Answer:
a) 19,600 fishes were originally put in the pond.
b) Population of fishes after 10 years = 3,863
Population of fishes after 20 years = 1,733
Population of fishes after 30 years = 1,403
Step-by-step explanation:
The population follows a logistic model
P(t) = d (1 + ke⁻ᶜᵗ)
For a fish pond,
d = 1400, k = 13, c = 0.2
Inserting the values of these constants
P(t) = 1400 (1 + 13 e⁻⁰•²ᵗ)
a) How many fish were originally put in the pond?
At the start of the whole thing, t = 0
P(t=0) = 1400 (1 + 13 e⁰) = 1400 × 14 = 19,600
Hence, 19,600 fishes were originally put in the pond.
b) Find the population after 10, 20, and 30 years.
P(t) = 1400 (1 + 13 e⁻⁰•²ᵗ)
At t = 10, 0.2t = 0.2 × 10 = 2
P(t=10) = 1400 (1 + 13e⁻²) = 1400 (1 + 1.759) = 3,863.1 = 3,863
At t = 20, 0.2t = 0.2 × 20 = 4
P(t=20) = 1400 (1 + 13e⁻⁴) = 1400 (1 + 0.238) = 1,733.3 = 1,733
At t = 30, 0.2t = 0.2 × 30 = 6
P(t = 30) = 1400 (1 + 13e⁻⁶) = 1400 (1 + 0.00248) = 1,403.47 = 1,403
Hope this Helps!!
Can someone help me on this question please
Answer:
about 11.11%, or 4/36
Step-by-step explanation:
Answer:
1/9
Step-by-step explanation:
There are 4 possible ways to get a sum of 5 out of total ways of 6*6= 36
So the probability of getting sum of 5 is 4/36= 1/9
Harper's Index reported that the number of (Orange County, California) convicted drunk drivers whose sentence included a tour of the morgue was 506, of which only 1 became a repeat offender.
a. Suppose that of 1056 newly convicted drunk drivers, all were required to take a tour of the morgue. Let us assume that the probability of a repeat offender is still p= 1/596. Explain why the Poisson approximation to the binomial would be a good choice for r = number of repeat offenders out of 963 convicted drunk drivers who toured the morgue.
The Poisson approximation is good because n is large, p is small, and np < 10.The Poisson approximation is good because n is large, p is small, and np > 10. The Poisson approximation is good because n is large, p is large, and np < 10.The Poisson approximation is good because n is small, p is small, and np < 10. What is λ to the nearest tenth?
b. What is the probability that r = 0? (Use 4 decimal places.)
c. What is the probability that r > 1? (Use 4 decimal places.)
d. What is the probability that r > 2? (Use 4 decimal places.)
e. What is the probability that r > 3? (Use 4 decimal places.)
Answer:
a. The Poisson approximation is good because n is large, p is small, and np < 10.
The parameter of thr Poisson distribution is:
[tex]\lambda =np\approx1.6[/tex]
b. P(r=0)=0.2019
c. P(r>1)=0.4751
d. P(r>2)=0.2167
e. P(r>3)=0.0789
Step-by-step explanation:
a. The Poisson distribution is appropiate to represent binomial events with low probability and many repetitions (small p and large n).
The approximation that the Poisson distribution does to the real model is adequate if the product np is equal or lower than 10.
In this case, n=963 and p=1/596, so we have:
[tex]np=963*(1/596)\approx1.6[/tex]
The Poisson approximation is good because n is large, p is small, and np < 10.
The parameter of thr Poisson distribution is:
[tex]\lambda =np\approx1.6[/tex]
We can calculate the probability for k events as:
[tex]P(r=k)=\dfrac{\lambda^ke^{-\lambda}}{k!}[/tex]
b. P(r=0). We use the formula above with λ=1.6 and r=0.
[tex]P(0)=1.6^{0} \cdot e^{-1.6}/0!=1*0.2019/1=0.2019\\\\[/tex]
c. P(r>1). In this case, is simpler to calculate the complementary probability to P(r<=1), that is the sum of P(r=0) and P(r=1).
[tex]P(r>1)=1-P(r\leq1)=1-[P(r=0)+P(r=1)]\\\\\\P(0)=1.6^{0} \cdot e^{-1.6}/0!=1*0.2019/1=0.2019\\\\P(1)=1.6^{1} \cdot e^{-1.6}/1!=1.6*0.2019/1=0.3230\\\\\\P(r>1)=1-(0.2019+0.3230)=1-0.5249=0.4751[/tex]
d. P(r>2)
[tex]P(r>2)=1-P(r\leq2)=1-[P(r=0)+P(r=1)+P(r=2)]\\\\\\P(0)=1.6^{0} \cdot e^{-1.6}/0!=1*0.2019/1=0.2019\\\\P(1)=1.6^{1} \cdot e^{-1.6}/1!=1.6*0.2019/1=0.3230\\\\P(2)=1.6^{2} \cdot e^{-1.6}/2!=2.56*0.2019/2=0.2584\\\\\\P(r>2)=1-(0.2019+0.3230+0.2584)=1-0.7833=0.2167[/tex]
e. P(r>3)
[tex]P(r>3)=1-P(r\leq2)=1-[P(r=0)+P(r=1)+P(r=2)+P(r=3)]\\\\\\P(0)=1.6^{0} \cdot e^{-1.6}/0!=1*0.2019/1=0.2019\\\\P(1)=1.6^{1} \cdot e^{-1.6}/1!=1.6*0.2019/1=0.3230\\\\P(2)=1.6^{2} \cdot e^{-1.6}/2!=2.56*0.2019/2=0.2584\\\\P(3)=1.6^{3} \cdot e^{-1.6}/3!=4.096*0.2019/6=0.1378\\\\\\P(r>3)=1-(0.2019+0.3230+0.2584+0.1378)=1-0.9211=0.0789[/tex]
Determine whether b can be written as a linear combination of Bold a Subscript Bold 1a1, Bold a Subscript Bold 2a2, and Bold a Subscript Bold 3a3. In other words, determine whether weights x 1x1, x 2x2, and x 3x3 exist, such that x 1x1Bold a Subscript Bold 1a1plus+x 2x2Bold a Subscript Bold 2a2plus+x 3x3Bold a Subscript Bold 3a3equals=b. Determine the weights x 1x1, x 2x2, and x 3x3 if possible.
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
b can be written as a linear combination of [tex]a_1 \ and \ a_2[/tex]
b
The values of [tex]x_1 = 4 \ and \ x_2 = 2[/tex]
Step-by-step explanation:
From the question we are told that
[tex]x_1 a_1 +x_2 a_2 = b[/tex]
Where [tex]a_ 1 = (4, 5,-4)[/tex], [tex]a_2 = (-4 , 3, 3)[/tex] and [tex]b = (8,26 , -10)[/tex]
So
[tex]x_1 ( 4, 5,-4) + x_2 (-4 , 3, 3) = (8,26 , -10)[/tex]
[tex]4x_1, 5x_1,-4x_1 + -4x_2 , 3x_2, 3x_2 = (8,26 , -10)[/tex]
=> [tex]4x_1 -4x_2 =8[/tex]
[tex]x_1 -x_2 =2 ---(1)[/tex]
=> [tex]5x_1 + 3x_2 = 26 --- (2)[/tex]
=> [tex]-4x_1 + 3x_2 = -10 ---(3)[/tex]
Now multiplying equation 1 by 3 and adding the product to equation 2
[tex].\ \ \ 3x_1 -3x_2 = 6\\+ \ \ 5x_1 + 3x_2 = 26 \\=> \ \ \ 8x_1 = 32[/tex]
=> [tex]x_1 = 4[/tex]
substituting [tex]x_1[/tex] into equation 1
[tex]4 - x_2 =2[/tex]
[tex]x_2 =2[/tex]
Now to test substitute [tex]x_1 \ and \ x_2[/tex] into equation 3
[tex]-4(4) + 3(2) = -10[/tex]
[tex]-10 = -10[/tex]
Since LHS = RHS then there exist values [tex]x_1 = 4 \ and \ x_2 = 2[/tex] such that
[tex]x_1 a_1 +x_2 a_2 = b[/tex]
Hence b can be written as a linear combination of [tex]a_1 \ and \ a_2[/tex]
Convert 8.25% to a decimal
Answer: 0.0825
Step-by-step explanation:
8.25% to decimal is simply divide 8.25 by 100
8.25/100= 0.0825
The required solution after convert it into decimal is,
⇒ 0.0825
We have to given that,
Convert 8.25% into a decimal.
We can change it as,
⇒ 8.25%
Since, 1% = 1/100
Hence,
⇒ 8.25/100
⇒ 0.0825
Therefore, After convert it into decimal number solution is,
⇒ 0.0825
Learn more about the percent visit:
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A manufacturer of detergent claims that the contents of boxes sold weigh on average at least 20 ounces. The distribution of weight is known to be normal, with a standard deviation of 0.4 ounces. A random sample of 25 boxes yielded a sample mean weight of 19.87 ounces. Test at the 1% significance level the null hypothesis that the population mean weight is at least 20 ounces.
Answer:
[tex]z=\frac{19.87-20}{\frac{0.4}{\sqrt{25}}}=-1.625[/tex]
The p value for this case would be given by:
[tex]p_v =P(z<-1.625)=0.052[/tex]
Since the p value is higher than the significance level provided we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 20 ounces.
Step-by-step explanation:
Information provided
[tex]\bar X=19.87[/tex] represent the sample mean
[tex]\sigma=0.4[/tex] represent the population deviation
[tex]n=25[/tex] sample size
[tex]\mu_o =68[/tex] represent the value that we want to test
[tex]\alpha=0.01[/tex] represent the significance level
z would represent the statistic
[tex]p_v[/tex] represent the p value
Hypothesis to test
We want to test if the true mean is at least 20 ounces, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 20[/tex]
Alternative hypothesis:[tex]\mu < 20[/tex]
The statistic is given by:
[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)
Replacing the info given we got:
[tex]z=\frac{19.87-20}{\frac{0.4}{\sqrt{25}}}=-1.625[/tex]
The p value for this case would be given by:
[tex]p_v =P(z<-1.625)=0.052[/tex]
Since the p value is higher than the significance level provided we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 20 ounces.
A consumer agency is investigating the blowout pressures of Soap Stone tires. A Soap Stone tire is said to blow out when it separates from the wheel rim due to impact forces usually caused by hitting a rock or a pothole in the road. A random sample of 29 Soap Stone tires were inflated to the recommended pressure, and then forces measured in foot-pounds were applied to each tire (1 foot-pound is the force of 1 pound dropped from a height of 1 foot). The customer complaint is that some Soap Stone tires blow out under small-impact forces, while other tires seem to be well made and don't have this fault. For the 29 test tires, the sample standard deviation of blowout forces was 1358 foot-pounds.
(a) Soap Stone claims its tires will blow out at an average pressure of 26,000 foot-pounds, with a standard deviation of 1020 foot-pounds. The average blowout force is not in question, but the variability of blowout forces is in question. Using a 0.1 level of significance, test the claim that the variance of blowout pressures is more than Soap Stone claims it is.
Classify the problem as being a Chi-square test of independence or homogeneity, Chi-square goodness-of-fit, Chi-square for testing or estimating σ2 or σ, F test for two variances, One-way ANOVA, or Two-way ANOVA, then perform the following.
(ii) Find the sample test statistic. (Use two decimal places.)
(iii) Find the P-value of the sample test statistic. (Use four decimal places.)
(b) Find a 99% confidence interval for the variance of blowout pressures, using the information from the random sample. (Use one decimal place.)
Answer:
Step-by-step explanation:
Hello!
The variable of interest is:
X: Impact force needed for Sap Stone tires to blow out. (foot-pounds)
n= 29, S= 1358 foot-pound
a)
Soap Stone claims that "The tires will blow out at an average pressure of μ= 26000 foot-pounds with a standard deviation of σ= 1020 foot-pounds.
According to the consumer's complaint, the variability of the blown out forces is greater than the value determined by the company.
I)
Then the parameter of interest is the population variance (or population standard deviation) and to test the consumer's complaint you have to conduct a Chi-Square test for σ².
σ²= (1020)²= 1040400 foot-pounds²
H₀: σ² ≤ 1040400
H₁: σ² > 1040400
α: 0.01
II)
[tex]X^2= \frac{(n-1)S^2}{Sigma^2} ~~X^2_{n-1}[/tex]
[tex]X^2_{H_0}= \frac{(n-1)S^2}{Sigma^2}= \frac{(29-1)*(1358)^2}{1040400} = 49.63[/tex]
III)
This test is one-tailed to the right and so is the p-value. This distribution has n-1= 29-1= 28 degrees of freedom, so you can calculate the p-value as:
P(X²₂₈≥49.63)= 1 - P(X²₂₈<49.63)= 1 - 0.99289= 0.00711
⇒ The p-value is less than the significance level so the test is significant at 1%. You can conclude that the population variance of the blowout forces is less than 1040400 foot-pounds², at the same level the population standard deviation of the blow out forces is less than 1020 foot-pounds.
b)
99% CI for the variance. Using the X² statistic you can calculate it as:
[tex][\frac{(n-1)S^2}{X^2_{n-1;1-\alpha /2}} ;\frac{(n-1)S^2}{X^2_{n-1;\alpha /2}} ][/tex]
[tex]X^2_{n-1;\alpha /2}= X^2_{28; 0.005}= 13.121[/tex]
[tex]X^2_{n-1;1-\alpha /2}= X^2_{28; 0.995}= 49.588[/tex]
[tex][\frac{28*(1358)^2}{49.588} ;\frac{28*(1358)^2}{13.121} ][/tex]
[1041312.253; 3935415.898] foot-pounds²
I hope this helps!
In a study, 37% of adults questioned reported that their health was excellent. A researcher wishes to study the health of people living close to nuclear plant. Among 12 adults randomly selected from this area, only 3 reported that their health was excellent. Find the probability that when 12 adults are randomly selected, 3 or fewer are in excellent health.
Answer:
The probability that when 12 adults are randomly selected, 3 or fewer are in excellent health is 0.2946
Step-by-step explanation:
BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 12 * 0.37
= 4.44
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 12 * 0.37 * 0.63
= 2.7972
III.
standard deviation = sqrt( variance ) = sqrt(2.7972)
=1.6725
the probability that when 12 adults are randomly selected, 3 or fewer are in excellent health
P( X < = 3) = P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 12 3 ) * 0.37^3 * ( 1- 0.37 ) ^9 + ( 12 2 ) * 0.37^2 * ( 1- 0.37 ) ^10 + ( 12 1 ) * 0.37^1 * ( 1- 0.37 ) ^11 + ( 12 0 ) * 0.37^0 * ( 1- 0.37 ) ^12
= 0.2947
According to the image whats the answer? 80 points brainliest
Answer:
Hey!
Your answer is 150 square meters!
Step-by-step explanation:
13*3=39 (the tilted face)
12*3=36 (the flat face)
5*3=15 (the base)
TO FIND THE TRINGLULAR AREA:
1/2 base x height...
1/2 x 5 x 12 = 30
(WE DOUBLE THIS BECAUSE THERE ARE TWO TRIANGLES)
SO 60...
ADD THESE TOGETHER...
39 + 36 + 15 + 30 + 30...
GIVES US 150 square metres
HOPE THIS HELPS!!Answer:
150 square meters
Step-by-step explanation:
What is the range of g?
Answer:
{-7, -4, -1, 3, 7}
Step-by-step explanation:
The range is the list of y-coordinates of the points:
range = {-7, -4, -1, 3, 7}
Benson asked a group of 9th, 10th, 11th, and 12th graders the
number of hours they work per week in the summer. The means of
each group are provided below:
9th: 14.2 hours
10th: 18.8 hours
11th: 21.2 hours
12th: 24.9 hours
Which conclusion can be made from this data?
Answer:
YlThe higher the grader the more hours they can work.
Step-by-step explanation:
We can see that the higher the grader the more hours they can work ; which could mean less academic work if the work defined is that with which to earn money but if the work defined is academic it means more hours of academic work
Suppose the average driving distance for last year's Player's Champion Golf Tournament in Ponte Vedra, FL, was 292.5 yards with a standard deviation of 14.2 yards. A random sample of 60 drives was selected from a total of 4,244 drives that were hit during this tournament. What is the probability that the sample average was 289 yards or less?
Answer:
The Probability that the sample average was 289 yards or less
P(x⁻≤ 289) = P( Z≤ -1.909) = 0.0287
Step-by-step explanation:
step(i):-
Mean of the Population = 292.5 yards
Standard deviation of the Population = 14.2 yards
sample size 'n' =60 drives
N = 4244 drives
Step(ii):-
Let X⁻ be random sample average
[tex]Z = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }[/tex]
Let X⁻ = 289
[tex]Z = \frac{289 -292.5}{\frac{14,2}{\sqrt{60} } }[/tex]
Z = - 1.909
The Probability that the sample average was 289 yards or less
P(x⁻≤ 289) = P( Z≤ -1.909)
= 0.5 -A(1.909)
= 0.5 -0.4713
= 0.0287
Conclusion:-
The Probability that the sample average was 289 yards or less = 0.0287
1. Divide 6/13 by 6/12
A. 12/13
B. 1/12
C. 13/12
D.916
Answer:
C
Step-by-step explanation:
[tex]\frac{6}{13}[/tex]÷[tex]\frac{6}{12}[/tex] can also be [tex]\frac{6}{13}[/tex]×[tex]\frac{12}{3}[/tex] and [tex]\frac{6}{13}[/tex]×[tex]\frac{12}{3}[/tex]=13/12
Step-by-step explanation:
Answer:
C
Step-by-step explanation:
\frac{6}{13}136 ÷\frac{6}{12}126 can also be \frac{6}{13}136 ×\frac{12}{3}312 and \frac{6}{13}136 ×\frac{12}{3}312 =13/12
what is the product a-3/7 ÷ 3-a/21
Answer:
the sum shown is (20/21)a -(1/7)
Step-by-step explanation:
As written, the sum is ...
[tex]a-\dfrac{\frac{3}{7}}{3}-\dfrac{a}{21}=\boxed{\dfrac{20}{21}a-\dfrac{1}{7}}[/tex]
__
We wonder if you mean the quotient ...
((a-3)/7)/((3-a)/21)
[tex]\dfrac{\left(\dfrac{a-3}{7}\right)}{\left(\dfrac{3-a}{21}\right)}=\dfrac{a-3}{7}\cdot\dfrac{21}{3-a}=\dfrac{-21(3-a)}{7(3-a)}=\boxed{-3}[/tex]
_____
Comment on the problem presentation
Parentheses are required when plain text is used to represent fractions. The symbols ÷, /, and "over" all mean the same thing: "divided by." The denominator is the next item in the expression. If arithmetic of any kind is involved in the denominator, parentheses are needed. This is the interpretation required by the Order of Operations.
When the expression is typeset, fraction bars and text formatting (superscript) serve to group items that require parentheses in plain text.
__
Please note that some authors make a distinction between the various forms of division symbol. Some use ÷ to mean ...
(everything to the left)/(everything to the right)
and they reserve / solely for use in fractions. Using this interpretation, your expression would be ...
(a -(3/7))/(3 -(a/21)) = (21a -9)/(63 -a)
That distinction is not supported by the Order of Operations.
What is the solution to the following system?
a
x+y+z= 6
X-Y+Z= 8
X+y-Z=0
Answer:
x= 4
Y= -1
Z= 3
step by step procedure:
let x+y+z=6, x-y+z=8, x+y-z=0 be equations 1, 2, and 3 respectively.
subtract equation 2 from equation 1 ; x+y+z-(x-y+z) = 6 - 8 y+y = -2 , 2y = -2, therefore y = -1substituting the value of y into equations 1, 2, and 3 . we get x+z=7---eqn 4, x+z=7---eqn 5 and x-z=1---eqn 6now add equation 6 to equation 5 ; x+z+(x-z)=7+1 ,x+x=7+1, 2x=8, therefore x=4put the value of x (x=4) back into equation 5; 4+z=7, z=7-4, therefore z=3.you are welcome bro....
What is the volume of a cube with a side length of 14 cm.?
Answer:
V =2744 cm^3
Step-by-step explanation:
The volume of a cube is given by
V = s^3 where s is the side length
V = 14^3
V =2744 cm^3
Answer: 2,744 cm³
Step-by-step explanation: Since the length, width, and height of a cube are all equal, we can find the volume of a cube by multiplying side × side × side.
So we can find the volume of a cube using the formula s³.
Notice that we have a side length of 14cm.
So plugging into the formula, we have (14 cm)³ or
(14 cm)(14 cm)(14 cm) which is 2,744 cm³.
So the volume of the cube is 2,744 cm³.
The estimator Yis a random variable that varies with different random samples; it has a probability distribution function that represents its sampling distribution, and mean and variance. Using the properties on expected values and variances of linear functions of random variables and sum operators, show that:
A. E(Y) = μ
B. Var(Y) σ2/N.
Answer:
Check Explanation
Step-by-step explanation:
According to the Central limit theorem, the population mean (μ) is approximately equal to the mean of sampling distribution (μₓ).
And the standard deviation of the sampling distribution (σₓ) is related to the population standard deviation (σ) through
Standard deviation of the sampling distribution = (Population standard deviation)/(√N)
where N = Sample size
σₓ = (σ/√N)
So, population mean (μ) = Mean of sampling distribution (μₓ)
Population Standard deviation = (Standard deviation of the sampling distribution) × √N
= σ × √N
A) The expected value of a given distribution is simply equal to the mean of that distribution.
Hence, the expected value of random variable Y thay varies with different samples is given as
E(Y) = Mean of sampling distribution = μₓ
But μₓ = μ
Hence, E(Y) = μ (Proved)
B) Var (Y) is given as the square of the random distribution's standard deviation.
Var (Y) = (standard deviation of the sampling distribution)²
= (σ/√N)²
= (σ²/N) (Proved)
Hope this Helps!!!
What is the quadratic formula?
Answer:
The quadratic formula is x= -b±√b²-4ac all over 2a
Step-by-step explanation:
The -b±√b²-4ac is part of the numerator of the fraction. And b²-4ac is all under the radical sign. The denominator is 2a so you divide -b±√b²-4ac by 2a to get the solutions of the quadratic function. Hope this helps!!
what measurement do you use for surface area
Answer:
Step-by-step explanation:
Surface Area is the combined area of all two-dimensional surfaces of a shape. Just like ordinary area, the units are the squares of the units of length (that is, if a shape's sides are measured in meters, then the shape's area is measured in square meters
Pedro is selling candy bars for his school baseball team each candy bar cost 1.25 and a box contains 20 bars
Answer:
$25
Step-by-step explanation:
You do 1.25. Y 20
A tank is filled with 1000 liters of pure water. Brine containing 0.04 kg of salt per liter enters the tank at 9 liters per minute. Another brine solution containing 0.05 kg of salt per liter enters the tank at 7 liters per minute. The contents of the tank are kept thoroughly mixed and the drains from the tank at 16 liters per minute. A. Determine the differential equation which describes this system. Let S(t)S(t) denote the number of kg of salt in the tank after tt minutes. Then
Answer:
The differential equation which describes the mixing process is [tex]\frac{dc_{salt,out}}{dt} + \frac{2}{125}\cdot c_{salt,out} = \frac{71}{100000}[/tex].
Step-by-step explanation:
The mixing process within the tank is modelled after the Principle of Mass Conservation, which states that:
[tex]\dot m_{salt,in} - \dot m_{salt,out} = \frac{dm_{tank}}{dt}[/tex]
Physically speaking, mass flow of salt is equal to the product of volume flow of water and salt concentration. Then:
[tex]\dot V_{water, in, 1}\cdot c_{salt, in,1} + \dot V_{water, in, 2} \cdot c_{salt,in, 2} - \dot V_{water, out}\cdot c_{salt, out} = V_{tank}\cdot \frac{dc_{salt,out}}{dt}[/tex]
Given that [tex]\dot V_{water, in, 1} = 9\,\frac{L}{min}[/tex], [tex]\dot V_{water, in, 2} = 7\,\frac{L}{min}[/tex], [tex]c_{salt,in,1} = 0.04\,\frac{kg}{L}[/tex], [tex]c_{salt, in, 2} = 0.05\,\frac{kg}{L}[/tex], [tex]\dot V_{water, out} = 16\,\frac{L}{min}[/tex] and [tex]V_{tank} = 1000\,L[/tex], the differential equation that describes the system is:
[tex]0.71 - 16\cdot c_{salt,out} = 1000\cdot \frac{dc_{salt,out}}{dt}[/tex]
[tex]1000\cdot \frac{dc_{salt, out}}{dt} + 16\cdot c_{salt, out} = 0.71[/tex]
[tex]\frac{dc_{salt,out}}{dt} + \frac{2}{125}\cdot c_{salt,out} = \frac{71}{100000}[/tex]
A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second.(a) What is the velocity of the top of the ladder when the base is given below?7 feet away from the wall ft/sec20 feet away from the wall ft/sec24 feet away from the wall ft/sec(b) Consider the triangle formed by the side of the house, ladder, and the ground. Find the rate at which the area of the triangle is changing when the base of the ladder is 7 feet from the wall. ft2/sec(c) Find the rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 7 feet from the wall. rad/sec
Answer:
a) The height decreases at a rate of [tex]\frac{7}{12}[/tex] ft/sec.
b) The area increases at a rate of [tex]\frac{527}{24}[/tex] ft^2/sec
c) The angle is increasing at a rate of [tex]\frac{1}{12}[/tex] rad/sec
Step-by-step explanation:
Attached you will find a sketch of the situation. The ladder forms a triangle of base b and height h with the house. The key to any type of problem is to identify the formula we want to differentiate, by having in mind the rules of differentiation.
a) Using pythagorean theorem, we have that [tex] 25^2 = h^2+b^2[/tex]. From here, we have that
[tex]h^2 = 25^2-b^2[/tex]
if we differentiate with respecto to t (t is time), by implicit differentiation we get
[tex]2h \frac{dh}{dt} = -2b\frac{db}{dt}[/tex]
Then,
[tex]\frac{dh}{dt} = -\frac{b}{h}\frac{db}{dt}[/tex].
We are told that the base is increasing at a rate of 2 ft/s (that is the value of db/dt). Using the pythagorean theorem, when b = 7, then h = 24. So,
[tex]\frac{dh}{dt} = -\frac{2\cdot 7}{24}= \frac{-7}{12}[/tex]
b) The area of the triangle is given by
[tex]A = \frac{1}{2}bh[/tex]
By differentiating with respect to t, using the product formula we get
[tex] \frac{dA}{dt} = \frac{1}{2} (\frac{db}{dt}h+b\frac{dh}{dt})[/tex]
when b=7, we know that h=24 and dh/dt = -1/12. Then
[tex]\frac{dA}{dt} = \frac{1}{2}(2\cdot 24- 7\frac{7}{12}) = \frac{527}{24}[/tex]
c) Based on the drawing, we have that
[tex]\sin(\theta)= \frac{b}{25}[/tex]
If we differentiate with respect of t, and recalling that the derivative of sine is cosine, we get
[tex] \cos(\theta)\frac{d\theta}{dt}=\frac{1}{25}\frac{db}{dt}[/tex] or, by replacing the value of db/dt
[tex]\frac{d\theta}{dt}=\frac{2}{25\cos(\theta)}[/tex]
when b = 7, we have that h = 24, then [tex]\cos(\theta) = \frac{24}{25}[/tex], then
[tex]\frac{d\theta}{dt} = \frac{2}{25\frac{24}{25}} = \frac{2}{24} = \frac{1}{12}[/tex]
Type the correct answer in the box. Spell all words correctly.
Which term describes the solution to a multiplication problem?
The
describes the solution to a multiplication problem.
Answer:
product is right
Step-by-step explanation:
what is 1+1, this is for you to get points.
Answer:
2
Step-by-step explanation:
1+1 = 2
Hope this helps;)
Answer:
1 + 1 =2
Step-by-step explanation:
Hi, how are you?
Do u want to be my friend?
Also, I hope this answered your question?
The graph to the right is the uniform density function for a friend whoThe graph to the right is the uniform density function for a friend who is x minutes late. Find the probability that the friend is at least 21 minutes late. is x minutes late. Find the probability that the friend is at least 21 minutes late.
Answer:
0.30
Step-by-step explanation:
Data provided in the question
Uniform density function for a friend = x minutes late
The Friend is at least 21 minutes late
Based on the above information, the probability that the friend is at least 21 minutes late is
[tex]= \frac{Total\ minutes - minimum\ minutes}{Total\ minutes}[/tex]
[tex]= \frac{30 - 21}{30}[/tex]
= 0.30
Based on the above formula we can easily find out the probability for the friend who is at least 21 minutes late
The probability of the friend to be at least 21 minutes late for the uniform density function shown in the graph is 0.30.
What is probability?Probability of an event is the ratio of number of favorable outcome to the total number of outcome of that event.
The graph is attached below shows the uniform density function for a friend who is x minutes late.
The probability that the friend is at least 21 minutes late-The friend is at least 21 minutes late. This means that the friend is 21 minutes late or more than it. 21 or more minutes goes from 21 to 30. Thus, the difference is,
[tex]d=30-21\\d=9[/tex]
The density of the graph is 1/30. The probability will be equal to the area under the curve.
In this, the length of the rectangle will be 9 and width will be 1/30 for the probability of at least 21 minutes late. The probability is,
[tex]P=9\times\dfrac{1}{30}\\P=0.30[/tex]
Thus, the probability of the friend to be at least 21 minutes late for the uniform density function shown in the graph is 0.30.
Learn more about the probability here;
https://brainly.com/question/24756209
looking for the 2 correct answers please
Answer:
ROQQOPStep-by-step explanation:
The sine of an angle is equal to the cosine of its complement, and vice versa.
sin ∠QOP = cos ∠ROQ
cos ∠ROQ = sin ∠QOP
In November Hillary drove 580 miles in her car the car travelled 33.5 miles for each gallon of petrol used
Petrol cost £1.09 per litre
1 gallon = 4.55 litres
Work out the cost of the petrol the car used in November
Answer:
T = £85.87
the cost of the petrol the car used in November is £85.87
Step-by-step explanation:
Given;
In November Hillary drove 580 miles in her car;
Distance travelled d = 580 miles
the car travelled 33.5 miles for each gallon of petrol used;
Fuel consumption rate r = 33.5 miles per gallon
Number of gallons N consumed by the car is;
N = distance travelled/fuel consumption rate
N = d/r = 580/33.5 = 17.3134 gallons
Given that;
Petrol cost £1.09 per litre
Cost per litre c = £1.09
1 gallon = 4.55 litres
Converting the amount of fuel used to litres;
N = 17.3134 gallons × 4.55 litres per gallon
N = 78.77612 litres
The total cost T = amount of fuel consumption N × fuel cost per litre c
T = N × c
T = 78.77612 litres × £1.09 per litre
T = £85.87
the cost of the petrol the car used in November is £85.87
The box plots show the weights, in pounds, of the dogs in two different animal shelters.
Weights of Dogs in Shelter A
6
8
10
12
14
16
18
20
22
24
26
28
30
Weights of Dogs in Shelter B
Which describes the spread of the data in the two box plots?
O
The data in shelter A show more spread than the data in shelter B.
The data in shelter B show more spread than the data in shelter A
The data in shelter A range from about 17 to 28, while the data in shelter B range from about 16 to 20
The data in shelter A range from about 21 to 28, while the data in shelter B range from about 18 to 20.
Answer:
The data in shelter A show more spread than the data in shelter B.
Step-by-step Explanation:
Judging from the data set given for shelter A and shelter B in the diagram of the box plots attached below, shelter A has a data set that ranges from 8 pounds to 30 pounds, while shelter B ranges from e can infer that the data in shelter A has more spread than the data in shelter B ranges from 10 to 28.
Taking into consideration the range of the data set of each shelter, shelter A obviously has a larger range compared to shelter B. This implies that, the data in shelter A show more spread than the data in shelter B.
Answer:
A
Step-by-step explanation:
what is an equation of the line that passes through the point (-3,-1) and is perpendicular to the line y=1/2x - 3
Answer:
Step-by-step explanation: gradient of perpendicular line is
-2
y+1/x+3=-2
y+1=-2x-6
y=-2x-7
If h = 12 units and r = 4 units, what is the volume of the cone shown above? Use 3.14 for .
Answer:
200.96 units
Step-by-step explanation:
Use the formula for the volume of a cone [tex]V=\pi r^{2} \frac{h}{3}[/tex]
Plug in the values ([tex]\pi[/tex]=3.14) and multiply them all out
Answer:
≈ 201
Step-by-step explanation:
V= πr²h/3
V= 3.14*4²*12/3= 200.96 ≈ 201
A padaria Beira Baixa vem perdendo dinheiro com a sua política de produção de venda do pãozinho francês. Atualmente, a cada dia é feita uma quantidade igual à procura do dia anterior. O proprietário, Sr. Joaquim, recolheu os seguintes dados:
Lucro na venda= $ 0,20 por unidade
Perda no pão não vendido = $ 0,05
(os pães não vendidos são moídos e vendidos como farinha de rosca: cada pão equivale a $0,15 de farinha de rosca)
Distribuição de frequência da demanda diária. Quantos pães devem ser fabricados diariamente para maximizar o lucro médio do Sr. Joaquim
1000 0,10
1100 0,10
1200 0,10
1300 0,15
1400 0,20
1500 0,15
1600 0,10
1700 0,05
1800 0,05
Answer:
The number of loaves to be made daily to maximize Mr. Joaquim's average profit is 1400.
O número de pães a serem feitos diariamente para maximizar o lucro médio do Sr. Joaquim é 1400.
Step-by-step explanation:
The table gives the daily demand of bread and the average profit that corresponds to each demand.
The daily demand values recorded ranged from 1000 to 1800.
And the corresponding average profit for the demands for each daily demand provided showed a steady increase from $0.10 as at a daily demand of 1000, through $0.15, then peaking at $0.20 at a daily demand of 1400.
After this point, the average profit begins to decline again as we move further away from the peak obtained at a daily demand of 1400.
Hence, it is evident that the daily demand that corresponds to the maximum average profit is 1400.
A daily demand of 1400 is where the peak average profit of $0.20 is obtained.
Any other demand lower or higher than 1400 corresponds to a departure from the peak profit obtained at that point according to the table of daily demand and average profits provided in the question.
In portugese/Em português
A tabela fornece a demanda diária de pão e o lucro médio que corresponde a cada demanda.
Os valores diários de demanda registrados variaram de 1000 a 1800.
E o lucro médio correspondente às demandas de cada demanda diária fornecida mostrou um aumento constante de US $ 0,10, com uma demanda diária de 1000, passando por US $ 0,15, atingindo um pico de US $ 0,20 com uma demanda diária de 1400.
Após esse ponto, o lucro médio começa a declinar novamente à medida que nos afastamos do pico obtido com uma demanda diária de 1400.
Portanto, é evidente que a demanda diária que corresponde ao lucro médio máximo é de 1.400.
Uma demanda diária de 1400 é onde é obtido o lucro médio máximo de US $ 0,20.
Qualquer outra demanda menor ou maior que 1400 corresponde a um afastamento do pico de lucro obtido nesse momento, de acordo com a tabela de demanda diária e lucros médios fornecidos na pergunta.
Hope this Helps!!!
Espero que isto ajude!!!