Answer with Step-by-step explanation:
We are given that two functions f and g are decreasing function for all real numbers .
We have to prove that fog is increasing.
Increasing Function: If [tex]x_1 \leq x_2[/tex]
Then, [tex]f(x_1)\leq f(x_2)[/tex]
Decreasing function: if [tex]x_1\leq x_2[/tex]
Then, [tex]f(x_1)\geq f(x_2)[/tex]
Suppose [tex]f(x)=-x[/tex]
[tex]g(x)=-2x[/tex]
fog(x)=f(g(x))=f(-2x)=-(-2x)=2x
Therefore, f and g are decreasing function for all real numbers then fog is increasing.
Hence, proved
A dress cost twice as much as a skirt. Mom bought 3 dresses and 2 skirts. She gave the cashier $1000 and received $300 in change.How much did a dress cost?
Answer:
$175 = Cost of single dress
Step-by-step explanation:
It is provided that cost of dress is twice that of a skirt.
Let say, cost of skirt = x
Total things bought 3 dresses and 2 skirts.
Cost of single dress = 2x
Cost of all items = 3 [tex]\times[/tex] 2x + 2 [tex]\times[/tex] x = $1,000 - $300 = $700
6x + 2x = $700
x = 700/8 = $87.50
Cost of single dress = 2x = $87.50 [tex]\times[/tex] 2 = $175
What is the range of the relation?
1. {3, 9, 12}
2. {−6, −1, −0.5}
3. {−6, 3, 9}
4. {−1, −0.5, 9, 12}
Answer:
the range of the relation is 3
The eye of a hurricane passes over Grand Bahama Island in a direction 60.0° north of west with a speed of 43.5 km/h. Three hours later, the course of the hurricane suddenly shifts due north, and its speed slows to 23.5 km/h. How far from Grand Bahama is the hurricane 4.95 h after it passes over the island?
Answer:
D = 170.6Km
Step-by-step explanation:
First of all, we set the reference (origin) at Grand Bahama.
Nw, from the first displacement of 3h we calculate the distance:
D1 = V1*t = 43.5 * 3 = 130.5 Km
The coordinates of this new location is given by:
r1 = ( -D1*cos(60°); D1*sin(60°)) = (-62.5; 158.835) Km
For the second displacement, the duration was of 1.95 hours (4.95 -3), so the distance traveled was:
D2 = V2*t = 23.5 * 1.95 = 45.825 Km
The coordinates of this new location is given by:
r2 = r1 + ( 0; D2) = (-62.25; 158.835) Km
Now we just need to calculate the magnitude of that vector to know the distance to Grand Bahama:
[tex]Dt = \sqrt{D_{2x}^{2}+D_{2y}^{2}}=170.6 Km[/tex]
u + 3b - 2a + 2 solve for a
Answer:
a=2
Step-by-step explanation:
u=3b-2(2) or u=3b-4
Final answer:
To solve the equation for a, rearrange the terms and isolate 'a' by dividing by the coefficient.
Explanation:
Step 1: Start with the equation u + 3b - 2a + 2 = 0.
Step 2: Rearrange the terms to isolate 'a', which gives -2a = -u - 3b - 2.
Step 3: Divide by -2 to solve for 'a', resulting in a = (u + 3b + 2) / 2.
Yusuf has between 10 and 20 toffees. If he counts his toffees in 4s, he has 3 left over. If he counts his toffees in 5s, he has 4 left over. How many toffees has Yusuf got?
Answer:
Yusuf got 19 toffees
Step-by-step explanation:
The problem tells us that if Yusuf counts his toffees in 4's, he has 3 left. Let's call the number of toffees "t", then we know that [tex]t[/tex]≡[tex]3(mod4)[/tex]
We also know that [tex]t[/tex]≡[tex]4(mod 5)[/tex]
Between 10 and 20, there are only 2 numbers t≡[tex]4(mod 5)[/tex], these numbers are 14 and 19.
However, 14≡[tex]2(mod 4)[/tex] so it doesn't add up.
On the other hand, 19 ≡3[tex](mod4)[/tex] and therefore, this is the answer.
Answer:
19 toffees.
Step-by-step explanation:
It is given that Yusuf has between 10 and 20 toffees.
Let n be the number of toffees.
If he counts his toffees in 4s, he has 3 left over.
[tex]n\equiv 3(mod4)[/tex] ... (1)
So, the possible values of n between 10 and 20 are 11, 15 and 19.
If he counts his toffees in 5s, he has 4 left over.
[tex]n\equiv 4(mod5)[/tex] ... (2)
So, the possible values of n between 10 and 20 are 14 and 19.
From (1) and (2) the possible value of n between 10 and 20 is 19.
Therefore, Yusuf has 19 toffees.
Find equations of the line that is parallel to the z-axis and passes through the midpoint between the two points (0, −4, 9) and (−8, 5, 1). (Enter your answers as a comma-separated list of equations.) x=−8, y=5+9t, z=1−8t
The equations of the line are:
x = -4,
y = 0.5,
z = 5 + t
A line that is parallel to the z-axis lies in the xy-plane. Since it's parallel to the z-axis, its direction in the xy-plane is determined by the coefficients of x and y in its direction vector.
Let's first find the midpoint between the two given points:
Midpoint = [(x₁ + x₂) / 2, (y₁ + y₂) / 2, (z₁ + z₂) / 2]
Midpoint = [(-8 + 0) / 2, (5 - 4) / 2, (1 + 9) / 2]
Midpoint = [-4, 0.5, 5]
So, the midpoint is (-4, 0.5, 5).
Now, let's create a line that passes through the midpoint and is parallel to the z-axis. The equation of such a line in vector form is:
r(t) = Midpoint + t Direction
Where r(t) is the position vector of a point on the line, t is a scalar parameter, Midpoint is the midpoint we calculated, and Direction is the direction vector.
Since the line is parallel to the z-axis, its direction vector is (0, 0, 1). Thus, the equation of the line is:
r(t) = (-4, 0.5, 5) + t (0, 0, 1)
r(t) = (-4, 0.5, 5 + t)
In component form, the equations of the line are:
x = -4
y = 0.5
z = 5 + t
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The midpoint of the two points is (-4, 0.5, 5). The line parallel to the z-axis passing through this point has the equations: x=-4, y=0.5, z=t.
Explanation:Firstly, we need to find out the midpoint between the two points (0, -4, 9) and (-8, 5, 1).
The formula to calculate the midpoint of two points in three dimensions is ((x1+x2)/2, (y1+y2)/2, (z1+z2)/2).
Applying this formula to our points, we get the midpoint as ((0-8)/2, (-4+5)/2, (9+1)/2) which is (-4, 0.5, 5).
Lines parallel to the z-axis in three-dimensional space have equations of the form x=a, y=b, z=t, where 'a' and 'b' are constants representing any particular point through which the line passes, and 't' represents a variable that can take any real value.
Since the line we want passes through the point (-4, 0.5, 5), our equations for the desired line become: x=-4, y=0.5, z=t.
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6r+7=13+7r
steps too pls
Answer:
r = -6
Step-by-step explanation:
We presume you want to find the value of r that satisfies the equation.
Subtract 6r+13 from both sides:
(6r+7) -(6r+13) = (13 +7r) -(6r+13)
-6 = r . . . . . simplify
_____
More detailed explanation
When we look at the equation we see the only variable is r, and that terms containing it appear on both sides of the equal sign. There is only one "r" term on each side, so we don't need to do any consolidation. We observe that the term with the smallest coefficient is 6r and that it is on the left side.
When we subtract 6r from the equation we will have the remaining "r" term on the right, but we will also have a constant there that we don't want. So, we can subtract that constant as well. That is why we choose to subtract 6r+13 from the equation. Doing so leaves the constants on the left and the "r" terms on the right.
As it happens, the difference between the "r" terms is plain "r", so we're done after we finish the subtraction.
__
When considering the "r" terms, we choose to subtract the term with the smallest coefficient so that the result has a positive coefficient for "r". This helps reduce mistakes in later steps, if there are any later steps.
___
Alternate "steps"
For a linear equation like this one, you can subtract one side from both sides. This might look like ...
0 = 6+r . . . . . after subtracting 6r+7 (left side) from both sides
Then you can divide by the coefficient of r (which does nothing to this equation), and subtract the resulting constant (on the side with the variable). Here, that would give ...
-6 = r
These three steps will work to solve any linear equation. Simplification steps may be required depending on the complexity. Again, it might be helpful, though is not essential, to subtract the side with the smallest coefficient of the variable.
___
Final note
The rules of equality say you can do anything you like to an equation, as long as you do the same thing to both sides. We can say "subtract the constant" because we are assured that you know it must be subtracted from both sides of the equation. Beware of any instruction that tells you to do something to one side of an equation and something different to the other side.
The following question has two parts. First, answer part A. Then, answer part B.
Part A
A high school athlete ran the 100 meter sprint in 13.245 seconds. Round the time to the nearest tenth. Enter the answer in the box.
_______ seconds
Part B
Explain how you arrived at the answer. Include any rules that you followed.
Answer:
Part A: 13.2
Part B: see below
Step-by-step explanation:
Part A:
13.245 rounded to tenths is 13.2
Part B:
The rule is ...
Add 1 in the number place you're rounding to if the digit to its right is 5 or more. Drop (or zero) all digits to the right of the place you're rounding to.
Here, the digit to the right of the tenths place is 4, so no action is taken other than dropping digits to the right of the tenths place.
Write an expression with (-1) as its base that will produce a positive product, and explain why your answer is valid
Here are some possible answers
A particular brand of dishwasher soap is sold in three sizes: 25 oz, 45 oz, and 60 oz. Twenty percent of all purchasers select a 25-oz box, 50% select a 45-oz box, and the remaining 30% choose a 60-oz box. Let X1 and X2 denote the package sizes selected by two independently selected purchasers. (a) Determine the sampling distribution of X. x 25 35 45 42.5 52.5 60 p(x) Calculate E(X). E(X) = oz Compare E(X) to μ. E(X) > μ E(X) < μ E(X) = μ
Answer:
(a) Sampling distribution
P(25) = 0,04
P(35) = 0.1 + 0.1 = 0,2
P(42,5) = 0.06 + 0.06 = 0,12
P(45) = 0,25
P(52,5) = 0.15 + 0.15 = 0,3
P(60) = 0,09
(b) E(X) = 45.5 oz
(c) E(X) = μ
Step-by-step explanation:
The variable we want to compute is
[tex]X=(X1+X2)/2[/tex]
For this we need to know all the possible combinations of X1 and X2 and the probability associated with them.
(a) Sampling distribution
Calculating all the 9 combinations (3 repeated, so we end up with 6 unique combinations):
P(25) = P(X1=25) * P(X2=25) = p25*p25 = 0.2 * 0.2 = 0,04
P(35) = p25*p45+p45*p25 = 0.2*0.5 + 0.5*0.2 = 0.1 + 0.1 = 0,2
P(42,5) = p25*p60 + p60*p25 = 0.2*0.3 + 0.3*0.2 = 0.06 + 0.06 = 0,12
P(45) = p45*p45 = 0.5 * 0.5 = 0,25
P(52,5) = p45*p60 + p60*p45 = 0.5*0.3 + 0.3*0.5 = 0.15 + 0.15 = 0,3
P(60) = p60*p60 = 0.3*0.3 = 0,09
(b) Using the sample distribution, E(X) can be expressed as:
[tex]E(X)=\sum_{i=1}^{6}P_{i}*X_{i}\\E(X)=0.04*25+0.2*35+0.12*42.5+0.3*52.5+0.09*60 = 45.5[/tex]
The value of E(X) is 45.5 oz.
(c) The value of μ can be calculated as
[tex]\mu=\sum_{i=1}^{3}P_{i}*X_{i}\\\mu=0.2*25+0.5*45+0.3*60=45.5[/tex]
We can conclude that E(X)=μ
We could have arrived to this conclusion by applying
[tex]E(X)=E((X1+X2)/2)=E(X1)/2+E(X2)/2\\\\\mu = E(X1)=E(X2)\\\\E(X)=\mu /2+ \mu /2 = \mu[/tex]
Given that [tex]z=x+iy[/tex], find the value of x and of y such that [tex]z+4iz^*=-3+18i[/tex] where z* is the complex conjugate of z.
Answer:
[tex]z= 5-2i[/tex]
Step-by-step explanation:
Start replacing!
[tex](x+iy) + 4i (x-iy) = -3+18i \\ x+iy +4ix -4i^2 y = -3 + 18i \\\\x + i (4x+y) +4y = -3 + 18i\\x+4y + (4x+y) i = -3 + 18i[/tex]
Now two complex numbers are equal if both the real and imaginary parts are equal, which gives you the system of equation [tex]\left \{x+4y=-3} \atop {4x+y=18} \right.[/tex]
Pick any method to solve it and you'll get [tex] x=5, y= -2 [/tex]
PLEASE HELP I AM ON A TIME LIMIT
Answer:
(a) even: J, K, M, O; odd: L, N
(b) L and O are connected to J
(c) N is of degree 3
Step-by-step explanation:
Count the edge ends that intersect each vertex. You get ...
J-2, K-2, L-3, M-2, N-3, O-4
These numbers are the degree of the vertex.
a) Vertices with even degree are J, K, M, O, since these have degrees of 2, 2, 2, and 4, respectively--all even numbers.
Vertices with odd degree are L and N, since these both have degree 3, an odd number.
__
b) The vertices that are adjacent to J are the ones at the other ends of the edges that intersect vertex J. One of those two edges connects to vertex L; the other to vertex O. J is adjacent to L and O.
__
c) When we counted edges in part (a), we found vertex N to be of degree 3.
Find the optimal solution for the following problem
Minimize C = 13x + 3y
subject to 12x + 14y ≥ 21
15x + 20y ≥ 37
and x ≥ 0, y ≥ 0.
1. What is the optimal value of x?
2. What is the optimal value of y?
3.What is the minimum value of the objective function?
Answer:
Minimize C =[tex]13x + 3y[/tex]
[tex]12x + 14y \geq 21[/tex]
[tex]15x + 20y \geq 37[/tex]
and x ≥ 0, y ≥ 0.
Plot the the lines on the graph and find the feasible region
[tex]12x + 14y \geq 21[/tex] -- Blue
[tex]15x + 20y \geq 37[/tex] --- Green
So, the boundary points of feasible region are (-3.267,4.3) , (0,1.85) and (2.467,0)
Substitute the value in Minimize C
Minimize C =[tex]13x + 3y[/tex]
At (-3.267,4.3)
Minimize C =[tex]13(-3.267) + 3(4.3)[/tex]
Minimize C =[tex]-29.571[/tex]
At (0,1.85)
Minimize C =[tex]13(0) + 3(1.85)[/tex]
Minimize C =[tex]5.55[/tex]
At (2.467,0)
Minimize C =[tex]13(2.467) + 3(0)[/tex]
Minimize C =[tex]32.071[/tex]
So, the optimal value of x is -3.267
So, the optimal value of y is 4.3
So, the minimum value of the objective function is -29.571
Preliminary data analyses indicate that it is reasonable to use a t-test to carry out the specified hypothesis test. Perform the t-test. Be sure to state the hypotheses and the P-Value. State your conclusion in a sentence. A test of sobriety involves measuring a subject's motor skills. The mean score for men who are sober is known to be 35.0. A researcher would like to perform a hypothesis test to determine whether the mean score for sober women differs from 35.0. Twenty randomly selected sober women take the test and produce a mean score of 41.0 with a standard deviation of 3.7. Perform the hypothesis test at the 0.01 level of significance.
By considering the given information, we have
Null hypothesis : [tex]H_0: \mu=35.0[/tex]
Alternative hypothesis : [tex]H_1: \mu\neq35.0[/tex]
Since the alternative hypothesis is two-tailed , so the test is a two-tailed test.
Given : Sample size : n= 20, since sample size is less than 30 so the test applied is a t-test.
[tex]\overline{x}=41.0[/tex] ; [tex]\sigma= 3.7[/tex]
Test statistic : [tex]t=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
i.e. [tex]t=\dfrac{41.0-35.0}{\dfrac{3.7}{\sqrt{20}}}=7.252112359\approx7.25[/tex]
Degree of freedom : n-1 = 20-1=19
Significance level = 0.01
For two tailed, Significance level [tex]=\dfrac{0.01}{2}=0.005[/tex]
By using the t-distribution table, the critical value of t =[tex]t_{19, 0.005}=2.861[/tex]
Since , the observed t-value (7.25) is greater than the critical value (2.861) .
So we reject the null hypothesis, it means we have enough evidence to support the alternative hypothesis.
We conclude that there is some significance difference between the mean score for sober women and 35.0.
Graph the system of equations. Then determine whether the system has no solution, one solution, or infinitely many solutions. If the system has one solution, name it.
y = -x + 5
y = x - 3
a) one solution; (1, 4)
b) infinitely many
c) no solution
d) one solution; (4, 1)
Answer:
d) one solution; (4, 1)
Step-by-step explanation:
It often works well to follow problem directions. A graph is attached, showing the one solution to be (4, 1).
_____
You know there will be one solution because the lines have different slopes. Each is in the form ...
y = mx + b
where m is the slope and b is the y-intercept.
The first line has slope -1 and y-intercept +5; the second line has slope 1 and y-intercept -3. The slope is the number of units of "rise" for each unit of "run", so it can be convenient to graph these by starting at the y-intercept and plotting points with those rise and run from the point you know.
In a fraternity with 32 members, 18 take mathematics, 5 take both mathematics and literature, and 8 take neither mathematics nor literature. How many take literature but not mathematics?
Answer:
1 member took literature but not mathematics.
Step-by-step explanation:
We can draw a Venn diagram for the given question.
In a fraternity total number of members = 32
Number of members who took mathematics M = 18
Number of members who took both mathematics and literature (M∩L) = 5
And number of members who took neither mathematics nor literature = 8
Therefore, number of members who took literature but not mathematics
= 32 - [(18 + 5) + 8]
= 32 - [23 + 8]
= 32 - 31
= 1
Therefore, 1 member took literature but not mathematics.
Answer:
1 member took literature but not mathematics.
Step-by-step explanation:
You have an order for Vasopressin 18 units/hr IV infusion. You have available vasopressin 200 units in 5000 mL D5W. Please calculate the drip rate in mL/hr.
Answer:
450 mL/hr
Step-by-step explanation:
Given:
Order for vasopressin = 18 units/hour
Available vasopressin = 200 units in 5000 mL
Now,
Volume of vasopressin per unit = [tex]\frac{\textup{Volume of vasopressin}}{\textup{Number of units}}[/tex]
or
Volume of vasopressin per unit = [tex]\frac{\textup{5000}}{\textup{200}}[/tex]
or
Volume of vasopressin per unit = 25 mL/unit
Thus,
Drip rate in mL/hr
= volume of vasopressin per unit × order for vassopressin
or
Drip rate in mL/hr = 25 × 18 = 450 mL/hr
The drip rate for an order of Vasopressin 18 units/hr, given a solution concentration of 200 units in 5000 mL, is calculated to be 450 mL/hr.
Explanation:To find the drip rate in mL/hr, we start by determining the concentration of the vasopressin solution. It is 200 units in 5000 mL D5W, so the concentration is 0.04 units/mL (200 units/5000 mL).
Next, we know the doctor prescribed 18 units/hr of vasopressin. To find out how many mL this corresponds to, we divide the order of 18 units/hr by the concentration in units/mL, which gives us 450 mL/hr (18 units/hr / 0.04 units/mL).
Therefore, the drip rate for the Vasopressin order is 450 mL/hr.
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Use the fact that the world population was 2560 million in 1950 and 3040 million in 1960 to model the population of the world in the second half of the 20th century. (Assume that the growth rate is proportional to the population size.) What is the relative growth rate? Use the model to estimate the world population in 1992 and to predict the population in the year 2030. SOLUTION We measure the time t in years and let t = 0 in the year 1950. We measure the population P(t) in millions of people. Then P(0) = 2560 and P(10) = 3040. Since we are assuming that dP/dt = kP, this theorem gives the following. (Round to six decimal places.)
Answer:
Let the growth function that shows the population in millions after x years,
[tex]P=P_0(1+r)^x[/tex]
Where,
[tex]P_0[/tex] = initial population,
r = growth rate per year,
Suppose the population is estimated since 1950,
Thus, if x = 0, P = 2560,
[tex]\implies 2560 = P_0 (1+r)^0\implies P_0 = 2560[/tex]
Now, if x = 10 ( that is, on 1960 ), P = 3040,
[tex]3040=2560(1+r)^{10}\implies r = 0.017[/tex]
Hence, the required function that shows the population after x years,
[tex]P=2560(1.017)^x[/tex]
If x = 42,
The population in 1992 would be,
[tex]P=2560(1.017)^{42}\approx 5196.608365\text{ millions}[/tex]
if x = 80,
The population in 2030 would be,
[tex]P=2560(1.017)^{80}\approx 9860.891929\text{ millions}[/tex]
Final answer:
The world population growth can be modeled using exponential growth with a relative growth rate calculated from given data. The relative growth rate, k, was found to be approximately 0.017355. Using this rate, we can estimate past and predict future populations; however, demographic trends show that actual growth rates are slowing.
Explanation:
To model the world population using a proportional growth rate, we employ the exponential growth model where dP/dt = kP. The world population was 2560 million in 1950 (P(0) = 2560) and 3040 million in 1960 (P(10) = 3040). We will use these data points to find the relative growth rate k.
Finding the Relative Growth Rate
The general solution of the differential equation is P(t) = P(0)e^(kt), where e is the base of the natural logarithm. Inserting our initial conditions:
P(0) = 2560, which implies C = 2560 where C is the initial population.
P(10) = 3040 yields 3040 = 2560e^(10k).
Dividing both sides by 2560 gives 3040/2560 = e^(10k), and then taking the natural logarithm of both sides we find ln(3040/2560) = 10k. Therefore, k ≈ ln(1.1875)/10. By calculating, k ≈ 0.017355 (rounded to six decimal places).
Estimating World Population for 1992 and Predicting for 2030
Using this growth rate, we can estimate the world population for any year t with P(t) = 2560e^(0.017355t):
For the year 1992 (t=42 years since 1950), P(42) ≈ 2560e^(0.017355*42).
For the year 2030 (t=80 years since 1950), P(80) ≈ 2560e^(0.017355*80).
Demographic trends suggest a slowing growth rate. Between 1965 and 1980, the annual rate was 2%, while predictions for 2005-2015 showed a decline to 1.1%. These numbers imply increasing doubling times and changing dynamics in world population growth.
Molecules of species A are reacting with molecules of species B. The concentration of A, given in moles per liter, follows the equation
CA = CA0 ( 1-e-k·t )
where t is time, given in minutes, CA is concentration in M, and CA0 is the initial concentration. What would be the units for the parameter k?
Answer:
Units for parameter k would be [tex]minutes^{-1}[/tex].
Step-by-step explanation:
The concentration of CA0 is given in M (moles per liter), which is the unit for CA; if we show units inside parenthesis in the equation, it would be:
[tex]CA (M)=CA0 (M) *(1-e-k(?)*t(minutes))[/tex]
For the concentration units of CA0 not to be affected by the units of the factor (1-e-k*t), this factor would have to be a number without units.
Since 1 is a constant without units, for the constant e to be able to subtract from 1 it would have to be a number without units, which also applies to the factor k*t.
For the factor k*t to be a number without units, k must have units that can be canceled when multiplied by t, which is given in minutes, so k must have units of [tex]\frac{1}{minutes} =minutes^{-1}[/tex]
This can be confirmed by operating the equation using only its units (units of parameter k are noted by a question mark):
[tex]M=M(0-0-?*minutes)[/tex]
[tex]\frac{M}{M} =?*minutes[/tex]
[tex]1=?*minutes[/tex]
[tex]\frac{1}{minutes}=?[/tex]
[tex]minutes^{-1}=?[/tex]
The population of a mining city grows at a rate proportional to that population, in two years the population has doubled and a year later there were 10,000 inhabitants.
What was the initial population?
Answer:
The initial population was approximatedly 3535 inhabitants.
Step-by-step explanation:
The population of the city can be given by the following differential equation.
[tex]\frac{dP}{dt} = Pr[/tex],
In which r is the rate of growth of the population.
We can solve this diffential equation by the variable separation method.
[tex]\frac{dP}{dt} = Pr[/tex]
[tex]\frac{dP}{P} = r dt[/tex]
Integrating both sides:
[tex]ln P = rt + c[/tex]
Since ln and the exponential are inverse operations, to write P in function of t, we apply ln to both sides.
[tex]e^{ln P} = e^{rt + C}[/tex]
[tex]P(t) = Ce^{rt}[/tex]
C is the initial population, so:
[tex]P(t) = P(0)e^{rt}[/tex]
Now, we apply the problem's statements to first find the growth rate and then the initial population.
The problem states that:
In two years the population has doubled:
[tex]P(2) = 2P(0)[/tex]
[tex]P(t) = P(0)e^{rt}[/tex]
[tex]2P(0) = P(0)e^{2r}[/tex]
[tex]2 = e^{2r}[/tex]
To isolate r, we apply ln both sides
[tex]e^{2r} = 2[/tex]
[tex]ln e^{2r} = ln 2[/tex]
[tex]2r = 0.69[/tex]
[tex]r = \frac{0.69}{2}[/tex]
[tex]r = 0.3466[/tex]
So
[tex]P(t) = P(0)e^{0.3466t}[/tex]
In two years the population has doubled and a year later there were 10,000 inhabitants.
[tex]P(3) = 10,000[/tex]
[tex]P(t) = P(0)e^{0.3466t}[/tex]
[tex]10,000= P(0)e^{0.3466*3}[/tex]
[tex]P(0) = \frac{10,000}{e^{1.04}}[/tex]
[tex]P(0) = 3534.55[/tex]
The initial population was approximatedly 3535 inhabitants.
State the converse, contrapositive, and inverse of each of these conditional statements a) If it snows tonight, then I will stay at home. b) I go to the beach whenever it is a sunny summer day. c) When I stay up late, it is necessary that I sleep until noon.
Step-by-step explanation:
Consider the provided information.
For the condition statement [tex]p \rightarrow q[/tex] or equivalent "If p then q"
The rule for Converse is: Interchange the two statements.The rule for Inverse is: Negative both statements.The rule for Contrapositive is: Negative both statements and interchange them.Part (A) If it snows tonight, then I will stay at home.
Here p is If it snows tonight, and q is I will stay at home.
Converse: If I will stay at home then it snows tonight.
[tex]q \rightarrow p[/tex]
Inverse: If it doesn't snows tonight, then I will not stay at home.
[tex]\sim p \rightarrow \sim q[/tex]
Contrapositive: If I will not stay at home then it doesn't snows tonight.
[tex]\sim q \rightarrow \sim p[/tex]
Part (B) I go to the beach whenever it is a sunny summer day.
Here p is I go to the beach, and q is it is a sunny summer day.
Converse: It is a sunny summer day whenever I go to the beach.
[tex]q \rightarrow p[/tex]
Inverse: I don't go to the beach whenever it is not a sunny summer day.
[tex]\sim p \rightarrow \sim q[/tex]
Contrapositive: It is not a sunny summer day whenever I don't go to the beach.
[tex]\sim q \rightarrow \sim p[/tex]
Part (C) When I stay up late, it is necessary that I sleep until noon.
P is I sleep until noon and q is I stay up late.
Converse: If I sleep until noon, then it is necessary that i stay up late.
[tex]q \rightarrow p[/tex]
Inverse: When I don't stay up late, it is necessary that I don't sleep until noon.
[tex]\sim p \rightarrow \sim q[/tex]
Contrapositive: If I don't sleep until noon, then it is not necessary that i stay up late.
[tex]\sim q \rightarrow \sim p[/tex]
The converse, contrapositive, and inverse of each of the specified conditional statements are shown below.
How to form converse, contrapositive, and inverse of conditional statement?Suppose the conditional statement given is:
[tex]p \rightarrow q[/tex] or 'if p then q'
Then, we get:
Converse: "if q then p" or [tex]q \rightarrow p[/tex]Contrapositive: "if not q then not p" [tex]\sim q \rightarrow \sim p[/tex]Inverse: "If not p then not q" [tex]\sim p \rightarrow \sim q[/tex]For the listed conditional statements, finding their converse, contrapositive, and inverse statements:
Case 1: If it snows tonight, then I will stay at home.Converse : If i will stay at home, then it snows tonight.Contrapositive : If i don't stay at home, it won't snow tonight.Inverse : If it doesn't snow tonight, then i will not stay at home.Case 2: I go to the beach whenever it is a sunny summer dayThis can be taken as: If it is a sunny summar day, then i go to the beach.
Converse : If i go to the beach, then it is a sunny summer day.Contrapositive : If i do not go to the beach, then it isn't a sunny summer dayInverse : If it's not a sunny summar day, then i do not go to the beach.Learn more about converse, contrapositive, and inverse statements here:
https://brainly.com/question/13245751?referrer=searchResults
A pharmacist attempts to weigh 0.375 g of morphine sulfate on a balance of dubious accuracy. When checked on a highly accurate balance, the weight is found to be 0.400 g. Calculate the percentage of error in the first weighing.
Answer: [tex]6.25\%[/tex]
Step-by-step explanation:
Given: A pharmacist attempts to weigh 0.375 g of morphine sulfate on a balance of dubious accuracy. When checked on a highly accurate balance, the weight is found to be 0.400 g.
i.e. Estimated weight = 0.375 g and Actual weight = 0.400 g
Now, the percentage of error in the first weighing is given by :-
[tex]\%\text{ Error}=\dfrac{|\text{Estimate-Actual}|}{\text{Actual}}\times100\\\\=\dfrac{|0.375-0.400|}{0.400}\times100\\\\=\dfrac{|-0.025|}{0.400}\times100\\\\=\dfrac{0.025}{0.4}\times100\\\\=\dfrac{25\times10}{4\times1000}\times100=\dfrac{25}{4}=6.25\%[/tex]
Hence, the percentage of error in the first weighing = [tex]6.25\%[/tex]
The image of (6, 9) under a dilation is (4, 6).
The scale factor is
0 -2
Answer:
The scale factor is 2/3.
Step-by-step explanation:
When the dilation is about the origin, the ratio of image coordinates to original coordinates is the scale factor:
4/5 = 6/9 = 2/3
The scale factor is 2/3.
Paco bought 3 CDs that cost d dollars each and a pack of gum for C cents. Write an expression for the total cost of his purchase. 3d + c 3C + d 3d/c
Answer:
The correct option is A) [tex]3d+C[/tex].
Step-by-step explanation:
Consider the provided information.
Paco bought 3 CDs that cost d dollars each.
Let d is the cost of each CDs.
The cost of 3 CDs will be 3 times of d.
This can be written as:
[tex]3d[/tex]
He bought a pack of gum for C cent. Thus, we can say that the cost of pack of gum is C.
Now add the cost of gum in above expression.
[tex]3d+C[/tex]
Hence, the required expression is [tex]3d+C[/tex].
Thus, the correct option is A) [tex]3d+C[/tex].
The proper interpretation of a 95% confidence interval is: A. We are 95% confident that the point estimate is between the lower confidence limit and upper confidence limit. B. We are 95% confident that the population parameter is between the lower confidence limit and upper confidence limit. C. We are 95% confident that the test statistic is between the lower confidence limit and upper confidence limit. D. We are 95% confident that the sample mean is between the lower confidence limit and upper confidence limit.
Answer: B. We are 95% confident that the population parameter is between the lower confidence limit and upper confidence limit.
Step-by-step explanation:
In statistics, we know that the interpretation of [tex](1-\alpha)\%[/tex] level of confidence interval is that we are [tex](1-\alpha)\%[/tex] sure that the true population parameter lies in it.
Therefore, the proper interpretation of a 95% confidence interval is that we are 95% confident that the population parameter is between the lower confidence limit and upper confidence limit.
Final answer:
The correct interpretation of a 95% confidence interval is that we are 95% confident that the population parameter is within the interval. The confidence interval represents the range within which the true population parameter lies, with a certain level of confidence, not the prediction of sample means or point estimates.
Explanation:
The correct interpretation of a 95% confidence interval is B: We are 95% confident that the population parameter is between the lower confidence limit and upper confidence limit. This means that if we were to take many samples from the population and construct a confidence interval from each sample, we would expect about 95% of those intervals to contain the actual population parameter, such as the population mean. It's important to note that the confidence interval does not predict where individual point estimates or sample means will fall. Rather, it expresses the reliability of the estimate of the population parameter based on the sampled data.
Jack has a collection of 10 pairs of gloves in his wardrobe. Before a business trip, he has to pack his luggage, and he selects 8 gloves, without looking at them. We assume that any set of 8 gloves is equally likely to be chosen. Find the probability that these 8 gloves do not include any matching pair of gloves, that is, that there are no two (left and right) gloves, coming from the same pair.
Answer:
[tex]\frac{6\cdot 8 \cdot 10 \cdot 12}{13 \cdot 15 \cdot 17\cdot 19}\approx 0.091 [/tex]
Step-by-step explanation:
To compute the probability of not including any matching pairs, we can compute the number of ways in which he could pick 8 gloves with no matching pairs, and divide it by the total number of ways in which he could pick 8 gloves.
The process of choosing 8 gloves with no matching pair can be seen as follows:
He first picks any random gloves out of the 20, in this 1st step he has then 20 available choices. Then he needs to pick some other glove, BUT it cannot be the other glove from the pair he already picked one from. So at this step there aren't 19 choices, but 18 available choices. Now he has 2 gloves from different pairs, and needs to pick another glove. There are 18 gloves left, BUT he cannot pick the remaining glove from any of the 2 pairs he already has chosen one from. Therefore he only has 16 choices left. The process continues like this, until he chooses 8 gloves in total. Hence the total number of ways to choose 8 gloves with no matching pair is
[tex] 20 \cdot 18 \cdot 16 \cdot 14 \cdot 12 \cdot 10 \cdot 8 \cdot 6[/tex]
Now, the total numer of ways in which he could pick any 8 gloves out of 20 can be seen as follows: At the start he has 20 available gloves, he chooses any of them, so having 20 available choices on that first step. He then needs to choose any other glove, so he has 19 choices. He then picks any other glove, so he has 18 choices. And so on, until he has chosen the 8 gloves. Hence the total number of ways to choose 8 gloves is
[tex] 20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13 [/tex]
Therefore the probability of choosing 8 gloves with no matching pair is
[tex]\frac{20 \cdot 18 \cdot 16 \cdot 14 \cdot 12 \cdot10 \cdot 8 \cdot 6}{20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13}=\frac{6\cdot 8 \cdot 10 \cdot 12}{13 \cdot 15 \cdot 17\cdot 19}\approx 0.091 [/tex]
A division of the Gibson Corporation manufactures bicycle pumps. Each pump sells for $10, and the variable cost of producing each unit is 60% of the selling price. The monthly fixed costs incurred by the division are $50,000. What is the break-even point for the division? (Round your answers to the nearest whole number.)
(x,y)=
Answer:
12,500 units
Step-by-step explanation:
Given:
Selling cost of the pump = $10
Variable cost of producing each unit = 60% of selling cost
or
Variable cost of producing each unit = 0.60 × $10 = $6
Monthly fixed cost = $50,000
Now,
The profit per unit = Selling cost - Variable cost
or
The profit per unit = $10 - $6 = $4
Therefore,
The breakeven point = [tex]\frac{\textup{Fixed cost}}{\textup{Profit per unit}}[/tex]
or
the breakeven point = [tex]\frac{\textup{50,000}}{\textup{4}}[/tex]
or
the breakeven point = 12,500 units
solve the lenear system by using the inverse of the coefficient matrix:
x + 2z = -1
2x - y = 2
3y + 4z = 1
Answer:
The solution of this system is x=9/4, y=5/2, and z=-13/8
Step-by-step explanation:
1. Writing the equations in matrix form
The system of linear equations given can be written in matrix form as
[tex]\left[\begin{array}{ccc}1&0&2\\2&-1&0\\0&3&4\end{array}\right]\left[\begin{array}{c}x&y&z\end{array}\right] = \left[\begin{array}{c}-1&2&1\end{array}\right][/tex]
Writing
A = [tex]\left[\begin{array}{ccc}1&0&2\\2&-1&0\\0&3&4\end{array}\right][/tex]
X = [tex]\left[\begin{array}{c}x&y&z\end{array}\right][/tex]
B = [tex]\left[\begin{array}{c}-1&2&1\end{array}\right][/tex]
we have
AX=B
This is the matrix form of the simultaneous equations.
2. Solving the simultaneous equations
Given
AX=B
we can multiply both sides by the inverse of A
[tex]A^{-1}AX=A^{-1}B[/tex]
We know that [tex]A^{-1}A=I[/tex], the identity matrix, so
[tex]X=A^{-1}B[/tex]
All we need to do is calculate the inverse of the matrix of coefficients, and finally perform matrix multiplication.
3. Calculate the inverse of the matrix of coefficients
A = [tex]\left[\begin{array}{ccc}1&0&2\\2&-1&0\\0&3&4\end{array}\right][/tex]
To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.
[tex]\left[\begin{array}{ccc|ccc}1&0&2&1&0&0\\2&-1&0&0&1&0\\0&3&4&0&0&1\end{array}\right][/tex]
Make zeros in column 1 except the entry at row 1, column 1. Subtract row 1 multiplied by 2 from row 2[tex]\left[\begin{array}{ccc|ccc}1&0&2&1&0&0\\0&-1&-4&-2&1&0\\0&3&4&0&0&1\end{array}\right][/tex]
Make zeros in column 2 except the entry at row 2, column 2. Add row 2 multiplied by 3 to row 3[tex]\left[\begin{array}{ccc|ccc}1&0&2&1&0&0\\0&-1&-4&-2&1&0\\0&0&-8&-6&3&1\end{array}\right][/tex]
Multiply row 2 by −1[tex]\left[\begin{array}{ccc|ccc}1&0&2&1&0&0\\0&1&4&2&-1&0\\0&0&-8&-6&3&1\end{array}\right][/tex]
Make zeros in column 3 except the entry at row 3, column 3. Divide row 3 by −8[tex]\left[\begin{array}{ccc|ccc}1&0&2&1&0&0\\0&1&4&2&-1&0\\0&0&1&3/4&-3/8&-1/8\end{array}\right][/tex]
Subtract row 3 multiplied by 2 from row 1[tex]\left[\begin{array}{ccc|ccc}1&0&0&-1/2&3/4&1/4\\0&1&4&2&-1&0\\0&0&1&3/4&-3/8&-1/8\end{array}\right][/tex]
Subtract row 3 multiplied by 4 from row 2[tex]\left[\begin{array}{ccc|ccc}1&0&0&-1/2&3/4&1/4\\0&1&0&-1&1/2&1/2\\0&0&1&3/4&-3/8&-1/8\end{array}\right][/tex]
As can be seen, we have obtained the identity matrix to the left. So, we are done.
[tex]A^{-1} = \left[\begin{array}{ccc}-1/2&3/4&1/4\\-1&1/2&1/2\\3/4&-3/8&-1/8\end{array}\right][/tex]
4. Find the solution [tex]X=A^{-1}B[/tex]
[tex]X= \left[\begin{array}{ccc}-1/2&3/4&1/4\\-1&1/2&1/2\\3/4&-3/8&-1/8\end{array}\right]\cdot \left[\begin{array}{c}-1&2&1\end{array}\right] = \left[\begin{array}{c}9/4&5/2&-13/8\end{array}\right][/tex]
A solution initially contains 200 bacteria. 1. Assuming the number y increases at a rate proportional to the number present, write down a differential equation connecting y and the time t. 2. If the rate of increase of the number is initially 100 per hour, how many bacteria are there after 2 hours? Solution:
Answer:
1.[tex]\frac{dy}{dt}=ky[/tex]
2.543.6
Step-by-step explanation:
We are given that
y(0)=200
Let y be the number of bacteria at any time
[tex]\frac{dy}{dt}[/tex]=Number of bacteria per unit time
[tex]\frac{dy}{dt}\proportional y[/tex]
[tex]\frac{dy}{dt}=ky[/tex]
Where k=Proportionality constant
2.[tex]\frac{dy}{y}=kdt[/tex],y'(0)=100
Integrating on both sides then, we get
[tex]lny=kt+C[/tex]
We have y(0)=200
Substitute the values then , we get
[tex]ln 200=k(0)+C[/tex]
[tex]C=ln 200[/tex]
Substitute the value of C then we get
[tex]ln y=kt+ln 200[/tex]
[tex]ln y-ln200=kt[/tex]
[tex]ln\frac{y}{200}=kt[/tex]
[tex]\frac{y}{200}=e^{kt}[/tex]
[tex]y=200e^{kt}[/tex]
Differentiate w.r.t
[tex]y'=200ke^{kt}[/tex]
Substitute the given condition then, we get
[tex]100=200ke^{0}=200 \;because \;e^0=1[/tex]
[tex]k=\frac{100}{200}=\frac{1}{2}[/tex]
[tex]y=200e^{\frac{t}{2}}[/tex]
Substitute t=2
Then, we get [tex]y=200e^{\frac{2}{2}}=200e[/tex]
[tex]y=200(2.718)=543.6=543.6[/tex]
e=2.718
Hence, the number of bacteria after 2 hours=543.6
36.25= 45.5% of ____
Answer:
36.25 is equal to the 45.5% of 79.67
Step-by-step explanation:
To solve this you just have to find the value that represents the 100%.
To do this, you can use the Rule of Three that allows you to solve problems of proportions.
In this case, you know that the 45.5% of the magnitude X is 36.25, now you have to find the value which corresponds to the 100%.
Mathematically it will be:
[tex]\frac{X}{36.25} =\frac{100}{45.5}[/tex]
Then you have to solve the equation to find X:
[tex]X=\frac{100}{45.5}*36.25[/tex]
And finally, the answer is:
[tex]X=79.67[/tex]