5800 - 10000
m = -4200/700
m = -6
Substituting m = -6 into eqn 1
10000 = (-6)1000 + C
C = 10000+ 6000 = 16000
N(P) = -6P + 16000
b) slope = -6 computers per dollar
That means the number of computer sold reduce by 6 per dollar increase in price.
Slope is the change in number of computer sold per unit Change in price.
c) since slope m = -6 computers per dollar
∆P = 110 dollars
∆N = m × ∆P
Substituting the values,
∆N = -6 computers/dollar × 110 dollars
∆N = -660 computers.
The number of computer sold reduce by 660 when the price increase by 110 dollars
Answer:
110
Step-by-step explanation:
Consider the angle -6 radians. Determine the quadrant in which the terminal side of the angle is found and find the corresponding reference angle ¯ θ . Round the reference angle to 4 decimal places.
Final answer:
The angle -6 radians places the terminal side in the third quadrant with a reference angle of 0.28 radians when expressed in the positive acute form and rounded to four decimal places.
Explanation:
The question involves determining the quadrant of an angle measured in radians and finding its corresponding reference angle, which is a common task in Mathematics, specifically in the study of trigonometry.
To find the quadrant in which the terminal side of the angle -6 radians lies, we need to recall that one full revolution around the unit circle is 2π or approximately 6.28 radians. Since the given angle is negative, we move in the clockwise direction from the positive x-axis. Dividing -6 by 6.28, we realize that it is just shy of a full negative revolution, thereby placing the terminal side in the third quadrant.
To find the reference angle ¯θ, which is the positive acute angle the terminal side makes with the x-axis, we need to subtract the given angle from one full revolution (if necessary) and find the absolute value. Hence, ¯θ = |2×π - (-6)| = |6.28 - (-6)| = |0.28|, which is the same as 0.28 radians when rounded to four decimal places.
An engineer in charge of water rationing for the U.S. Army wants to determine if the average male soldier spends less time in the shower than the average female soldier. Let μm represent the average time in the shower of male soldiers and μf represent the average time in the shower of female soldiers.
a) What are the appropriate hypotheses for the engineer?
H0: μm = μf versus Ha: μm > μf
H0: σm = σf versus Ha: σm > σf
H0: μm = μf versus Ha: μm ≠ μf
H0: μm = μf versus Ha: μm < μf
b) Among a sample of 66 male soldiers the average shower time was found to be 2.68 minutes and the standard deviation was found to be 0.65 minutes. Among a sample of 69 female soldiers the average shower time was found to be 2.7 minutes and the standard deviation was found to be 0.5 minutes. What is the test statistic? Give your answer to three decimal places.
c) What is the P-value for the test? Give your answer to four decimal places.
d) Using a 0.1 level of significance, what is the appropriate conclusion?
Reject the claim that the average shower times are different for male and female soldiers because the P-value is greater than 0.1.
Conclude that the average shower time for males is less than the average shower time for females because the P-value is less than 0.1.
Fail to reject the claim that the average shower times are the same for male and female soldiers because the P-value is greater than 0.1.
Conclude that the average shower time for males is equal to the average shower time for females because the P-value is less than 0.1.
Answer:
a) H0: μm = μf versus Ha: μm < μf
b) [tex]t=\frac{(2.68-2.7)-0}{\sqrt{\frac{0.65^2}{66}+\frac{0.5^2}{69}}}}=-0.200[/tex]
c) [tex]p_v =P(t_{133}<-0.200)=0.421[/tex]
d) Fail to reject the claim that the average shower times are the same for male and female soldiers because the P-value is greater than 0.1.
Step-by-step explanation:
Data given and notation
[tex]\bar X_{m}=2.68[/tex] represent the mean for the sample male
[tex]\bar X_{f}=2.7[/tex] represent the mean for the sample female
[tex]s_{m}=0.65[/tex] represent the sample standard deviation for the males
[tex]s_{f}=0.5[/tex] represent the sample standard deviation for the females
[tex]n_{m}=66[/tex] sample size for the group male
[tex]n_{f}=69[/tex] sample size for the group female
t would represent the statistic (variable of interest)
Part a
Concepts and formulas to use
We need to conduct a hypothesis in order to check if the average male soldier spends less time in the shower than the average female soldier, the system of hypothesis would be:
Null hypothesis:[tex]\mu_{m}-\mu_{f}\geq 0[/tex]
Alternative hypothesis:[tex]\mu_{m} - \mu_{f}< 0[/tex]
Or equivalently:
Null hypothesis:[tex]\mu_{m}-\mu_{f}= 0[/tex]
Alternative hypothesis:[tex]\mu_{m} - \mu_{f}< 0[/tex]
And the best option is:
H0: μm = μf versus Ha: μm < μf
Part b
We don't have the population standard deviation, so for this case is better apply a t test to compare means, and the statistic is given by:
[tex]t=\frac{(\bar X_{m}-\bar X_{f})-\Delta}{\sqrt{\frac{s^2_{m}}{n_{m}}+\frac{s^2_{f}}{n_{f}}}}[/tex] (1)
And the degrees of freedom are given by [tex]df=n_m +n_f -2=66+69-2=133[/tex]
t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.
What is the test statistic?
With the info given we can replace in formula (1) like this:
[tex]t=\frac{(2.68-2.7)-0}{\sqrt{\frac{0.65^2}{66}+\frac{0.5^2}{69}}}}=-0.200[/tex]
Part c What is the p-value?
Since is a left tailed test the p value would be:
[tex]p_v =P(t_{133}<-0.200)=0.421[/tex]
Part d
The significance level given is [tex] \alpha =0.1[/tex] since the p value is higher than the significance level we can conclude:
Fail to reject the claim that the average shower times are the same for male and female soldiers because the P-value is greater than 0.1.
The question relates to hypothesis testing in statistics, insights into average shower times for male and female soldiers. After formulating the hypotheses, we calculate the test statistic using the provided sample data, and then find the corresponding P-value. If the P-value is less than our significance level, we reject the null hypothesis and side with the alternative hypothesis.
Explanation:The subject of this question falls under Mathematics, specifically it deals with hypothesis testing statistics.
a) The appropriate hypotheses for the engineer to consider would be: H0: μm = μf versus Ha: μm > μf
b) To calculate the test statistic, we use the formula for the test statistic in a independent two-sample t-test which incorporates the sample sizes, means, and standard deviations from the two groups. The formula is (avg(male soldiers)-avg(female soldiers))/sqrt(((sd(male soldiers))^2/number(male soldiers))+((sd(female soldiers))^2/number(female soldiers))). Plug in given values, we can obtain the test statistic.
c) The P-value can be obtained by looking up the test statistic in the T distribution table.
d) If the P-value is greater than the level of significance (0.1), we would fail to reject the null hypothesis. If the P-value is less than the level of significance, we would reject the null hypothesis. The conclusion is based on the specific P-value we computed.
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Which equation shows this relationship?
Answer: y=x+2
Step-by-step explanation:
Help me plsss I need it by tonight
Answer:
[tex]y=2x+2[/tex]
Step-by-step explanation:
we know that
The equation of a linear function has no exponents higher than 1, and the graph of a linear function is a straight line.
Verify each case
case 1) we have
[tex]y=2x+2[/tex]
Is the equation of a line in slope intercept form
so
Is a straight line and has no exponents higher than 1
therefore
Is a linear equation
case 2) we have
[tex]y=2x^{2}+2[/tex]
Is a quadratic equation
Is a curved line and has at least one exponent higher than 1,
therefore
Is a non-linear equation
case 3) we have
[tex]y=2x^{3}+2[/tex]
Is a cubic equation
Is a curved line and has at least one exponent higher than 1,
therefore
Is a non-linear equation
case 4) we have
[tex]y=2x^{4}+2[/tex]
Is a quartic equation
Is a curved line and has at least one exponent higher than 1,
therefore
Is a non-linear equation
The average age in a sample of 190 students at City College is 22. As a result of this sample, it can be concluded that the average age of all the students at City College
Answer:
The true mean [tex]\mu[/tex] it probably could be larger, smaller, or equal to 22
Step-by-step explanation:
False.
By definition the sample mean is defined as:
[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]
For this case the value for the sample size is n =190 and the calculated sample mean is [tex] \bar X=22[/tex]. This value represent the sample and for this case we can't assume that this value represent at all the population as the population mean [tex]\mu[/tex] since we probably have variability from the data of the students at City College.
So we can conclude that the true mean [tex]\mu[/tex] it probably could be larger, smaller, or equal to 22
The true mean it probably could be larger, smaller, or equal to 22.
Given that:
Total student, n = 190.
Average age of 190 students, [tex]\bar X = 22\\[/tex].
By definition the sample mean is defined as:
[tex]\bar X =\dfrac{{\sum_{i=1}^nX_i}}{n}[/tex]
For this case, the sample size n =190 and the calculated sample mean is [tex]\bar X = 22\\[/tex]. This value represent the sample and for this case can't assume that this value represent at all the population as the population mean since probably have variability from the data of the students at city college.
Hence, conclude that the true mean it probably could be larger, smaller, or equal to 22
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there are 21 members of the soccer team on the us. if each player carries on 4 pieces of equipment, how many pieces of equipment are on the bus?
Answer:
84
Step-by-step explanation:
21x4=84
Answer:the total number of pieces of equipment in the bus is 84
Step-by-step explanation:
The total number of members of the soccer team on the bus is 21.
if each player carries on 4 pieces of equipment, then the total number of pieces of equipment in the bus would be
21 × 4 = 84
Marian went shopping and bought clothes for $76.17 and books for $44.98. She then had a meal at the mall for $19.15. Which is the best estimate of the total cost of her shopping trip?
A. $130
B. $120
C. $150
D. $140
Answer:
D. $140.
Step-by-step explanation:
Given:
Cost of clothes = $76.17
Cost of books = $44.98
Cost of meal = $19.15
We need to find the best estimate of total cost of her shopping trip.
Solution:
First we will find the total cost of her shopping trip.
total cost of her shopping trip is equal to sum of Cost of clothes, Cost of books and Cost of meal.
framing in equation form we get;
total cost of her shopping trip = [tex]76.17+44.98+19.15 = \$140.3[/tex]
Now we can say that;
140.3 is close to 140
Hence Best estimate of total cost of shopping trip is $140.
Calculate the fuel economy of your family or personal car in terms of gallons per mile and the total cost for one fill-up git hub.
Answer:
0.028 Gallons/mile
Step-by-step explanation:
1 Gallon = 3.78541 Litres
1 Mile = 1.60934 Km
Total Fuel Expenditure (Dec 2019) = PKR 7,000
Fuel cost / Litre = 114Litre/Km
Total Fuel consumption = (7000/114)
= 61.40 Litres or 16.22 Gallons
Total Distance Travelled in Dec 2019 = 921 Km or 572.28 miles
Fuel Economy = 16.22/572.28
= 0.028 Gallons/mile
One time tank fill up cost = PKR 1,500
In a certain country, the life expectancy for women in 1900 was 4949 years and in 2000 it was 8181 years. Assuming that life expectancy between 2000 and 2100 increases by the same percentage as it did between 1900 and 2000, what will the life expectancy be for women in 2100?
Answer:
The life expectancy for 2100 will be 134 years.
Step-by-step explanation:
Consider the provided information.
In 1900 women life expectancy was 49 years and in 2000 it was 81 years.
First calculate the difference of life expectancy between 1900 to 2000.
81-49=32 years
Percentage change is:
[tex]\text{Percentage change}=\frac{32}{49}\times 100\approx65.31[/tex]
It is given that the increase in percentage remains same.
Therefore, next year life expectancy will be:
[tex]\begin{aligned}81+65.31\% \times81&=81(1+\frac{65.31}{100})\\&=81(1.6531)\\&=133.9011\approx134\end{aligned}[/tex]
Hence, the life expectancy for 2100 will be 134 years.
To find the life expectancy for women in 2100 with the same percentage increase as from 1900 to 2000, we calculate the percentage increase between these years and apply it to the year 2000 figure, resulting in an estimated life expectancy of approximately 134 years in 2100.
Explanation:In 1900, the life expectancy for women was 49 years, and in 2000 it was 81 years. To calculate the life expectancy for women in 2100, assuming it increases by the same percentage as from 1900 to 2000, we must first determine the percentage increase from 1900 to 2000.
The percentage increase is calculated as follows:
Percentage Increase = ((Life Expectancy in 2000 - Life Expectancy in 1900) / Life Expectancy in 1900) * 100
This gives us:
Percentage Increase = ((81 - 49) / 49) * 100 = (32 / 49) * 100 ≈ 65.31%
Now, we apply this percentage increase to the life expectancy in 2000 to estimate the life expectancy for women in 2100:
Estimated Life Expectancy in 2100 = Life Expectancy in 2000 * (1 + Percentage Increase)
Estimated Life Expectancy in 2100 = 81 * (1 + 0.6531) ≈ 133.9
Therefore, if the life expectancy for women continues to increase by the same percentage, it will be approximately 134 years in 2100.