The value at a distance of +1 standard deviation from the mean of the normally distributed data set with a mean of 39 and a standard deviation of 6.2 is 45.2.
To calculate the value at a distance of +1 standard deviation from the mean of a normally distributed data set with a mean of 39 and a standard deviation of 6.2, we need to use the formula below;
Z = (X - μ) / σ
Where:
Z = the number of standard deviations from the mean
X = the value of interest
μ = the mean of the data set
σ = the standard deviation of the data set
We can rearrange the formula above to solve for the value of interest:
X = Zσ + μAt +1 standard deviation,
we know that Z = 1.
Substituting into the formula above, we get:
X = 1(6.2) + 39
X = 6.2 + 39
X = 45.2
Therefore, the value at a distance of +1 standard deviation from the mean of the normally distributed data set with a mean of 39 and a standard deviation of 6.2 is 45.2.
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A manufacturer of yeast finds that the culture grows exponentially at the rate of 13% per hour . a) if the initial mass is 3.7 , what mass will be present after: 7 hours and then 2 days
After 7 hours, the mass of yeast will be approximately 9.718 grams. After 2 days (48 hours), the mass of yeast will be approximately 128.041 grams.
To calculate the mass of yeast after a certain time using exponential growth, we can use the formula:
[tex]M = M_0 * e^{(rt)}[/tex]
Where:
M is the final mass
M0 is the initial mass
e is the base of the natural logarithm (approximately 2.71828)
r is the growth rate (expressed as a decimal)
t is the time in hours
Let's calculate the mass of yeast after 7 hours:
M = 3.7 (initial mass)
r = 13% per hour
= 0.13
t = 7 hours
[tex]M = 3.7 * e^{(0.13 * 7)}[/tex]
Using a calculator, we can find that [tex]e^{(0.13 * 7)[/tex] is approximately 2.628.
M ≈ 3.7 * 2.628
≈ 9.718 grams
Now, let's calculate the mass of yeast after 2 days (48 hours):
M = 3.7 (initial mass)
r = 13% per hour
= 0.13
t = 48 hours
[tex]M = 3.7 * e^{(0.13 * 48)][/tex]
Using a calculator, we can find that [tex]e^{(0.13 * 48)}[/tex] is approximately 34.630.
M ≈ 3.7 * 34.630
≈ 128.041 grams
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a) After 7 hours, the mass will be approximately 7.8272.
b) After 2 days, the mass will be approximately 69.1614.
The growth of the yeast culture is exponential at a rate of 13% per hour.
To find the mass present after a certain time, we can use the formula for exponential growth:
Final mass = Initial mass × [tex](1 + growth ~rate)^{(number~ of~ hours)}[/tex]
a) After 7 hours:
Final mass = 3.7 ×[tex](1 + 0.13)^7[/tex]
To calculate this, we can plug in the values into a calculator or use the exponent rules:
Final mass = 3.7 × [tex](1.13)^{7}[/tex] ≈ 7.8272
Therefore, the mass present after 7 hours will be approximately 7.8272.
b) After 2 days:
Since there are 24 hours in a day, 2 days will be equivalent to 2 × 24 = 48 hours.
Final mass = 3.7 × [tex](1 + 0.13)^{48}[/tex]
Again, we can use a calculator or simplify using the exponent rules:
Final mass = 3.7 ×[tex](1.13)^{48}[/tex] ≈ 69.1614
Therefore, the mass present after 2 days will be approximately 69.1614.
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How does the number 32.4 change when you multiply it by 10 to the power of 2 ? select all that apply.
a). the digit 2 increases in value from 2 ones to 2 hundreds.
b). each place is multiplied by 1,000
c). the digit 3 shifts 2 places to the left, from the tens place to the thousands place.
The Options (a) and (c) apply to the question, i.e. the digit 2 increases in value from 2 ones to 2 hundred, and, the digit 3 shifts 2 places to the left, from the tens place to the thousands place.
32.4×10²=32.4×100=3240
Hence, digit 2 moves from one's place to a hundred's. (a) satisfied
And similarly, digit 3 moves from ten's place to thousand's place. Now, 1000=10³=10²×10.
Hence, it shifts 2 places to the left.
Therefore, (c) is satisfied.
As for (b), where the statement: Each place is multiplied by 1,000; the statement does not hold true since each digit is shifted 2 places, which indicates multiplied by 10²=100, not 1000.
Hence (a) and (c) applies to our question.
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Find the general solution to the following differential equations:
16y''-8y'+y=0
y"+y'-2y=0
y"+y'-2y = x^2
The general solution of the given differential equations are:
y = c₁e^(x/4) + c₂xe^(x/4) (for 16y''-8y'+y=0)
y = c₁e^x + c₂e^(-2x) (for y"+y'-2y=0)
y = c₁e^x + c₂e^(-2x) + (1/2)x
(for y"+y'-2y=x²)
Given differential equations are:
16y''-8y'+y=0
y"+y'-2y=0
y"+y'-2y = x²
To find the general solution to the given differential equations, we will solve these equations one by one.
(i) 16y'' - 8y' + y = 0
The characteristic equation is:
16m² - 8m + 1 = 0
Solving this quadratic equation, we get m = 1/4, 1/4
Hence, the general solution of the given differential equation is:
y = c₁e^(x/4) + c₂xe^(x/4)..................................................(1)
(ii) y" + y' - 2y = 0
The characteristic equation is:
m² + m - 2 = 0
Solving this quadratic equation, we get m = 1, -2
Hence, the general solution of the given differential equation is:
y = c₁e^x + c₂e^(-2x)..................................................(2)
(iii) y" + y' - 2y = x²
The characteristic equation is:
m² + m - 2 = 0
Solving this quadratic equation, we get m = 1, -2.
The complementary function (CF) of this differential equation is:
y = c₁e^x + c₂e^(-2x)..................................................(3)
Now, we will find the particular integral (PI). Let's assume that the PI of the differential equation is of the form:
y = Ax² + Bx + C
Substituting the value of y in the given differential equation, we get:
2A - 4A + 2Ax² + 4Ax - 2Ax² = x²
Equating the coefficients of x², x, and the constant terms on both sides, we get:
2A - 2A = 1,
4A - 4A = 0, and
2A = 0
Solving these equations, we get
A = 1/2,
B = 0, and
C = 0
Hence, the particular integral of the given differential equation is:
y = (1/2)x²..................................................(4)
The general solution of the given differential equation is the sum of CF and PI.
Hence, the general solution is:
y = c₁e^x + c₂e^(-2x) + (1/2)x²..................................................(5)
Conclusion: Therefore, the general solution of the given differential equations are:
y = c₁e^(x/4) + c₂xe^(x/4) (for 16y''-8y'+y=0)
y = c₁e^x + c₂e^(-2x) (for y"+y'-2y=0)
y = c₁e^x + c₂e^(-2x) + (1/2)x
(for y"+y'-2y=x²)
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The particular solution is: y = -1/2 x². The general solution is: y = c1 e^(-2x) + c2 e^(x) - 1/2 x²
The general solution of the given differential equations are:
Given differential equation: 16y'' - 8y' + y = 0
The auxiliary equation is: 16m² - 8m + 1 = 0
On solving the above quadratic equation, we get:
m = 1/4, 1/4
∴ General solution of the given differential equation is:
y = c1 e^(x/4) + c2 x e^(x/4)
Given differential equation: y" + y' - 2y = 0
The auxiliary equation is: m² + m - 2 = 0
On solving the above quadratic equation, we get:
m = -2, 1
∴ General solution of the given differential equation is:
y = c1 e^(-2x) + c2 e^(x)
Given differential equation: y" + y' - 2y = x²
The auxiliary equation is: m² + m - 2 = 0
On solving the above quadratic equation, we get:m = -2, 1
∴ The complementary solution is:y = c1 e^(-2x) + c2 e^(x)
Now we have to find the particular solution, let us assume the particular solution of the given differential equation:
y = ax² + bx + c
We will use the method of undetermined coefficients.
Substituting y in the differential equation:y" + y' - 2y = x²a(2) + 2a + b - 2ax² - 2bx - 2c = x²
Comparing the coefficients of x² on both sides, we get:-2a = 1
∴ a = -1/2
Comparing the coefficients of x on both sides, we get:-2b = 0 ∴ b = 0
Comparing the constant terms on both sides, we get:2c = 0 ∴ c = 0
Thus, the particular solution is: y = -1/2 x²
Now, the general solution is: y = c1 e^(-2x) + c2 e^(x) - 1/2 x²
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