Final answer:
The convection heat transfer coefficient of air at the upper surface of the plate is found to be 133.33 W/m·K by using the Fourier's law of heat conduction in conjunction with Newton's law of cooling and rearranging the expression to solve for the convection heat transfer coefficient.
Explanation:
To determine the convection heat transfer coefficient of air at the upper surface of the plate, we apply Fourier's law of heat conduction for a steady-state situation and Newton's law of cooling:
Fourier's law gives us:
Q = -kA(dT/dx)
Where:
Q is the rate of heat transfer (Watts),
k is the thermal conductivity of the material (W/m·K),
A is the surface area (m²),
(dT/dx) is the temperature gradient through the material (K/m).
We then use Newton's law of cooling:
Q = hA(T_surface - T_fluid)
Where:
h is the convection heat transfer coefficient (W/m·K),
T_surface is the surface temperature (K),
T_fluid is the fluid temperature (K).
We know the temperature difference across the plate (dT) is 10°C (60°C - 50°C) and assuming the heat transfer rate (Q) is the same throughout the plate due to steady-state conditions, we can rearrange and solve for h:
h = (k/d) (dT/(T_surface - T_fluid))
Given the values:
k = 80 W/m·K,
d = 15 cm = 0.15 m,
50°C (T_surface) - 10°C (T_fluid) = 40 K.
Substituting these into the equation gives us:
h = (80/0.15) (10/40) = 133.33 W/m·K
The convection heat transfer coefficient of air at the upper surface is [tex]\( 133.33 \) W/m^\²\cdotK.[/tex]
[tex]\[ q = h_1 (T_{s1} - T_{\infty}) = \frac{k}{L} (T_{s1} - T_{s2}) = h_2 (T_{s2} - T_{\infty}) \][/tex]
where:
- [tex]\( q \)[/tex] is the heat transfer rate per unit area[tex](W/m^\²)[/tex],
- [tex]\( h_1 \)[/tex] and [tex]\( h_2 \)[/tex] are the convection heat transfer coefficients at the upper and lower surfaces, respectively [tex](W/m^\²K)[/tex],
- [tex]\( T_{s1} \)[/tex] and [tex]\( T_{s2} \)[/tex] are the temperatures at the upper and lower surfaces of the plate, respectively (°C),
-[tex]\( T_{\infty} \)[/tex] is the ambient air temperature (°C),
- [tex]\( k \)[/tex] is the thermal conductivity of the plate [tex](W/m\cdot K)[/tex],
- [tex]\( L \)[/tex] is the thickness of the plate (m).
Given:
- [tex]\( T_{s1} = 50 \)\°C[/tex] (temperature at the upper surface),
- [tex]\( T_{s2} = 60 \)\°C[/tex] (temperature at the lower surface),
- [tex]\( T_{\infty} = 10 \)\°C[/tex] (air temperature),
- [tex]\( k = 80 \) W/m\cdotK[/tex] (thermal conductivity of the plate),
- [tex]\( L = 0.15 \)[/tex] m (thickness of the plate).
Since we want to find [tex]\( h_1 \)[/tex], we can equate the expressions for [tex]\( q \)[/tex] that involve [tex]\( h_1 \)[/tex] and [tex]\( k \)[/tex]:
[tex]\[ h_1 (T_{s1} - T_{\infty}) = \frac{k}{L} (T_{s1} - T_{s2}) \][/tex]
Now, solve for [tex]\( h_1 \)[/tex]:
[tex]\[ h_1 = \frac{k}{L} \cdot \frac{(T_{s1} - T_{s2})}{(T_{s1} - T_{\infty})} \][/tex]
Substitute the given values:
[tex]\[ h_1 = \frac{80 \text{ W/m\cdotK}}{0.15 \text{ m}} \cdot \frac{(50 \text{ \°C} - 60 \text{ \°C})}{(50 \text{ \°C} - 10 \text{ \°C})} \][/tex]
[tex]\[ h_1 = \frac{80}{0.15} \cdot \frac{-10}{40} \][/tex]
[tex]\[ h_1 = 533.33 \cdot \frac{-1}{4} \][/tex]
[tex]\[ h_1 = -133.33 \text{ W/m}^2\text{\cdotK} \][/tex]
The negative sign indicates that the heat transfer is from the plate to the air, which is expected. However, the convection heat transfer coefficient is typically reported as a positive value, so we take the absolute value:
[tex]h_1 = -133.33 {(\frac{W}{m}})^2} { \cdot K}[/tex]
Wich values from the given replacement set make up the solution set of the inequality
Answer:
ljgiuipuiopu
Step-by-step explanation:
A standard number cube numbered one through six on each side is rolled five times. What is the probability of landing on a three or four on all five rules?
Answer:
1/243 ≈ 0.004 (which is a low probability)Explanation:
The event landing on a three or four on a rolling is independent of the event of landing on a three or four on any other rolling.
Thus, for independent events you calculate the joint probability as the product of the probabilities of each event:
P(A and B and C and D and E) = P(A) × P(B) × P(C) × P(D) × P(E)Here, P(A) = P(B) = P(C) = P(D) = P(E) = probability of landing on a three or four.
Now to calculate the probability of landing on a three or four, you follow this reasoning:
Positive outcomes = three or fourNumber of positive outcomes = 2 Possible outcomes = one, two, three, four, five, or sixNumber of possible outcomes = 6Probability = numer of positive outcomes / numer of possible outcomesProbability = 2/6 = 1/3Finally, probability of landing on a three or four on all five rolls: (1/3)⁵ = 1/243 ≈ 0.004 ← answer
An article reported the following data on oxygen consumption (mL/kg/min) for a sample of ten firefighters performing a fire-suppression simulation: 29.6 49.3 30.1 28.3 28.7 26.3 33.5 29.8 23.7 31.0 Compute the following. (Round your answers to four decimal places.)
(a) the sample range(b) the sample variance s2 from the definition(c) the sample standard deviation(d) s2 using the shortcut method
Answer:
a) Range = 25.6
b) Variance = 48.1446
c) s = 6.9386
d) See below
Step-by-step explanation:
(a) the sample range
To obtain the sample range, we must sort the data increasingly :
23.7, 26.3, 28.3, 28.7, 29.6, 29.8, 30.1, 31, 33.5, 49.3
Then, find the distance between the largest and the lowest
Range = 49.3 - 23.7 = 25.6
(b) the sample variance [tex]s^2[/tex] from the definition
In order to find the variance from the definition, we need first the mean. The mean is defined as the average
[tex]\bar x=\displaystyle\frac{\displaystyle\sum_{i=1}^{n}x_i}{n}[/tex]
where the [tex]x_i[/tex] are the values of the data collected and n=10 the size of the sample.
So, the mean is
[tex]\bar x=31.03[/tex]
Now, the variance of the sample is defined as
[tex]s^2=\displaystyle\frac{\displaystyle\sum_{i=1}^n(x_i-\bar x)^2}{n-1}[/tex]
and we have that the variance is
[tex]s^2=48.1446[/tex]
(c) the sample standard deviation
The sample standard deviation is nothing but the square root of the variance
[tex]s=\sqrt{48.1446}=6.9386[/tex]
d) [tex]s^2[/tex] using the shortcut method
The shortcut method figures out the variance without having to compute the mean. The formula is
[tex]s^2=\frac{n\sum x_i^2-(\sum x_i)^2}{n(n-1)}[/tex]
where n=10 is the sample size .
So, using the shortcut method,
[tex]s^2=\frac{10*10,061.91-96,286.09}{10*9}=48.1446[/tex]
Let X and Y be independent random variables representing the lifetime (in 100 hours) of Type A and Type B light bulbs, respectively.
Both variables have exponential distributions, and the mean of X is 2 and the mean of Y is 3.
Find the joint pdf f(x|y) of X and Y.
Find the conditional pdf f_2(y|x) of Y given X = x.
Find the probability that a Type A bulb lasts at least 300 hours and a Type B bulb lasts at least 400 hours.
Given that a Type B bulb fails at 300 hours, find the probability that a Type A bulb lasts longer than 300 hours.
What is the expected total lifetime of two Type A bulbs and one Type B bulb?
What is the variance of the total lifetime of two Type A bulbs and one Type B bulb?
Answer:
a. is below in the explanation
b. 1/3e^(-y/3)
c.0.05882
d. 0.2231
e. 7 f.25
Step-by-step explanation:
Let and be independent random variables representing the lifetime (in 100 hours) of Type A and Type B light bulbs, respectively. Both variables have exponential distributions, and the mean of X is 2 and the mean of Y is 3.
a can be solved as follows [tex]\frac{1}{\alpha\beta } e^{\frac{-x-y}{\alpha\beta } }[/tex]
[tex]\frac{1}{6} e^{\frac{-\alpha }{2} -\beta/3 }[/tex]
1/6[tex]e^{\frac{-(3x+2y}{6 }[/tex]
b.
f(y/x)=f(x).f(y)/{f(y)}=f(y)
1/3e^(-y/3)
c. Find the probability that a Type A bulb lasts at least 300 hours and a Type B bulb lasts at least 400 hours.
[tex]\int\limits^\alpha _4 {1/6e^{-(3x+2y)/6} } \, dy[/tex]
1/2e^-(3x+8)/6
[tex]\int\limits^\alpha _3 {e^{-(3x+800)} } \, dx \\[/tex]
0.05882
Given that a type B fails at 300 hours . we find the probability that type A bulb lasts longer than 300hr
f(x>y|y=3)=f(x>3)
[tex]\int\limits^a_3 {1/2e^{-x/2} } \, dx[/tex]
0.2231
e.e) What is the expected total lifetime of two Type A bulbs and one Type B bulb?
E(2A+B)
(2E(A)+E(y)
2*2+3
=7
f. variance of the total lifetime of two types A bulbs and one type B bulb
V(2x+y)=
4*4+9
=25
The joint pdf of X and Y is (1/6)e^(-x/2-y/3). The conditional pdf of Y given X = x is (1/2)e^(-y/3). The probability that a Type A bulb lasts at least 300 hours and a Type B bulb lasts at least 400 hours is 9/250. The probability that a Type A bulb lasts longer than 300 hours, given that a Type B bulb fails at 300 hours, is approximately 0.9999. The expected total lifetime of two Type A bulbs and one Type B bulb is 7 100 hours. The variance of the total lifetime of two Type A bulbs and one Type B bulb is 10 100 hours.
Explanation:To find the joint pdf f(x|y) of X and Y, we need to find the product of the individual pdfs of X and Y since they are independent. Since X and Y follow exponential distributions with means of 2 and 3 respectively, their pdfs can be written as f(x) = (1/2)e^(-x/2) and f(y) = (1/3)e^(-y/3). Therefore, the joint pdf f(x|y) is given by:
f(x|y) = f(x)*f(y) = (1/2)e^(-x/2) * (1/3)e^(-y/3) = (1/6)e^(-x/2-y/3).
To find the conditional pdf f_2(y|x) of Y given X = x, we use Bayes' theorem:
f_2(y|x) = (f(x|y)*f(y))/f(x) = [(1/6)e^(-x/2-y/3) * (1/3)e^(-y/3)] / [(1/2)e^(-x/2)] = (1/2)e^(-y/3).
To find the probability that a Type A bulb lasts at least 300 hours and a Type B bulb lasts at least 400 hours, we integrate the joint pdf over the given range:
P(X >= 300, Y >= 400) = ∫∫(1/6)e^(-x/2-y/3)dxdy = ∫(3/2)e^(-x/2)dx * ∫e^(-y/3)dy. Solving the integrals, we get P(X >= 300, Y >= 400) = 9/250.
Given that a Type B bulb fails at 300 hours, we need to find the probability that a Type A bulb lasts longer than 300 hours. This is equivalent to finding P(X > 300 | Y = 300). Using the conditional pdf f_2(y|x) = (1/2)e^(-y/3), we integrate from 300 to infinity:
P(X > 300 | Y = 300) = ∫(1/2)e^(-y/3)dy = 1 - ∫(1/2)e^(-y/3)dy = 1 - (1/2)e^(-y/3) = 1 - (1/2)e^(-300/3) = 1 - e^(-100) ≈ 0.9999.
The expected total lifetime of two Type A bulbs and one Type B bulb is obtained by summing the means of X and Y twice and the mean of Y once:
E[Total Lifetime] = 2*E[X] + E[Y] = 2*2 + 3 = 7.
The variance of the total lifetime of two Type A bulbs and one Type B bulb is obtained by summing the variances of X and Y twice and the variance of Y once:
Var[Total Lifetime] = 2*Var[X] + Var[Y] = 2*(2^2) + 3^2 = 10.
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The function f is differentiable and ∫x0(3f(t)+5t)dt=sin(x). Determine the value of f′(1π).
Answer:
f'(π) = (-5/3)
Step-by-step explanation:
for the equation
∫₀ˣ (3f(t)+5t)dt=sin(x)
3*∫₀ˣ f(t) dt+5*∫₀ˣ t*dt=sin(x)
then
∫₀ˣ t*dt = (t²/2) |₀ˣ = x²/2 - 0²/2 = x²/2
thus
3*∫₀ˣ f(t) dt + 5* x²/2 = sin(x)
∫₀ˣ f(t) dt = 1/3*sin(x) - 5/6*x²
then applying differentiation
d/dx ( ∫₀ˣ f(t) dt) = f(x) - f(0) ( from the fundamental theorem of calculus)
d/dx (1/3*sin(x) - 5/6*x²) = 1/3*cos(x) - 5/3*x
therefore
f(x) - f(0) = 1/3*cos(x) - 5/3*x
f(x) = 1/3*cos(x) - 5/3*x + f(0)
applying differentiation again (f'(x) =df(x)/dx)
f'(x) = -1/3*sin(x) - 5/3
then
f'(π) = -1/3*sin(π) - 5/3 = 0 - 5/3 = -5/3
f'(π) = (-5/3)
Given expression is:
→ [tex]\int x_0 (3f(t)+5t) dt = sin(x)[/tex]
or,
→ [tex]3\times \int x_0 f(t) dt+5\times \intx_0 t\times dt = sin(x)[/tex]
then,
→ [tex]\int x_0 t\times dt = (\frac{t^2}{2} ) x_0[/tex]
[tex]\frac{x^2}{2} -\frac{0^2}{2} = \frac{x^2}{2}[/tex]
thus,
→ [tex]3\times \int x_0 f(t) +5\times \frac{x^2}{2} = sin(x)[/tex]
[tex]\int x_0 f(t) dt = \frac{1}{3}\times sin(x) - \frac{5}{6}\times x^2[/tex]
By applying differentiation, we get
→ [tex]\frac{d}{dx} (\int x_0 f(t) dt) = f(x) - f(0)[/tex]
[tex]\frac{d}{dx}(\frac{1}{3} sin(x)) - \frac{5}{6}x^2= \frac{1}{3} cos(x) - \frac{5}{3} x[/tex]
Therefore,
→ [tex]f(x) -f(0) = \frac{1}{3}cos (x) - \frac{5}{3} x[/tex]
[tex]f(x) = \frac{1}{3}cos (x) -\frac{5}{3} x+f(0)[/tex]
By differentiating again, we get
→ [tex]f'(x) = -\frac{1}{3}sin(\pi) -\frac{5}{3}[/tex]
[tex]= 0-\frac{5}{3}[/tex]
[tex]= -\frac{5}{3}[/tex]
Thus the response above is correct.
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uppose that a certain fortunate person has a net worth of $71.0 billion ($7.10×1010). If her stock has a good year and gains $3.20 billion (3.20×109) in value, what is her new net worth
The new net worth after a gain in stock value is $74.2 billion, or in the scientific notation it is $7.42×10¹0.
Explanation:In this problem, we have a person with a starting net worth of $71.0 billion, or $7.10×1010 in scientific notation. This person's stock has increased by $3.20 billion, or 3.20×109 in scientific notation. To find the new net worth, these two amounts should be added together.
Step 1: Write the starting net worth and stock increase in standard form: $71.0×10⁹ + $3.2×10⁹. Step 2: Add these two amounts together to find the new net worth. $74.2×10⁹ This is the new net worth in standard form.
To express this amount in scientific notation, it becomes $7.42×10¹0, which is your final answer.
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If 400 mL of a 20% w/v solution were diluted to 2 L, what would be the final percentage strength?
Answer:
4%
Step-by-step explanation:
A 20% solution with a total volume of 400 mL has 20% * 400 mL of solute.
20% * 400 mL = 80 mL
When you dilute the solution to 2 L, you introduce additional water, but no additional solute, so now you have the same 80 mL of solute in 2 L of total solution.
The concentration is:
(80 mL)/(2 L) = (80 mL)/(2000 mL) = 0.04
As a percent it is:
0.04 * 100% = 4%
"4" will be the final percentage strength.
According to the question,
20% solution with 400 mL has 20% × 400 mL of solute
then,
→ [tex]20 \ percent\times 400=80[/tex]
hence,
The concentration will be:
→ [tex]\frac{80 \ mL}{2 \ L} = \frac{80 \ mL}{2000 \ mL}[/tex]
[tex]= 0.04[/tex]
or,
[tex]= 0.04\times 100[/tex]
[tex]= 4[/tex] (%)
Thus the answer above is right.
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Which of the following statements properly define a variable? Select all that apply. a. Let w represent Enrique's weight b. Let d represent Enrique's distance (in miles). c. Let d represent Enrique's distance from home (in miles). d. Let d represent Enrique's distance from home. e. Let w represent Enrique's weight in pounds
Answer: c. Let d represent Enrique's distance from home (in miles).
e. Let w represent Enrique's weight in pounds
Step-by-step explanation:
A variable is a particular type of value that denotes an unknown quantity.
It is denotes by using alphabets.It can change.To define a variable , we need define a amount or quantity with units and in case of length an exact position is required.
a. Let w represent Enrique's weight
It is not a variable because no units is defined.
b. Let d represent Enrique's distance (in miles).
It is not a variable because the points to establish distance traveled by Enrique is not given .
c. Let d represent Enrique's distance from home (in miles).
It is a variable as both units and position as mentioned.
d. Let d represent Enrique's distance from home.
It is not a variable because no units is defined.
e. Let w represent Enrique's weight in pounds.
It is a variable because unit of weight is mentioned.
Hence, the correct statements that properly define a variable are :
c. Let d represent Enrique's distance from home (in miles).
e. Let w represent Enrique's weight in pounds
The sampling distribution for the variance estimator in ANOVA is chi-square regarless of the assumptions of the data
Answer:
False
Step-by-step explanation:
The mentioned statement is not true because the F sampling distribution is used for variance estimator in the Analysis of the variance. The Analysis of variance technique used to compare two different variance estimate to check the equality of population mean. The chi-square statistic involve the comparison of variance with specified variance whereas F-statistic is calculated by taking the ratio of two variances.
The sampling distribution for the variance estimator in Analysis of Variance (ANOVA) is Chi-square if key assumptions such as random selection, equal standard deviations, and normally distributed populations are met. If not, further statistical analysis is required.
Explanation:In Analysis of Variance (ANOVA), the sampling distribution for the variance estimator is indeed Chi-square, but only if the data meets certain assumptions. These assumptions are:
If these assumptions are met, we can proceed to calculate the F ratio, which is our test statistic for Analysis of Variance (ANOVA). However, if any of these assumptions aren't met, further statistical testing is required. Even in such cases, it's important to know the distribution of sample means, the sums, and the F Distribution to make accurate statistical conclusions.
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Without solving the equation 3x2−12x+7=0 find the sum of the squares of its roots.
Answer:
Step-by-step explanation:
3x²-12x+7=0
x = (-b ± √ b²-4ac) / 2a
from these equation: a = 3 , b = 12 , c = 7
Accordingly, b² - 4ac = 12²- 4 (3 x 7) = 144 - 4(21) = 144 - 84 = 60
2a= 2 x 3 = 6
Applying the quadratic formula :
x = (-12 ± √ 60 )/6
√ 60 rounded to 2 decimal digit using calculator 7.75
x = (-12 ± 7.75 )/6
(-12 ± 7.75 ) = (-12 + 7.75 ) = 5.75
(-12 - 7.75 ) = -19.75
x = 5.75/6 = 0.96
-19.75/6 = 3.29
sum of the square of roots = 0.96² + 3.29²
0.9261 + 10.8241 = 11.75
The sum of the squares of its roots of the quadratic equation
3x² - 12x + 7 = 0 is p² + q² = 34/3.
What is a quadratic equaton?A quadratic equation is an algebraic expression in the form of variables and constants.
A quadratic equation has two roots as its degree is two.
We have a quadratic equation 3x² - 12x + 7 = 0.
We know a quadratic equation ax² - bx + c = 0 can be written as,
x² - (b/a)x + (c/a) = 0 ⇒ x² - (p + q)x + pq = 0 where p and q are the roots.
∴ 3x² - 12x + 7 = 0.
x² - (12/3)x + 7/3 = 0.
Hence (p + q) = 4 and pq = 7/3.
Now, (p + q)² = p² + q² + 2pq.
16 = p² + q² + 14/3.
p² + q² = 16 - 14/3.
p² + q² = (48 - 14)/3.
p² + q² = 34/3.
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Cooley et al. (2009) randomly assigned either of two treatments, naturopathic care (diet counseling, breathing techniques, vitamins, and a herbal medicine) or stan- dardized psychotherapy (psychotherapy with breathing techniques and a placebo added), to 81 individuals having moderate to severe anxiety. Anxiety scores decreased an average of 57% in the naturopathic group and 31% in the psychotherapy group. Is this an experimental or observational study?
Answer: Experimental study
Step-by-step explanation:
This is an experimental study because it involves determination of the effect of each treatment on separate individuals. It involves selecting the 81 individuals randomly to be treated with a particular treatment (either of the two treatment). In experimental study the researcher apply separate treatments on a different groups and measure its effect on each of the groups.
This is an experimental study where participants were randomly assigned to either naturopathic care or standardized psychotherapy to assess their effects on anxiety scores.
Explanation:This is an example of an experimental study. In an experimental study, researchers assign participants to different groups and manipulate variables to determine cause-and-effect relationships. In this case, Cooley et al. randomly assigned individuals to either naturopathic care or standardized psychotherapy, which are the two treatments being compared. They then measured the effects of these treatments on anxiety scores.
The fact that participants were randomly assigned to the two treatments suggests that this study was experimental rather than observational.
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Let t^2y''+10ty'+8y=0Find all values of r such that y = tr satisfies the differential equation for t > tr0. If there is more than one correct answer, enter your answers as a comma separated list.
Answer:
The correct question is
Let [tex] t^2y''+10ty'+8y=0 [/tex]. Find all values of r such that [tex] y = t^r [/tex] satisfies the differential equation for t > 0. If there is more than one correct answer, enter your answers as a comma separated list.
The answer is r = 8, r = 1
Step-by-step explanation:
By substituting [tex] y = t^r [/tex] into the given equation [tex] t^2y''+10ty'+8y=0 [/tex].
That is substituting [tex] y = t^r [/tex] , [tex] y' = rt^{r-1} [/tex], [tex] y'' = r(r-1)t^{r-2} [/tex] into the given equation, we have
[tex] t^2 r(r-1)t^{r-2} +10t rt^{r-1}+8 t^r =0 [/tex]
Implies that [tex] r(r-1)t^r +10rt^r + 8 t^r =0 [/tex]
Implies that [tex] [r(r-1) +10r + 8]t^r = 0[/tex]
Implies that [tex] [r^2 - r +10r + 8]t^r = 0[/tex]
Implies that [tex] [r^2 - 9r + 8]t^r = 0[/tex]
Implies that [tex] r^2 - 9r + 8 = 0[/tex], assuming [tex] t^r \ne 0[/tex] being the assumed non- trivial solution.
The by solving this quadratic equation [tex] r^2 - 9r + 8 = 0[/tex] using factorization method, we have
[tex] r^2[/tex] – r – 8r + 8 = 0
Implies that r(r – 1) – 8(r – 1) = 0
Implies that (r – 8)(r – 1) = 0
Implies that r – 8 = 0 or r – 1 = 0
Implies that r = 8 or r = 1
Therefore, the value of r that satisfies the given equation is r = 8, r = 1.
Final answer:
The values of r such that y = t^r satisfies the given differential equation are r = -8 and r = -1.
Explanation:
The student is tasked with finding values of r such that y = t^r satisfies the given differential equation t^2y'' + 10ty' + 8y = 0 for t > 0. To solve this, we will substitute y = tr into the equation and find the characteristic equation that the values of r must satisfy.
Using the proposed solution y = t^r, we calculate the first and second derivatives of y:
y' = rt^{r-1}
y'' = r(r-1)t^{r-2}
Substituting these derivatives back into the original differential equation, we get:
t^2(r(r-1)t^{r-2}) + 10t(rt^{r-1}) + 8t^r = 0
Simplifying, we find the characteristic equation:
r(r-1) + 10r + 8 = 0
Which factors to:
(r + 8)(r + 1) = 0
Thus, the possible values of r are:
r = -8
r = -1
There is a mound of g pounds of gravel in a quarry. Throughout the day, 300 pounds of gravel are added to the mound. Two orders of 800 pounds are sold and the gravel is removed from the mound. At the end of the day, the mound has 1,500 pounds of gravel. Write the equation that describes the situation. Then solve for g
Final answer:
The equation that describes the situation is g + 300 - 1600 = 1500. Simplifying this equation gives g = 2800.
Explanation:
To write the equation that describes the situation, we start with the initial weight of the mound as g pounds. Throughout the day, 300 pounds are added to the mound, so the new weight is g + 300 pounds. Two orders of 800 pounds each are sold, so 1600 pounds are removed from the mound. At the end of the day, the weight of the mound is 1500 pounds. The equation that describes the situation is:
g + 300 - 1600 = 1500
Simplifying this equation, we have:
g - 1300 = 1500
Adding 1300 to both sides, the equation becomes:
g = 2800
Suppose that the probability that a person is killed by lightning, in a year is, independently of other people, 1/(500 million). Assume that the U.S. population is 300 million. Write down the exact expression for the probability P(3 or more people will be killed by lightning in the U.S. next year). Evaluate this expression to 6 decimal places. Write down a relevant approximate expression for the probability from (a). Justify briefly the approximation. Evaluate this expression to 6 decimal places.
Answer:
a) [tex] P(X \geq 3) = 1- P(X<3) = 1-P(X \leq 2) = 1-[P(X=0)+P(X=1) +P(X=2)][/tex]
The individual pprobabilities are:
[tex]P(X=0)=(300000000C0)(\frac{1}{500000000})^3 (1-\frac{1}{500000000})^{300000000-0}=0.548812[/tex]
[tex]P(X=1)=(300000000C0)(\frac{1}{500000000})^3 (1-\frac{1}{500000000})^{300000000-0}=0.329287[/tex]
[tex]P(X=2)=(300000000C0)(\frac{1}{500000000})^3 (1-\frac{1}{500000000})^{300000000-0}=0.098786[/tex]
And if we replace we got:
[tex] P(X \geq 3) = 1- [0.548812+0.329287+0.098786]=0.023115[/tex]
b) [tex] P(X \geq 3) = 1- P(X<3) = 1-P(X \leq 2) = 1-[P(X=0)+P(X=1) +P(X=2)][/tex]
And we can calculate this with the following excel formula:
[tex] =1-BINOM.DIST(2;300000000;(1/500000000);TRUE)[/tex]
And we got:
[tex] P(X\geq 3) = 0.023115[/tex]
Step-by-step explanation:
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Solution to the problem
Let X the random variable of interest, on this case we now that:
[tex]X \sim Binom(n=300000000, p=\frac{1}{500000000})[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
(a) Write down the exact expression for the probability P(3 or more people will be killed by lightning in the U.S. next year). Evaluate this expression to 6 decimal places.
For this case we want this probability, we are going to use the complement rule:
[tex] P(X \geq 3) = 1- P(X<3) = 1-P(X \leq 2) = 1-[P(X=0)+P(X=1) +P(X=2)][/tex]
The individual probabilities are:
[tex]P(X=0)=(300000000C0)(\frac{1}{500000000})^3 (1-\frac{1}{500000000})^{300000000-0}=0.548812[/tex]
[tex]P(X=1)=(300000000C0)(\frac{1}{500000000})^3 (1-\frac{1}{500000000})^{300000000-0}=0.329287[/tex]
[tex]P(X=2)=(300000000C0)(\frac{1}{500000000})^3 (1-\frac{1}{500000000})^{300000000-0}=0.098786[/tex]
And if we replace we got:
[tex] P(X \geq 3) = 1- [0.548812+0.329287+0.098786]=0.023115[/tex]
(b) Write down a relevant approximate expression for the probability from (a). Justify briefly the approximation. Evaluate this expression to 6 decimal places.
For this case we want this probability, we are going to use the complement rule:
[tex] P(X \geq 3) = 1- P(X<3) = 1-P(X \leq 2) = 1-[P(X=0)+P(X=1) +P(X=2)][/tex]
And we can calculate this with the following excel formula:
[tex] =1-BINOM.DIST(2;300000000;(1/500000000);TRUE)[/tex]
And we got:
[tex] P(X\geq 3) = 0.023115[/tex]
A wireless service provider has a support center customers can call to get questions answered about their cell phone accounts. The manager in charge of the support center has recently conducted a study in which she surveyed 2,300 customers. The Customers who called the support center were transferred to a third party who asked the customers a series of questions.
a. Indicate whether the data generated from this study will be considered cross-sectional or time series. Explain Why.
b. One of the questions asked customers was approximately how many minutes they had been on hold waiting to get through to a support person. What level of data measurement is obtained from this question? Explain.
c. Another question asked the customer to rate the service on a scale of 1-7, with 1 being the worst possible service and 7 being the best possible service. Which level of data measurement is achieved from this question? Will the data be quantitative or qualitative? Explain.
Answer:
A) The data generated is a cross-sectional data because time is not a factor .Secondly. 2,300(variables) people are selected at a particular time.
B) It is a ratio level of measurement because zero has meaning and it is clearly defined that is if the customer did not wait and was attended to immediately then the value is zero. Secondly there is a greater than and less than relationship that exits between the data for example if a customer wait for 20 minutes and another customer waits for 25 minutes, 25 is greater than 20. Hence the greater than and less than relation between data.
C)The level of measurement achieve from the data collected from this question is ordinal and the data is a qualitative. It’s ordinal because the relationship between the data set is focused on the order at which the data value are arranged. Secondly it’s a qualitative data set in the data value represents the difference in quality and not necessarily how much the difference is.
Step-by-step explanation:
Step One: Definition Of Cross-sectional data,
This can be defined as a type of data that is been collected by noting or observing many variables at a given period of time
Step Two:Definition Of Time series data
This can be defined as a type of that that is been collected by noting or observing a particular variable at different points of time. They are usually collected at fixed intervals
Step Three: Difference between cross-sectional data and time series data
The main difference between time series data and cross-sectional data is that for time series data the focus of observation for collecting the data s on the same variable over a period of time while for cross sectional data the focus is on the different variable at the same point in time
Step Four: Definition Of Level of measurement
This defines the relationship that exits among data values. The of level of measurement includes nominal, ordinal, interval, ratio
Step Five:Definition Of Ratio level of measurement
The ratio level of measurement or the ratio scale gives a description of the order and the exact difference that exit between data values, it also has a clear definition for zero value.
Step Six:Definition Of Ordinal level of measurement
For this level of measurement the main focus is the order of the data values, but the exact difference between each data values is not really specified looking at our question part c we know that a #7 is better than a #6 but we cannot say how much better it is.
The data from the wireless service provider's survey is cross-sectional, as it captures data at a single point in time. The question about hold time collects quantitative continuous data, while the service rating question provides ordinal data, which is qualitative.
The data generated from the study conducted by the wireless service provider's support center would be considered cross-sectional because it captures data at a single point in time rather than over multiple periods. Cross-sectional data is collected by observing many subjects at the same point in time or without regard to differences in time.
The level of data measurement obtained from the question asking customers about how long they had been on hold is quantitatively continuous. This is because the time spent on hold can be measured in units (such as minutes) that can take on any value within a given range, including fractions of a minute.
Regarding the customer service rating on a scale of 1-7, the level of data measurement is ordinal. This type of data is qualitative, as it categorizes data into different levels with a meaningful order, but the differences between the levels are not mathematically uniform or meaningful. It captures the order of the ratings but does not quantify the difference between each level.
A parcel of land is 8 ft longer than it is wide. Each diagonal from one corner to the opposite corner is 232 ft long.
What are the dimensions of the parcel?
width?
length?
The dimensions of the parcel are 80 ft (width) and 88 ft (length).
Explanation:To find the dimensions of the parcel, we can set up a system of equations using the given information. Let's denote the width of the parcel as x ft. Since the parcel is 8 ft longer than it is wide, the length can be represented as x + 8 ft. We can use the Pythagorean theorem to find the length of the diagonal: x^2 + (x + 8)^2 = 232^2. Solving this equation will give us the width and length of the parcel.
Expanding the equation, we get: x^2 + (x^2 + 16x + 64) = 53824. Combining like terms, we have: 2x^2 + 16x + 64 = 53824. Rearranging the equation, we have a quadratic equation: 2x^2 + 16x + (-53760) = 0. To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula.
By factoring, we find that the dimensions of the parcel are 80 ft (width) and 88 ft (length).
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01:39:07
The cost, c, of a ham sandwich at a deli varies directly with the number of sandwiches, n. If c = $54 when n is 9, what is the
cost of the sandwiches when n is 3?
$18
$21
$27
$48
Mark this and return
Next
Submit
Answer:
$18
Step-by-step explanation:
The cost, c, of a ham sandwich at a deli varies directly with the number of sandwiches, n. c = $54 when n is 9
This means that to buy 9 sandwiches, it costs $54. So one sandwich costs 54/9 = $6.
What is the cost of the sandwiches when n is 3?
To buy three sandwiches, it costs $6*3 = $18.
So the correct answer is:
$18
Answer:A
Step-by-step explanation:
$18
What is the difference between relative frequency and cumulative frequency?
A. Relative frequency of a class is the percentage of the data that falls in that class, while cumulative frequency of a class is the sum of the frequencies of that class and all previous classes.
B. Relative frequency of a class is the proportion of the data in that class, while cumulative frequency of a dass is the number of observations in that class.
C. Relative frequency of a class is the number of observations in that dass, while cumulative frequency of a dass is the sum of all the frequencies.
D. There is no difference between the two.
Answer:
A
Step-by-step explanation:
The relative frequency is calculated by dividing the frequency of that class to the sum of frequencies. It can be represented as
[tex]Relative frequency=\frac{f}{sum(f)}[/tex]
Hence, the relative frequency of class is the percentage or proportion of data lies in that class.
The cumulative frequency of a class is computed by adding the frequency of the respective class to the frequencies of all previous classes. The cumulative frequency of first class will always be equal to the frequency of first class.
Hence, the cumulative frequency for a class is the sum of frequency for that class and the frequencies of all previous classes.
Relative frequency is the proportion of total data that falls into a class, while cumulative frequency is the sum of frequencies of that class and all previous classes. Therefore, relative frequency illustrates the percentage of data in a certain class, while cumulative frequency shows the accumulation of data up to that point.
Explanation:The terms relative frequency and cumulative frequency both pertain to statistics; however, they represent different concepts. Relative frequency of a class is the percentage or proportion of the whole set of data that falls in that class. It's calculated by dividing the frequency of that class by the total number of data points.
On the other hand, cumulative frequency of a class is the sum of the frequencies of that class and all previous classes in a dataset. Essentially, it accumulates counts as you proceed through the dataset.
For instance, if you have five classes with frequencies of 1, 2, 3, 4 and 5, the relative frequencies would be 0.0667, 0.1333, 0.2, 0.2667, and 0.3333 respectively (assuming we divide each frequency by the total data points, which is 15 in this case). In contrast, the cumulative frequencies would be 1, 3, 6, 10 and 15, indicating the total count up to each class.
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A sample is a subset of all possible data values for a given subject under consideration.
(A) True
(B) False
Answer: True
Step-by-step explanation:
A subset is a term used in the sets topic In mathematics to describe a Part of A larger values being considered.
Sample In experimental procedures is known as Part of what is to be used to determine the reaction of a larger mass of specimen.
Taking these two definitions by side, it is quite evident that both of them represent taking part of a larger picture to form a smaller picture which does not totally deviate from the larger picture being considered.
Find the mean of the data summarized in the given frequency distribution. A company had 110 employees whose salaries are summarized in the frequency distribution below. Find the mean salary. Round to the nearest hundredth. Salery ($) Employees 5,001-10,000 27 10,001-15,000 23 15,001-20,000 12 20,001-25,000 18 25,001-30,000 30
Answer:
[tex] \bar X= \frac{1930055}{110}=17545.95[/tex]
Step-by-step explanation:
For this case we can construct the following table in order to find the mean for the grouped data:
Interval Frequency (fi) Midpoint (Xi) Xi *fi
5001-10000 27 7500.5 202513.5
10001-15000 23 12500.5 287511.5
15001-20000 12 17500.5 210006
20001-25000 18 22500.5 405009
25001-30000 30 27500.5 825015
Total 110 1930055
And the mean is calculated with the following formula:
[tex] \bar X= \frac{\sum_{i=1}^n x_i f_i}{n}[/tex]
Where [tex] n = \sum_{i=1}^n f_i = 110[/tex]
So then if we replace into the formula we got:
[tex] \bar X= \frac{1930055}{110}=17545.95[/tex]
Find values for θ that make each statement true (show steps)
1) sinθ = cos(25)
θ =
2) sin(θ/3 + 10 ) = cos θ
θ =
Answer:
Step-by-step explanation:
1. sinФ = cos 25
25 ° is in the between 0 and 90°
therefore it can simply represent
cosФ= sin (90-Ф) = sin (90 - 25) = sin 65
2. sin(Ф/3 + 10) = cos Ф
cos Ф = sin (90 -Ф)
sin(Ф/3 + 10) = cos Ф
sin(Ф/3 + 10) = cos Ф = sin(90-Ф)
Ф/3+10=90-Ф
10Ф+3/30 = 90-Ф
10Ф+3 = 30(90-Ф)
10Ф+3 = 2700-30Ф
10Ф+30Ф=2700-3
40Ф = 2697
Ф = 2697 / 40 = 67.425 ≅ 67.4°
Prove the following theorems algebraically:
(a) X(X′ + Y) = XY
(b) X + XY = X
(c) XY + XY′ = X
(d) (A + B)(A + B′) = A
Answer:
Algebraic Theorem Proofs
Step-by-step explanation:
We have to prove the following in the question:
Theorems used:
[tex]XX' = 0\\X + 1 = X\\X + 0 = X\\X+X'=1\\XX = X\\X+X=X[/tex]
(a) X(X′ + Y) = XY
[tex]X(X'+ Y) \\=XX'+XY\\=0+XY\\=XY[/tex]
(b) X + XY = X
[tex]X + XY \\=X(1+Y)\\=XY[/tex]
(c) XY + XY′ = X
[tex]XY + XY'\\=X(Y+Y')\\=X(1)\\=X[/tex]
(d) (A + B)(A + B′) = A
[tex](A + B)(A + B')\\=AA + AB' +BA + BB'\\=A + A(B+B')+0\\=A+A(1)\\=A[/tex]
I would be happy to provide step-by-step solutions to each of these theorems.
(a) Suppose we have X(X′ + Y) = XY. If we were to simplify using the law of distribution, we would have X(X′) + XY. However, since X′ is the complement of X or the negation of X, X(X′) becomes zero and the equation simplifies to 0 + XY which is XY. If we equate the two expressions X(X′ + Y) and XY, they are not equivalent.
(b) Now we have X + XY = X. On the left side, the expression can be rewritten as X(1 + Y). But observing it closely, the factor of X is multiplied by the bracket. However, bringing X out in front results in an equation that isn't equivalent to X.
(c) Then we move on to XY + XY′ = X. First, we attempt to simplify the left side of the equation using the law of distribution. It becomes X(Y + Y′). It is important to remember that Y + Y′ equates to 1 based on Boolean algebra. Hence, our equation simplifies to X(1), or simply "X". But considering the fact that Y' means the complement or the negation of Y, XY' would be zero (because anything multiplied by zero is zero). Therefore, the equation is not equivalent to X.
(d) Finally, (A + B)(A + B′) = A. The left-hand side of the equation can be expanded using distributive law into AA′ + AB + AB' + BB′. Looking at it closely, AA′ and BB′ become zero because of the property that states an element and its complement's product is always zero. Coming to AB and AB', since A' is the complement of A, AB + AB' simplifies to 0 + 0 which is equal to 0. But 0 doesn't equate to A making the equation non-equivalent.
In summary, all four theorems failed to prove equivalent under Boolean algebra rules.
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When 258 college students are randomly selected and surveyed, it is found that 106 own a car. Find a 99% confidence interval for the true proportion of all college students who own a car. 0.351 < p < 0.471 0.332 < p < 0.490 0.339 < p < 0.482 0.360 < p < 0.461
Answer: 0.332 < p < 0.490
Step-by-step explanation:
We know that the confidence interval for population proportion is given by :-
[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
, where n= sample size
[tex]\hat{p}[/tex] = sample proportion
z* = critical z-value.
As per given , we have
n= 258
Sample proportion of college students who own a car = [tex]\hat{p}=\dfrac{106}{258}\approx0.411[/tex]
Critical z-value for 99% confidence interval is 2.576. (By z-table)
Therefore , the 99% confidence interval for the true proportion(p) of all college students who own a car will be :[tex]0.411\pm (2.576)\sqrt{\dfrac{0.411(1-0.411)}{258}}\\\\=0.411\pm (2.576)\sqrt{0.00093829}\\\\= 0.411\pm (2.576)(0.0306315197142)\\\\=0.411\pm 0.0789=(0.411-0.0789,\ 0.411+0.0789)\\\\=(0.3321,\ 0.4899)\approx(0.332,\ 0.490)[/tex]
Hence, a 99% confidence interval for the true proportion of all college students who own a car : 0.332 < p < 0.490
At a certain college, 30% of the students major in engineering, 20% play club sports, and 10% both major in engineering and play club sports. A student is selected at random. What is the probability that the student is majoring in engineering?
The probability that a student selected at random majors in engineering is 30% which is 0.3.
The probability that the student both majors in engineering and play club sports is 10% which is 0.1.
For a student who is selected at random to be one who majors in engineering, there are two possible ways.
The student majors in engineering OR the Student both majors in engineering and plays club
The Probabilty that the student majors in Enginnering =
The probability that the student majors in engineering plus the probability that Student both majors in engineering and plays club sport
= 0.3+0.1= 0.4
If the sun is 68° above the horizon, find the length of the shadow cast by a building 88 ft tall. Round your answer to the nearest tenth.
Answer:
Length of shadow = 35.56 ft
Step-by-step explanation:
∠ BAC = 68°
Height of building = BC = 88 ft
Length of shadow = AC = ?
[tex]tan(68^{o})=\frac{BC}{AC}\\\\tan(68^{o})=\frac{88}{b}\\\\b=\frac{88}{tan(68^{o})}\\\\b=\frac{88}{2.475}\\\\b=35.56 \,ft[/tex]
The Length of shadow is 35.56 ft.
Given that,
If the sun is 68° above the horizon.
Based on the above information, the calculation is as follows:
[tex]tan 68 ^{\circ} = BC \div AC\\\\tan 68 ^{\circ} = 88 \div b\\\\b = 88 \div tan 68 ^{\circ} \\\\= 88 \div 2.475[/tex]
= 35.56 ft.
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(Anderson, 1.14) Assume that P(A) = 0.4 and P(B) = 0.7. Making no further assumptions on A and B, show that P(A ∩ B) satisfies 0.1 ≤ P(A ∩ B) ≤ 0.
Answer with Step-by-step explanation:
We are given that
P(A)=0.4 and P(B)=0.7
We know that
[tex]P(A)+P(B)+P(A\cap B)=P(A\cup B)[/tex]
We know that
Maximum value of [tex]P(A\cup B)[/tex]=1 and minimum value of [tex]P(A\cup B)[/tex]=0
[tex]0\leq P(A\cup B )\leq 1[/tex]
[tex]0\leq P(A)+P(B)-P(A\cap B)\leq 1[/tex]
[tex]0\leq 0.4+0.7-P(A\cap B)\leq 1[/tex]
[tex]0\leq 1.1-P(A\cap B)\leq 1[/tex]
[tex]0\leq 1.1-P(A\cap B)[/tex]
[tex]P(A\cap B)\leq 1.1[/tex]
It is not possible that [tex]P(A\cap B)[/tex] is equal to 1.1
[tex]1.1-P(A\cap B)\leq 1[/tex]
[tex]-P(A\cap B)\leq 1-1.1=-0.1[/tex]
Multiply by (-1) on both sides
[tex]P(A\cap B)\geq 0.1[/tex]
Again, [tex]P(A\cup B)\geq P(B)[/tex]
[tex]0.4+0.7-P(A\cap B)\geq 0.7[/tex]
[tex]1.1-P(A\cap B)\geq 0.7[/tex]
[tex]-P(A\cap B)\geq -1.1+0.7=-0.4[/tex]
Multiply by (-1) on both sides
[tex]P(A\cap B)\leq 0.4[/tex]
Hence, [tex]0.1\leq P(A\cap B)\leq 0.4[/tex]
What is the simple interest on a loan of $16 500 for 2 years at 8 per year ?
Answer:
$2 640
Step-by-step explanation:
Principal = $16 500; Time = 2 years Rate = 8%
Simple Interest = P × R × T/100
= $16 500 × 8 × 2/100
=$264 000/100
Simple Interest = $2 640
What is 4^150 divided by 2^150?
The sales of digital cameras (in millions of units) in year t is given by the function
f(t) = 3.06t + 6.84 (0 ≤ t ≤ 3)
where t = 0 corresponds to the end of 2001. Over that same period, the sales of film cameras (in millions of units) is given by
g(t) = −1.85t + 16.48 (0 ≤ t ≤ 3).
(a) Show that more film cameras than digital cameras were sold in 2001.
digital cameras
million
film cameras
million
(b) When did the sales of digital cameras first exceed those of film cameras?
Explanation on determining which type of cameras sold more in 2001 and when digital camera sales surpassed film camera sales.
Explanation:(a) To show that more film cameras than digital cameras were sold in 2001, we need to substitute t = 0 into both functions:
For digital cameras: f(0) = 3.06(0) + 6.84 = 6.84 million units
For film cameras: g(0) = -1.85(0) + 16.48 = 16.48 million units
Therefore, more film cameras (16.48 million units) were sold in 2001 than digital cameras (6.84 million units).
(b) To find when digital camera sales first exceeded film camera sales, we need to set the two functions equal to each other and solve for t:
3.06t + 6.84 = -1.85t + 16.48
4.91t = 9.64
t ≈ 1.96 years, which is approximately at the end of 2003.
Convert 7 kl 345 L to liters. (Enter a number as an exact integer or decimal.)
Answer:
The answer to your question is 7345 l
Step-by-step explanation:
Data
Convert 7kl to 345 L
Process
- To solve this question we must know the equivalence of kl to L.
1 kl ----------------- 1000 l
- Now, use proportions and cross multiplication to solve it
1 kl ---------------- 1000 l
7 kl---------------- x l
x = (7 x 1000) / 1
x = 7000 / 1
x = 7000 l
- Calculate the result
7000 + 345 = 7345 l