Answer:
A(t) = 300 -260e^(-t/50)
Step-by-step explanation:
The rate of change of A(t) is ...
A'(t) = 6 -6/300·A(t)
Rewriting, we have ...
A'(t) +(1/50)A(t) = 6
This has solution ...
A(t) = p + qe^-(t/50)
We need to find the values of p and q. Using the differential equation, we ahve ...
A'(t) = -q/50e^-(t/50) = 6 - (p +qe^-(t/50))/50
0 = 6 -p/50
p = 300
From the initial condition, ...
A(0) = 300 +q = 40
q = -260
So, the complete solution is ...
A(t) = 300 -260e^(-t/50)
___
The salt in the tank increases in exponentially decaying fashion from 40 grams to 300 grams with a time constant of 50 minutes.
The number of grams of salt in the tank at any time t is 40 grams. The inflow and outflow of brine do not affect the amount of salt in the tank because the solution is well-mixed, and the salt concentration remains constant.
To solve this problem, we need to set up a differential equation that describes the rate of change of salt in the tank over time. Let A(t) represent the number of grams of salt in the tank at time t.
Let's break down the components affecting the rate of change of salt in the tank:
Salt inflow rate: The brine is being pumped into the tank at a constant rate of 6 liters per minute, and it contains 1 gram of salt per liter. So, the rate of salt inflow is 6 grams per minute.
Salt outflow rate: The solution in the tank is being pumped out at the same rate of 6 liters per minute, which means the rate of salt outflow is also 6 grams per minute.
Mixing of the solution: Since the tank is well-mixed, the concentration of salt remains uniform throughout the tank.
Now, let's set up the differential equation for A(t):
dA/dt = Rate of salt inflow - Rate of salt outflow
dA/dt = 6 grams/min - 6 grams/min
dA/dt = 0
The above equation shows that the rate of change of salt in the tank is constant and equal to zero. This means the number of grams of salt in the tank remains constant over time.
Now, let's find the constant value of A(t) using the initial condition where the tank initially contains 40 grams of salt.
When t = 0, A(0) = 40 grams
Since the rate of change is zero, A(t) will be the same as the initial amount of salt in the tank at any time t:
A(t) = 40 grams
So, the number of grams of salt in the tank at any time t is 40 grams. The inflow and outflow of brine do not affect the amount of salt in the tank because the solution is well-mixed, and the salt concentration remains constant.
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How does the Binomial Theorem’s use Pascal’s triangle to expand binomials raised to positive integer powers?
Answer:
There are ways for quickly multiply out a binomial that's being raised by an exponent. Like
(a + b)0 = 1
(a + b)1 = a + b
(a + b)2 = a2 + 2ab + b2
(a + b)3 = (a + b)(a + b)2 = (a + b)(a2 + 2ab + b2) = a3 + 3a2b + 3ab2 + b3
and so on and so on
but there was this mathematician named Blaise Pascal and he found a numerical pattern, called Pascal's Triangle, for quickly expanding a binomial like the ones from earlier. It looks like this
1 1
2 1 2 1
3 1 3 3 1
4 1 4 6 4 1
5 1 5 10 10 5 1
Pascal's Triangle gives us the coefficients for an expanded binomial of the form (a + b)n, where n is the row of the triangle.
Hope this helps!
The Binomial Theorem uses Pascal's triangle for expanding binomials raised to positive integer powers. Pascal's triangle provides the coefficients for each term in the binomial expansion, simplifying the expansion process.
Explanation:
The Binomial Theorem uses Pascal's triangle to expand binomials that are raised to positive integer power. Pascal's triangle is a triangular array of binomial coefficients. Each line of the triangle represents the coefficients of the terms of a binomial expansion. Let's take for instance binomial expansion of (a + b)n = an+nan-1b+…from Pascal's triangle, the coefficients are 1,n, and so on. The role of Pascal's triangle in this case is pivotal in knowing the coefficients of each term in the binomial expansion and thus facilitates the expansion process.
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Verify that the function satisfies the three hypotheses of Rolle's Theorem on the given interval. Then find all numbers c that satisfy the conclusion of Rolle's Theorem. (Enter your answers as a comma-separated list.) f(x) = x3 − x2 − 12x + 7, [0, 4]
Rolle's theorem works for a function [tex]f(x)[/tex] over an interval [tex][a,b][/tex] if:
[tex]f(x)[/tex] is continuous on [tex][a,b][/tex][tex]f(x)[/tex] is differentiable on [tex](a,b)[/tex][tex]f(a)=f(b)[/tex]This is our case: [tex]f(x)[/tex] is a polynomial, so it is continuous and differentiable everywhere, and thus in particular it is continuous and differentiable over [0,4].
Also, we have
[tex]f(0)=7=f(4)[/tex]
So, we're guaranteed that there exists at least one point [tex]c\in(a,b)[/tex] such that [tex]f'(c)=0[/tex].
Let's compute the derivative:
[tex]f'(x)=3x^2-2x-12[/tex]
And we have
[tex]f'(x)=0 \iff x= \dfrac{1\pm\sqrt{37}}{3}[/tex]
In particular, we have
[tex]\dfrac{1+\sqrt{37}}{3}\approx 2.36[/tex]
so this is the point that satisfies Rolle's theorem.
The number C that satisfies the conclusion of Rolle's Theorem on the interval [0, 4] is: [tex]\[ c = \frac{1 + \sqrt{37}}{3} \][/tex]
To verify that the function [tex]\( f(x) = x^3 - x^2 - 12x + 7 \)[/tex] satisfies the three hypotheses of Rolle's Theorem on the interval [0, 4] and then to find all numbers c that satisfy the conclusion of Rolle's Theorem, follow these steps:
1. The function f is continuous on the closed interval [a, b]:
- [tex]\( f(x) = x^3 - x^2 - 12x + 7 \)[/tex] is a polynomial, and polynomials are continuous everywhere.
- Therefore, f is continuous on [0, 4].
2. The function f is differentiable on the open interval (a, b):
- Again, [tex]\( f(x) = x^3 - x^2 - 12x + 7 \)[/tex] is a polynomial, and polynomials are differentiable everywhere.
- Therefore, f is differentiable on (0, 4).
3. f(a) = f(b) :
- Calculate [tex]\( f(0) \)[/tex] and f(4):
[tex]\[ f(0) = 0^3 - 0^2 - 12 \cdot 0 + 7 = 7 \] \[ f(4) = 4^3 - 4^2 - 12 \cdot 4 + 7 = 64 - 16 - 48 + 7 = 7 \][/tex]
- Therefore, f(0) = f(4) = 7 .
Since all three hypotheses are satisfied, by Rolle's Theorem, there exists at least one number c in (0, 4) such that f'(c) = 0 .
Finding c
1. Compute the derivative of f:
[tex]\[ f(x) = x^3 - x^2 - 12x + 7 \] \[ f'(x) = 3x^2 - 2x - 12 \][/tex]
2. Set the derivative equal to zero and solve for x:
[tex]\[ f'(x) = 3x^2 - 2x - 12 = 0 \][/tex]
Solve the quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where a = 3, b = -2 , and c = -12 :
[tex]\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-12)}}{2 \cdot 3} \] \[ x = \frac{2 \pm \sqrt{4 + 144}}{6} \] \[ x = \frac{2 \pm \sqrt{148}}{6} \] \[ x = \frac{2 \pm 2\sqrt{37}}{6} \] \[ x = \frac{1 \pm \sqrt{37}}{3} \][/tex]
3. Check which solutions are in the interval (0, 4):
- For [tex]\( x = \frac{1 + \sqrt{37}}{3} \)[/tex]:
[tex]\[ \frac{1 + \sqrt{37}}{3} \approx \frac{1 + 6.08}{3} \approx \frac{7.08}{3} \approx 2.36 \][/tex]
- For [tex]\( x = \frac{1 - \sqrt{37}}{3} \)[/tex]:
[tex]\[ \frac{1 - \sqrt{37}}{3} \approx \frac{1 - 6.08}{3} \approx \frac{-5.08}{3} \approx -1.69 \][/tex]
- This solution is not in the interval (0, 4).
The rate of recipt of income from the sales of vases from 1988 to 1993 can be approximated by R(t)= 100/(t+0.87)^2 billion dollars per year, where t is time in years since January 1988. Estimate to the nearest $1 billion, the total change in income from January 1988 to January 1993.
Answer choices are: $43, $53, $137, $98, $117
Answer:
The correct option is 4.
Step-by-step explanation:
It is given that the rate of recipt of income from the sales of vases from 1988 to 1993 can be approximated by
[tex]R(t)=\frac{100}{(t+0.87)^2}[/tex]
billion dollars per year, where t is time in years since January 1988.
We need to estimate the total change in income from January 1988 to January 1993.
[tex]I=\int_{0}^{5}R(t)dt[/tex]
[tex]I=\int_{0}^{5}\frac{100}{(t+0.87)^2}dt[/tex]
[tex]I=100\int_{0}^{5}\frac{1}{(t+0.87)^2}dt[/tex]
On integration we get
[tex]I=-100[\frac{1}{(t+0.87)}]_{0}^{5}[/tex]
[tex]I=-100(\frac{1}{5+0.87}-\frac{1}{0+0.87})[/tex]
[tex]I=-100(-0.979)[/tex]
[tex]I=97.9[/tex]
[tex]I\approx 98[/tex]
The total change in income from January 1988 to January 1993 is $98. Therefore the correct option is 4.
Given the expression A ∩ (B − C), can you use the distributive law to say:
A ∩ (B − C) = (A ∩ B) – (B ∩ C)
Why or why not?
Answer:
Step-by-step explanation:
We know that X-Y = X∩Y'
Using it ,we get
A ∩(B∩C') which can be written as (A∩B)∩C' or (A∩B) - C
And right hand side is
(A∩B)-(B∩C) =B ∩(A-C) = B∩(A∩C') = A∩B∩C'
Since both left and right side both leads to same expression A∩B∩C'
Therefore both are equal.
A scientist mixes water (containing no salt) with a solution that contains 35% salt. She wants to obtain 175 ounces of a mixture that is 20% salt. How many ounces of water and how many ounces of the 35% salt solution should she use?
Answer: There is 100 ounces of 35% salt solution and 100 ounces of water.
Step-by-step explanation:
Since we have given that
Percent of salt in a solution = 35%
Percent of salt in a mixture = 20%
Number of ounces of a mixture = 175 ounces
We need to find the number of ounces of water and salt as well as .
We would use "Mixture and Allegation":
Salt Water
35% 0%
20%
---------------------------------------------------------------
20% - 0% : 35% - 20%
20% : 15%
4 : 3
So, Ratio of salt and water in the mixture is 4 : 3.
So, Number of ounces of salt in the mixture is given by
[tex]\dfrac{4}{7}\times 175\\\\=100\ ounces[/tex]
Number of ounces of water in the mixture is given by
[tex]\dfrac{3}{7}\times 175\\\\=75\ ounces[/tex]
Hence, there is 100 ounces of 35% salt solution and 100 ounces of water.
Final answer:
To make a 175-ounce mixture with 20% salt, the scientist should mix 75 ounces of water with 100 ounces of the 35% salt solution.
Explanation:
The student is asking for help with a typical mixture problem in algebra that involves determining the amounts of two different concentrations in order to create a mixture with a desired concentration. To solve this, we can set up two equations, one based on the total volume of the mixture and one based on the total amount of salt.
Let x be the amount of water (0% salt) and y be the amount of the 35% salt solution. The total volume should be 175 ounces, so we have:
Equation 1: x + y = 175
The total amount of salt in the solution must be 20% of 175 ounces, which is 35 ounces. So for the salt amount, we have:
Equation 2: 0.35y = 35
Solving Equation 2 gives us y = 100 ounces for the 35% solution. Substituting y in Equation 1, we get x = 75 ounces for the water. Therefore, the scientist should mix 75 ounces of water with 100 ounces of the 35% salt solution to obtain 175 ounces of a 20% salt mixture.
An auto license plate consists of 6 digits; the first three are any letter (from the 26 alphabets), and the last three are any number from 0 to 9. For example, AAA 000, ABC 123, and ZZZ 999 are three possible license plate numbers. How many different license plate numbers may be created?
Answer: There are 17576000 ways to generate different license plates.
Step-by-step explanation:
Since we have given that
Numbers are given = 0 to 9 = 10 numbers
Number of letters = 26
We need to generate the license plate numbers.
Since there are repetition allowed.
We would use "Fundamental theorem of counting".
So, the number of different license numbers may be created as given as
[tex]26\times 26\times 26\times 10\times 10\times 10\\\\=26^3\times 10^3\\\\=17576\times 1000\\\\=17576000[/tex]
Hence, there are 17576000 ways to generate different license plates.
You take out a simple interest loan for $ 922 to pay for tuition. If the annual interest rate is 6 % and the loan must be repaid in 6 months, find the amount that you, the borrower, will have to repay. Round your answer to the nearest cent.
Answer:
The total amount to be repaid is equal to $949.66
Step-by-step explanation:
Simple interest is a type of interest which is usually applied on short term loans, where when a payment is made towards this kind of interest the payment first goes towards monthly interest and then the remainder is reverted towards the principal.
FORMULA FOR CALCULATING SIMPLE INTEREST =
[tex]\frac{PRINCIPAL \times RATE OF INTEREST \times TIME PERIOD}{100}[/tex]
Here principal = $922
interest rate = 6%
time period = 6 months (when made per annum it will be 6/12)
[tex]\frac{\$ 922 \times 6 \times 1}{100\times 2}[/tex]
SIMPLE INTEREST IS EQUAL TO $27.66
The total amount that is to be repaid is equal to
PRINCIPAL + SIMPLE INTEREST
= $922 + $27.66
= $949.66
A researcher testing the effects of two treatments for anxiety computed a 95% confidence interval for the difference between the mean of treatment 1 and the mean of treatment 2. If this confidence interval includes the value of zero, then she should reject the null hypothesis that the two population means are equal: true or false?
Answer:
False
Step-by-step explanation:
Given that a researcher testing the effects of two treatments for anxiety computed a 95% confidence interval for the difference between the mean of treatment 1 and the mean of treatment 2
Also that confidence interval includes zero.
When confidence interval includes zero, we need not reject the null hypothesis since null hypothesis claims that difference =0
When confidence interval includes 0 it confirms that there is no difference and hence null hypothesis should be accepted.
Final answer:
The statement is false; if a 95% confidence interval for the difference between two treatment means includes zero, it indicates no significant difference, implying the null hypothesis cannot be rejected.
Explanation:
If a researcher testing the effects of two treatments for anxiety computed a 95% confidence interval for the difference between the mean of treatment 1 and the mean of treatment 2, and this confidence interval includes the value of zero, the correct interpretation is false regarding the statement that she should reject the null hypothesis that the two population means are equal. A confidence interval that contains zero indicates that the difference between the two treatments could be zero, suggesting there is no significant difference between the two treatments.
Therefore, there is insufficient evidence to reject the null hypothesis, and it is retained.
Understanding confidence intervals is crucial in hypothesis testing. A 95% confidence interval includes the true mean 95% of the time if the same experiment is repeated under the same conditions. Including zero in this interval suggests that the effect of the treatments could be negligible, meaning we cannot confidently claim there is a difference between the treatments based on the data provided.
You invested a total of $9,000 at 4 1/2 % and 5% simple interest. During one year, the two accounts earned $435. How much did you invest in each account
Answer:
The amount invested at 4.5% was [tex]\$3,000[/tex]
The amount invested at 5% was [tex]\$6,000[/tex]
Step-by-step explanation:
we know that
The simple interest formula is equal to
[tex]I=P(rt)[/tex]
where
I is the Final Interest Value
P is the Principal amount of money to be invested
r is the rate of interest
t is Number of Time Periods
Let
x -----> the amount invested at 4.5%
9,000-x -----> the amount invested at 5%
in this problem we have
[tex]t=1\ year\\ P1=\$x\\ P2=\$(9,000-x)\\I=\$435\\r1=0.045\\r2=0.05[/tex]
substitute
[tex]435=x(0.045*1)+(9,000-x)(0.05*1)[/tex]
[tex]435=0.045x+450-0.05x[/tex]
[tex]0.05x-0.045x=450-435[/tex]
[tex]0.005x=15[/tex]
[tex]x=\$3,000[/tex]
so
[tex]9,000-x=\$6,000[/tex]
therefore
The amount invested at 4.5% was [tex]\$3,000[/tex]
The amount invested at 5% was [tex]\$6,000[/tex]
show that {(1,1,0),(1,0,1),(0,1,1)} is linearly independent subset of r^3
Answer: Yes, the given set of vectors is a linearly independent subset of R³.
Step-by-step explanation: We are given to show that the following set of three vectors is a linearly independent subset of R³ :
B = {(1, 1, 0), (1, 0, 1), (0, 1, 1)} .
Since the given set contains three vectors which is equal to the dimension of R³, so it is a subset of R³.
To check the linear independence, we will find the determinant formed by theses three vectors as rows.
If the value of the determinant is non zero, then the set of vectors is linearly independent. Otherwise, it is dependent.
The value of the determinant can be found as follows :
[tex]D\\\\\\=\begin{vmatrix}1 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 1\end{vmatrix}\\\\\\=1(0\times1-1\times1)+1(1\times0-1\times1)+0(1\times1-0\times0)\\\\=1\times(-1)+1\times(1)+0\\\\=-1-1\\\\=-2\neq0.[/tex]
Since the determinant is not equal to 0, so the given set of vectors is a linearly independent subset of R³.
Thus, the given set is a linearly independent subset of R³.
Eliminate the parameter to find a Cartesian equation of the following curve: x(t) = cos^2 (6t), y(t) = sin^2(6t) Choose the answer from the following: y(x) = 1 + x y(x) = 1 - x y(x) = 1 - 6x
Answer:
y(x) = 1 - x
Step-by-step explanation:
Given the two parametric equations:
[tex] x(t)=cos^{2}(6t) [/tex] ---(1)
[tex] sin^{2}(6t) [/tex] ----(2)
We can add eq (1) and eq (2) and consider the trigonometric identity:
[tex] cos^{2}(6t)+sin^(6t) = 1 [/tex]
so,
[tex] x+y=1 [/tex]
in other way we can express this like:
[tex] y(x)=1-x [tex].
Find all the zeros of the polynomial function. x^3 + 2x^2 -5x-6 f(x) a) (-3) b) (-2, 1, 3 c) (-3, -1, 2) d) -1) e) none
Answer:1,2,3
Step-by-step explanation:
F(x)=[tex]x^{3}[/tex]+2[tex]x^2[/tex]-[tex]5x[/tex]-[tex]6[/tex]=0
disintegrating 2[tex]x^2[/tex] to [tex] x^2[/tex] + [tex]x^2[/tex]
[tex]x^{3}[/tex]+[tex]x^2[/tex]+[tex]x^2[/tex]-[tex]5x-6[/tex]=0
[tex]x^2[/tex][tex]\left ( x+1\right )[/tex]+[tex]x^2[/tex]-5x-6=0
[tex]x^2[/tex][tex]\left ( x+1\right )[/tex]+[tex]x^2[/tex]-6x+x-6=0
[tex]x^2[/tex][tex]\left ( x+1\right )[/tex]+[tex]\left (x-6 \right )[/tex][tex]\left ( x+1\right )[/tex]=0
[tex]\left ( x+1\right )[/tex][tex]\left ( x^2+x-6\right )[/tex]=0
[tex]\left ( x+1\right )[/tex][tex]\left ( x^2+3x-2x-6\right )[/tex]=0
[tex]\left ( x+1\right )[/tex][tex]\left ( x+3\right )[/tex][tex]\left ( x-2\right )[/tex]=0
Express the given expanded numeral as a Hindu-Arabic numeral. (8x102) +(4x10)(2x1)
Answer:
The Hindu-Arabic numeral form of the given expanded numeral is 842.
Step-by-step explanation:
The given expanded numeral is
[tex](8\times 10^2)+(4\times 10)+(2\times 1)[/tex]
We need to express the given expanded numeral as a Hindu-Arabic numeral.
According to Hindu-Arabic numeral the given expanded numeral is written as
[tex](8\times 10^2)+(4\times 10)+(2\times 1)=(8\times 100)+(4\times 10)+(2\times 1)[/tex]
On simplification we get,
[tex](8\times 10^2)+(4\times 10)+(2\times 1)=(800)+(40)+(2)[/tex]
[tex](8\times 10^2)+(4\times 10)+(2\times 1)=842[/tex]
Therefore the Hindu-Arabic numeral form of the given expanded numeral is 842.
Any equation or inequality with variables in it is a predicate in the domain of real numbers. For the following statement, tell whether the statement is true or false. (∀x)(x4> x)
The statement is given by:
∀ x , [tex]x^4>x[/tex]
This statement is false
Since, if we consider,
[tex]x=\dfrac{1}{2}[/tex]
then we have:
[tex]x^4=(\dfrac{1}{2})^4\\\\i.e.\\\\x^4=\dfrac{1}{2^4}\\\\i.e.\\\\x^4=\dfrac{1}{16}[/tex]
Also, we know that:
[tex]\dfrac{1}{16}<\dfrac{1}{2}[/tex]
( Since, two number with same numerator; the number with greater denominator is smaller than the number with the smaller denominator )
Hence, we get:
[tex]x^4<x[/tex]
when [tex]x=\dfrac{1}{2}[/tex]
Hence, the result :
[tex]x^4>x[/tex] is not true for all x belonging to real numbers.
Hence, the given statement is a FALSE statement.
Final answer:
The statement (∀x)(x⁴ > x) is false, as it does not hold true for all real numbers. For instance, when x is a negative number like -1, the inequality x⁴ > x is false.
Explanation:
False
Explanation:
Given statement: (∀x)(x⁴ > x)
This statement asserts that for all real numbers x, x⁴ will be greater than x. However, this statement is false because it doesn't hold for all real numbers. For instance, when x is a negative number such as -1, (-1)4 is greater than -1, which means the inequality x⁴ > x is false.
One common system for computing a grade point average (GPA) assigns 4 points to an A, 3 points to a B, 2 points to a C, 1 point to a D, and 0 points to an F. What is the GPA of a student who gets an A in a 3-credit course, a B in each of three 4-credit courses, a C in a 3-credit course, and a D in a 2-credit course?
Answer:
2.8
Step-by-step explanation:
The weighted average is found by dividing the total number of points by the total number of credits.
GPA = (4×3 + 3×4 + 3×4 + 3×4 + 2×3 + 1×2) / (3 + 4 + 4 + 4 + 3 + 2)
GPA = 56 / 20
GPA = 2.8
The GPA of the student will be 2.8.
What is Algebra?Algebra is the study of mathematical symbols, and the rule is the manipulation of those symbols.
One common system for computing a grade point average (GPA) assigns 4 points to an A, 3 points to a B, 2 points to a C, 1 point to a D, and 0 points to an F.
Then the GPA of a student who gets an A in a 3-credit course, a B in each of three 4-credit courses, a C in a 3-credit course, and a D in a 2-credit course will be
GPA = (4×3 + 3×4 + 3×4 + 3×4 + 2×3 + 1×2) / (3 + 4 + 4 + 4 + 3 + 2)
GPA = 56 / 20
GPA = 2.8
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A lawyer has found 60 investors for a limited partnership to purchase an inner-city apartment building, with each contributing either $3,000 or $6,000. If the partnership raised $258,000, then how many investors contributed $3,000 and how many contributed $6,000?
Answer:
There are 26 investors which contributed 6,000
And 34 investors which contributed 3,000
Step-by-step explanation:
You need to set up a two equation problem
[tex]\left \{ {{258,000=6,000y+3,000x} \atop {60=y+x}} \right.[/tex]
Now you have to clear "x" or "y" from the second equation:
[tex]y = 60 - x[/tex]
And replace on the first equation:
[tex]258,000 = 6,000 (60 - x) + 3,000x\\258,000 = 360,000 - 6,000x + 3,000x\\3,000x = 360,000 - 258,000\\x = 102,000/3,000\\x = 34[/tex]
And now you use this "x" vale on the second equation
[tex]60 = y + x\\60 = y + 34\\60 - 34 = y\\y = 26[/tex]
There are 26 investors which contributed 6,000
And 34 investors which contributed 3,000
An environmentalist wants to find out the fraction of oil tankers that have spills each month.Step 1 of 2:Suppose a sample of 474tankers is drawn. Of these ships, 318 did not have spills. Using the data, estimate the proportion of oil tankers that had spills. Enter your answer as a fraction or a decimal number rounded to three decimal places.
Answer: The proportion of oil tankers that had spills is [tex]\dfrac{156}{474}[/tex] or 0.329.
Step-by-step explanation:
Since we have given that
Number of tankers is drawn = 474
Number of tankers did not have spills = 318
Number of tankers have spills = 474 - 318 = 156
Proportion of oil tankers that had spills is given by
[tex]\dfrac{Containing\ spill}{Total}=\dfrac{156}{474}=0.329[/tex]
Hence, the proportion of oil tankers that had spills is [tex]\dfrac{156}{474}[/tex] or 0.329.
Six Hatfields and two McCoys are up for 3 construction jobs in Williamson. What is the probability that all 3 jobs go to Hatfields? Answer in decimal form. Round to 3 decimal places as needed. Your Answer:
Answer: 0.357
Step-by-step explanation:
Given : The number of Hatfield = 6
The number of McCoys = 2
The number of companies = 8
The number of construction jobs -3
Now, the required probability is given by :-
[tex]\dfrac{^6C_3\times^2C_0}{^8C_3}\\\\=\dfrac{\dfrac{6!}{3!(6-3)!}}{\dfrac{8!}{3!(8-3)!}}=0.357142857143\approx0.357[/tex]
Hence, the probability that all 3 jobs go to Hatfields =0.357
An object with weight W is dragged along a horizontal plane by a force acting along a rope attached to the object. If the rope makes an angle θ with a plane, then the magnitude of the force is F = μW μ sin(θ) + cos(θ) where μ is a constant called the coefficient of friction. For what value of θ is F smallest?
Answer:[tex]\theta =\arctan \mu [/tex]
Step-by-step explanation:
we know force sin component would oppose the weight of object thus normal reaction will not be W rather it would be
[tex]N=W-Fsin\theta [/tex]
therefore force cos component will balance the friction force
F[tex]cos\theta[/tex] =[tex]\left ( \mu N\right )[/tex]
F[tex]cos\theta[/tex] =[tex]\left ( \mu \left ( W-Fsin\theta \right )\right )[/tex]
F=[tex]\frac{\mu W}{cos\theta +\mu sin\theta}[/tex]
F will be smallest when [tex]cos\theta +\mu sin\theta[/tex] will be maximum
and it will be maximum when we differentiate it to get
[tex]\theta =\arctan \mu[/tex]
The magnitude of the force, F, is varies with the angle the rope makes
with the plane according to the given equations.
F will be smallest when [tex]\underline{\theta \ is \ arctan (\mu)}[/tex].
Reason:
The given parameters are;
Angle the rope makes with the plane = θ
The magnitude of the force is, [tex]F = \dfrac{ \mu \cdot W}{\mu \cdot sin(\theta) +cos(\theta) }[/tex]
The value of θ for which the value of F is smallest.
Solution;
When, F is smallest, we have;
[tex]\dfrac{dF}{d \theta} = \dfrac{d}{d\theta} \left(\dfrac{ \mu \cdot W}{\mu \cdot sin(\theta) +cos(\theta) } \right) = \dfrac{-\mu \cdot W \cdot (\mu \cdot cos(\theta) -sin(\theta))}{\left( \mu \cdot sin(\theta) +cos(\theta) \right)^2} = 0[/tex]
Therefore;
-μ·W·(μ·cos(θ) - sin(θ))
μ·cos(θ) = sin(θ)
By symmetric property, we have;
sin(θ) = μ·cos(θ)
[tex]\mathbf{\dfrac{sin(\theta)}{cos(\theta)} = tan (\theta) = \mu}[/tex]
Which gives;
θ = arctan(μ)
Therefore;
F, will be smallest when [tex]\underline{\theta = arctan (\mu)}[/tex].
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Question; The given equation of the magnitude of the force in relation to the angle the rope makes with the plane, θ, is presented as follows;
[tex]F = \dfrac{ \mu \cdot W}{\mu \cdot sin(\theta) +cos(\theta) }[/tex]
On a single roll of a pair of dice, what are the odds against rolling a sum of 12?
Answer:
[tex]\frac{1}{35}[/tex]
Step-by-step explanation:
On a single roll of a pair of dice. When a pair of dice are rolled the possible outcomes are as follows:
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
The number of outcomes that gives us 12 are (6,6). There is only one outcome that gives us sum 12.
Total outcomes = 36
Odd against favor = [tex]\frac{non \ favorable\ outcomes}{favorable \ outcomes}[/tex]
Number of outcomes of getting sum 12 is 1
Number of outcomes of not getting sum 12 is 36-1= 35
odds against rolling a sum of 12= [tex]\frac{1}{35}[/tex]
Final answer:
A detailed explanation of the odds against rolling a sum of 12 on a pair of dice.
Explanation:
On a single roll of a pair of dice, the odds against rolling a sum of 12 are:
There is only one way to roll a 12, which is by getting a 6 on each die.
The probability of rolling a 6 on one die is 1/6 or approximately 0.166.
The probability of rolling a 12 on both dice is (1/6) * (1/6) = 1/36, which is about 2.8%.
A cylindrical package to be sent by a postal service can have a maximum combined length and girth (perimeter of a cross section) of 144 inches. Find the dimensions of the package of maximum volume that can be sent. (The cross section is circular.)
Answer:
The dimensions of the package is [tex]r=\frac{48}{\pi}\ \text{and} \ h=48[/tex].
Step-by-step explanation:
Consider the provided information.
As it is given that, cylindrical package to be sent by a postal service can have a maximum combined length and girth is 144 inches.
Therefore,
144 = 2[tex]\pi[/tex]r + h
144-2[tex]\pi[/tex]r = h
The volume of a cylindrical package can be calculated as:
[tex]V=\pi r^{2}h[/tex]
Substitute the value of h in the above equation.
[tex]V=\pi r^{2}(144-2\pi r)[/tex]
Differentiate the above equation with respect to r.
[tex]\frac{dV}{dr}=2\pi r(144-2\pi r)+\pi r^{2}(-2\pi)[/tex]
[tex]\frac{dV}{dr}=288\pi r-4{\pi}^2 r^{2}-2{\pi}^2 r^{2}[/tex]
[tex]\frac{dV}{dr}=288\pi r-6{\pi}^2 r^{2}[/tex]
[tex]\frac{dV}{dr}=-6\pi r(-48+\pi r)[/tex]
Substitute [tex]\frac{dV}{dr}=0[/tex] in above equation.
[tex]0=-6\pi r(-48+\pi r)[/tex]
Therefore,
[tex]0=-48+\pi r[/tex]
[tex]r=\frac{48}{\pi}[/tex]
Now, substitute the value of r in 144-2[tex]\pi[/tex]r = h.
[tex]144-2\pi\frac{48}{\pi}=h[/tex]
[tex]144-96=h[/tex]
[tex]48=h[/tex]
Therefore the dimensions of the package should be:
[tex]r=\frac{48}{\pi}\ \text{and} \ h=48[/tex]
This is about optimization problems in mathematics.
Dimensions; Height = 48 inches; Radius = 48/π inches
We are told the combined length and girth is 144 inches.Girth is same as perimeter which is circumference of the circular side.
Thus; Girth = 2πr
If length of cylinder is h, then we have;2πr + h = 144
h = 144 - 2πr
Now, to find the dimensions at which the max volume can be sent;Volume of cylinder; V = πr²h
Let us put 144 - 2πr for h to get;
V = πr²(144 - 2πr)
V = 144πr² - 2π²r³
Differentiating with respect to r gives;
dV/dr = 288πr - 6π²r²
Radius for max volume will be when dV/dr = 0Thus; 288πr - 6π²r² = 0
Add 6π²r² to both sides to get;
288πr = 6π²r²
Rearranging gives;
288/6 = (π²r²)/πr
48 = πr
r = 48/π inches
Put 48/π for r in h = 144 - 2πr to get;h = 144 - 2π(48/π)
h = 144 - 96
h = 48 inches
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A textbook store sold a combined total of 440 physics and sociology textbooks in a week. The number of sociology textbooks sold was 54 less than the number of physics textbooks sold. How many textbooks of each type were sold?
Step-by-step explanation:
If p is the number of physics books and s is the number of sociology books, then:
p + s = 440
s = p - 54
Substituting:
p + (p - 54) = 440
2p - 54 = 440
2p = 494
p = 247
Solving for s:
s = p - 54
s = 247 - 54
s = 193
The store sold 247 physics books and 193 sociology books.
g 4. Determine which of the following functions are even, which are odd, and which are neither. (a) f(x) = x 3 + 3x (b) f(x) = 4 sin 2x (c) f(x) = x 2 + |x| (d) f(x) = e x (e) f(x) = 1 x (f) f(x) = 1 2 (e x + e −x ) (g) f(x) = x cos x (h) f(x) = 1 2 (e x − e −x ).
Answer:Given below
Step-by-step explanation:
A function is said to be odd if
[tex]F\left ( x\right )=F\left ( -x\right )[/tex]
(a)[tex]F\left ( x\right )=x^3+3x[/tex]
[tex]F\left ( -x\right )=-x^3-3x=-\left ( x^3+3x\right )[/tex]
odd function
(b)[tex]F\left ( x\right )=4sin2x[/tex]
[tex]F\left ( -x\right )=-4sin2x[/tex]
odd function
(c)[tex]F\left ( x\right )=x^2+|x|[/tex]
[tex]F\left ( -x\right )=\left ( -x^2\right )+|-x|=x^2+|x|[/tex]
even function
(d)[tex]F\left ( x\right )=e^x[/tex]
[tex]F\left ( -x\right )=e^{-x}[/tex]
neither odd nor even
(e)[tex]F\left ( x\right )=\frac{1}{x}[/tex]
[tex]F\left ( -x\right )=-\frac{1}{x}[/tex]
odd
(f)[tex]F\left ( x\right )=\frac{1}{2}\left ( e^x+e^{-x}\right )[/tex]
[tex]F\left ( -x\right )=\frac{1}{2}\left ( e^{-x}+e^{x}\right )[/tex]
even function
(g)[tex]F\left ( x\right )=xcosx(h)[/tex]
[tex]F\left ( -x\right )=-xcosx(h)[/tex]
odd function
(h)[tex]F\left ( x\right )=\frac{1}{2}\left ( e^x-e^{-x}\right )[/tex]
[tex]F\left ( -x\right )=\frac{1}{2}\left ( e^{-x}+e^{x}\right )[/tex]
odd function
Consider a periodic review system. The target inventory level is 1000 units. It is time to review the item, and the on-hand inventory level is 200 units. How many units should be ordered?
a) 800
b) 1000
c) 1200
d) the EOQ amount
e) the safety stock amount
Answer:
The answer is - a) 800
Step-by-step explanation:
In a periodic review system, we calculate the quantity of an item, a company has on hand at specified and fixed interval of time.
Given is : The target inventory level is 1000 units and the on-hand inventory level is 200 units.
So, the quantity will be =[tex]1000-200=800[/tex] units.
The answer is option A.
Question 9 (7 points) Adam borrows $6000 for two years at an 8% annual interest rate, what is his monthly payment? A. $250 D. $522 B. $271 C.$290 E. None of these
Answer:
B. $271
Step-by-step explanation:
Given,
Present value of the loan, PV = $ 6000,
Annual rate of interest = 8 % = 0.08,
So, the monthly rate of interest, r = [tex]\frac{0.08}{12}[/tex],
Also, time = 2 years,
So, the total number of months, n = 24,
Hence, the monthly payment would be,
[tex]A=\frac{PV(r)}{1-(1+r)^{-n}}[/tex]
[tex]=\frac{6000(\frac{0.08}{12})}{1-(1+\frac{0.08}{12})^{-24}}[/tex]
[tex]=\$271.363748737[/tex]
[tex]\approx \$271[/tex]
Option B is correct.
Answer 2. Ella was born early in the morning on a Monday. She got married on the 9074th day of her life. What day of the week was it? 9074 24 66 Answer
Answer:
Ella got married on a Wednesday.
Step-by-step explanation:
Let's solve this problem by understanding the following:
Each week is composed by 7 days: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, and Sunday.
So 1 week = 7 days;
Because the married day was on the 9074th day of her life, we can find the number of weeks that 9074 days represent:
[tex]\frac{1 week}{7 days} * 9074 days = 1296.285714 weeks[/tex]
This means that 9074 days represent 1296.285714 weeks, which can be interpreted as 1296 entire weeks and a fraction of a week (0.285714).
Now let's calculate how many days 0.285714 weeks represent:
[tex]\frac{7 days}{1 week} * 0.285714 weeks = 2 days[/tex]
This means that 9074 days are actually 1296 weeks and 2 days, because Ella was born on a Monday, and because after 7 days (1 week) it is Monday again, after 1296 weeks it is Monday, but as we also calculated 2 extra days, then the married day is two days after a Monday, that is a Wednesday.
In conclusion, Ella got married on a Wednesday.
Find the roots of the parabola given by the following equation.
2x^2+ 5x - 9 = 2x
Show work please!
ANSWER
[tex]x = \frac{3}{2} \: or \: x = - 3[/tex]
EXPLANATION
We want to find the roots of the parabola with equation:
[tex]2 {x}^{2} + 5x - 9 = 2x[/tex]
We need to write this in the standard quadratic equation form.
We group all terms on the left to get:
[tex]2 {x}^{2} + 5x - 2x - 9 = 0[/tex]
We simplify to get:
[tex]2 {x}^{2} +3x- 9 = 0[/tex]
We now compare to:
[tex]a {x}^{2} + bx + c = 0[/tex]
[tex] \implies \: a = 2 , \: \: b = 3 \: \: and \: c=- 9[/tex]
[tex] \implies ac = 2 \times - 9 = - 18[/tex]
The factors of -18 that sums up to 3 are -3, 6.
We split the middle term with these factors to get:
[tex]2 {x}^{2} +6x - 3x- 9 = 0[/tex]
Factor by grouping:
[tex]2x(x + 3) -3(x + 3) = 0[/tex]
Factor again to obtain:
[tex](2x - 3)(x + 3) = 0[/tex]
Apply the zero product principle to get:
[tex]2x - 3 = 0 \: or \: x + 3 = 0[/tex]
[tex] \implies \: x = \frac{3}{2} \: or \: x = - 3[/tex]
Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview
Answer: The required solution is
[tex]y=(-2+t)e^{-5t}.[/tex]
Step-by-step explanation: We are given to solve the following differential equation :
[tex]y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
Let us consider that
[tex]y=e^{mt}[/tex] be an auxiliary solution of equation (i).
Then, we have
[tex]y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.[/tex]
Substituting these values in equation (i), we get
[tex]m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.[/tex]
So, the general solution of the given equation is
[tex]y(t)=(A+Bt)e^{-5t}.[/tex]
Differentiating with respect to t, we get
[tex]y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.[/tex]
According to the given conditions, we have
[tex]y(0)=-2\\\\\Rightarrow A=-2[/tex]
and
[tex]y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.[/tex]
Thus, the required solution is
[tex]y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.[/tex]
A student guesses on every question of a multiple-choice test that has 6 questions, each with 3 possible answers. What is the probability that the student will get at least 4 of the questions right?
Answer:
The probability that the student will get at least 4 of the questions right is 0.0823044.
Step-by-step explanation:
For each question we have 3 choices. So,total choices will be :
[tex]3\times3\times3\times3\times3\times3=729[/tex]
Getting 4 correct means, 4 corrects and two wrongs
Now, as there are 3 answer choices, out of which only one will be correct, so 2/3 is the probability if a question is answered wrong.
And 1/3 is the probability if a question is answered correctly.
Hence, we can consider this probability :
[tex]P=(2/3)*(2/3)*(1/3)*(1/3)*(1/3)*(1/3)[/tex] = 4/729
=> P = 0.00548696
We can select any combination of 2 from 6 for being wrong, so we will multiply P by (6,2)=6!/(2!*4!) = 15
So the answer is P*15 =[tex]0.00548696*15=0.0823044[/tex]
The probability that the student will get at least 4 of the questions right is 0.0823044.
With 3 choices per question, the probability of getting at least 4 out of 6 questions correct is approximately 0.0823044
1: Total Choices
Each question has 3 possible answers.
So, the total choices for 6 questions would be 3 raised to the power of 6 (3^6).
2: Probability of Getting 4 Correct and 2 Wrong
Getting 4 correct and 2 wrong means selecting 4 correct answers out of 6 questions.
The probability of a question being answered correctly is 1/3, and the probability of being answered incorrectly is 2/3.
So, the probability of getting 4 correct and 2 wrong is calculated using combinations (6 choose 4) multiplied by (1/3)^4 multiplied by (2/3)^2.
3: Calculate Probability
(6 choose 4) is the number of ways to choose 4 correct answers out of 6 questions, which is 15.
The probability (P) is then calculated as 15 multiplied by (1/3)^4 multiplied by (2/3)^2.
4: Multiply by Number of Combinations
Since there are 15 ways to choose 4 correct answers out of 6 questions, multiply the probability by 15.
So, the probability that the student will get at least 4 of the questions right is approximately 0.0823044.
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Find all values of x that are NOT in the domain of h.
If there is more than one value, separate them with commas.
h(x) = x + 1 / x^2 + 2x + 1
Answer:
if x=-1 then its is NOT in the domain of h.
Step-by-step explanation:
Domain is the set of values for which the function is defined.
we are given the function
h(x) = x + 1 / x^2 + 2x + 1
h(x) = x+1 /x^2+x+x+1
h(x) = x+1/x(x+1)+1(x+1)
h(x) = x+1/(x+1)(x+1)
h(x) = x+1/(x+1)^2
So, the function h(x) is defined when x ≠ -1
Its is not defined when x=-1
So, if x=-1 then its is NOT in the domain of h.
Answer: [tex]x=-1[/tex]
Step-by-step explanation:
Given the function h(x):
[tex]h(x)=\frac{x+1}{ x^2 + 2x + 1}[/tex]
The values that are not in the domain of this function are those values that make the denominator equal to zero.
Then, to find them, you can make the denominator equal to zero and solve for "x":
[tex]x^2 + 2x + 1=0\\\\(x+1)(x+1)=0\\\\(x+1)^2=0\\\\x=-1[/tex]