Answer:
[tex]y=exp(\int\limits^x_4 {e^{-t^{2} } } \, dt)[/tex]
Step-by-step explanation:
This is a separable equation with an initial value i.e. y(3)=1.
Take y from right hand side and divide to left hand side ;Take dx from left hand side and multiply to right hand side:
[tex]\frac{dy}{y} =e^{-x^{2} }dx[/tex]
Take t as a dummy variable, integrate both sides with respect to "t" and substituting x=t (e.g. dx=dt):
[tex]\int\limits^x_3 {\frac{1}{y} } \, \frac{dy}{dt} dt=\int\limits^x_3 {e^{-t^{2} } } dt[/tex]
Integrate on both sides:
[tex]ln(y(t))\left \{ {{t=x} \atop {t=3}} \right. =\int\limits^x_3 {e^{-t^{2} } } \, dt[/tex]
Use initial condition i.e. y(3) = 1:
[tex]ln(y(x))-(ln1)=\int\limits^x_3 {e^{-t^{2} } } \, dt\\ln(y(x))=\int\limits^x_3 {e^{-t^{2} } } \, dt\\[/tex]
Taking exponents on both sides to remove "ln":
[tex]y=exp (\int\limits^x_3 {e^{-t^{2} } } \, dt)[/tex]
The teacher recorded the mean and median of the hourly wage for each student. Unfortunately, he forgot to label them. The numbers he wrote down were: $11.25/hour and $9.38/hour. Which would be the mean and which would be the median
Step-by-step explanation:
We can not exactly predict the values of mean and median of the data un till and unless we know about the skewness of the data.
Skewness represents the asymmetry or tapering in the distribution of data sample. If skewness is
Negative skew: median > mean:
Positive skew: mean > median :
Although this generalization is not always true.
Teacl Smolts are young salmon at a stage when their skin becomes covered with silvery scales and they start to migrate from freshwater to the sea. The reflectance of a light shined on a smolt's skin is a measure of the smolt's readiness for the migration. Here are the reflectances, in percents, for a sample of 50 smolts. 57.6 54.9 63.4 57.1 54.8 42.4 63.5 55.5 33.6 63.2 58.3 42.0 56.0 47.7 56.2 56.0 38.7 49.6 42.2 45.7 69.1 50.5 53.0 38.3 60.4 49.2 42.8 44.6 46.3 44.2 59.0 42.0 47.6 47.9 69.2 46.6 68.0 42.9 45.5 474 59.5 37.7 53.9 43.1 51.3 64.6 43.7 42.7 50.9 43.8 (a) Find the standard deviation of the reflectance for these smolts. (Round your answer to two decimal places.) (b) Find the quartiles of the reflectance for these smolts (c) Do you prefer the standard deviation or the quartiles as a measure of spread for these data? Give reasons for your preference.
Answer:
a) s = 8.81
b) Q1 = 43.4, Q2 = 49.4, Q3 = 57.35
c) See below
Step-by-step explanation:
(a) Find the standard deviation of the reflectance for these smolts. (Round your answer to two decimal places.)
In order to find the standard deviation, we need the mean first. The mean is defined as
[tex]\bar x=\displaystyle\frac{\displaystyle\sum_{i=1}^{n}x_i}{n}[/tex]
where the [tex]x_i[/tex] are the values of the data collected and n=50 the size of the sample.
So, the mean is
[tex]\bar x=50.882[/tex]
Now, the standard deviation of the sample is defined as
[tex]s=\sqrt{\displaystyle\frac{\displaystyle\sum_{i=1}^n(x_i-\bar x)^2}{n-1}}[/tex]
and we have that our standard deviation is
s = 8.81
(b) Find the quartiles of the reflectance for these smolts
To find the quartiles, we must sort the data from lowest to largest:
33.6, 37.7, 38.3, 38.7, 42, 42, 42.2, 42.4, 42.7, 42.8, 42.9, 43.1, 43.7, 43.8, 44.2, 44.6, 45.5, 45.7, 46.3, 46.6, 47.4, 47.6, 47.7, 47.9, 49.2, 49.6, 50.5, 50.9, 51.3, 53, 53.9, 54.8, 54.9, 55.5, 56, 56, 56.2, 57.1, 57.6, 58.3, 59, 59.5, 60.4, 63.2, 63.4, 63.5, 64.6, 68, 69.1, 69.2
The first quartile is the number between the 12th and the 13th data (so 25% of the data are below it and 75% above it)
So the 1st quartile is
[tex]Q_1=\displaystyle\frac{43.1+43.7}{2}=43.4[/tex]
The 2nd quartile is the median, the point between the 25th and 26th data, it splits the data in two halves.
[tex]Q_2=\displaystyle\frac{49.2+49.6}{2}=49.4[/tex]
The 3rd quartile is the point between the 38th and 39th data (so 75% of the data are below it and 25% above it)
[tex]Q_3=\displaystyle\frac{57.1+57.6}{2}=57.35[/tex]
(c) Do you prefer the standard deviation or the quartiles as a measure of spread for these data? Give reasons for your preference.
In this case, we prefer the quartiles as a measure of spread since the data are very scattered around the mean and there is no a central tendency.
Exercise 3.23 introduces a husband and wife with brown eyes who have 0.75 probability of having children with brown eyes, 0.125 probability of having children with blue eyes, and 0.125 probability of having children with green eyes
(a) What is the probability that their first child will have green eyes and the second will not?
(b) What is the probability that exactly one of their two children will have green eyes?
(c) If they have six children, what is the probability that exactly two will have green eyes?
(d) If they have six children, what is the probability that at least one will have green eyes?
Answer:
a) There is a 10.9375% probability that their first child will have green eyes and the second will not.
b) There is a 21.875% probability that exactly one of their two children will have green eyes.
c) There is a 13.74% probability that exactly two will have green eyes.
d) There is a 55.12% probability that at least one will have green eyes.
Step-by-step explanation:
In this problem, the binomial probability distribution is going to be important for itens b,c and d.
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
We have these following probabilities:
0.75 probability of having children with brown eyes, 0.125 probability of having children with blue eyes, and 0.125 probability of having children with green eyes.
(a) What is the probability that their first child will have green eyes and the second will not?
There is a 0.125 probability a child will have green eyes and an 1-0.125 = 0.875 probability a child will not have green eyes.
So
0.125*0.875 = 0.109375
There is a 10.9375% probability that their first child will have green eyes and the second will not.
(b) What is the probability that exactly one of their two children will have green eyes?
Here we use the binomial probability distribution, with [tex]n = 2, p = 0.125[/tex].
We want P(X = 1).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 1) = C_{2,1}*(0.125)^{1}*(0.875)^{1} = 0.21875[/tex]
There is a 21.875% probability that exactly one of their two children will have green eyes.
(c) If they have six children, what is the probability that exactly two will have green eyes?
Again the binomial probability distribution, with [tex]n = 6, p = 0.125[/tex]
We want P(X = 2)
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{6,2}*(0.125)^{2}*(0.875)^{4} = 0.1374[/tex]
There is a 13.74% probability that exactly two will have green eyes.
(d) If they have six children, what is the probability that at least one will have green eyes?
[tex]n = 6, p = 0.125[/tex]
Either none has green eyes, or at least one has. The sum of the probabilities of these events is decimal 1. So
[tex]P(X = 0) + P(X \geq 1) = 1[/tex]
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
In which
[tex]P(X = 0) = C_{6,0}*(0.125)^{0}*(0.875)^{6} = 0.4488[/tex]
Finally
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.4488 = 0.5512[/tex]
There is a 55.12% probability that at least one will have green eyes.
The probability that their first child will have green eyes and the second will not is 0.109375. The probability that exactly one of their two children will have green eyes is 0.21875. If they have six children, the probability that exactly two will have green eyes is 0.19140625. The probability that at least one of the six children will have green eyes is 0.6499367.
Explanation:a) Probability that the first child will have green eyes and the second will not:Given that the parents have a probability of 0.125 of having a child with green eyes, the probability of the first child having green eyes is 0.125.
The probability that the second child does not have green eyes is 1 - 0.125 = 0.875.
Therefore, the probability that the first child has green eyes and the second child does not is 0.125 * 0.875 = 0.109375.
b) Probability that exactly one of their two children will have green eyes:There are two possible scenarios: (1) the first child has green eyes but not the second child, or (2) the first child does not have green eyes but the second child does.
The probability of the first scenario is the same as in part (a), which is 0.109375.
The probability of the second scenario is also 0.109375.
The total probability is the sum of the probabilities of the two scenarios, which is 0.109375 + 0.109375 = 0.21875.
c) Probability that exactly two out of six children will have green eyes:This can be calculated using the binomial probability formula.
The probability of having two children with green eyes and four children without green eyes is:
P(2 green, 4 not green) = C(6, 2) * (0.125)^2 * (0.875)^4 = 0.19140625
d) Probability that at least one out of six children will have green eyes:The probability that none of the six children have green eyes is (1 - 0.125)^6 = 0.3500633.
Therefore, the probability that at least one child will have green eyes is 1 - 0.3500633 = 0.6499367.
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It is known that x1 and x2 are roots of the equation 6x2+7x+k=0, where 2x1+3x2=−4.
Find k.
Answer:
1
Step-by-step explanation:
For a quadratic equation, the roots are expressed by the quadratic formula.
x=(-b+/- Sqrt[b^2-4ac])/2a
In this case a=6, b=-7 and c=k
So,
x=(7 +/- √[(-7)^2-4(6)(k)]/2(6))
Simplifying gives:
x=(7 +/- √[49-24k])/12
For k=0 the square root simplifies to √[49]=7 which yields roots of 7/6 and 0
For k=1 the square root simplifies to √[49-24]=√[25]=5 which yields roots of 1 and 1/6
For k=2 the square root simplifies to √[49-48]=√[1]=1 which yields roots of 2/3 and 1/2
k= 1 as other roots are fractions
Answer:
k=-5
Step-by-step explanation:
6x^2+7x+k=0 is a quadratic equation.
a= 6; b=7; c=k
Let the roots of the equation be R1 and R2
R1+R2 = -b/a = -7/6 ---------1
R1xR2 = c/a =k/6 or (R1xR2)6=k------------2
From Equation 1:
R1=-7/6-R2
We know 2R1+3R2=-4 Substituting for R1, we get
3 (-7/6-R2)+3R2=-4
R2=-5/3
R1=-7/6-(-5/3)= 1/2
Substituting these values in Eq2,
k= (-5/3 x 1/2) 6
k=-5
A sample of 15 from a normal population yields a sample mean of 43 and a sample standard deviation of 4.7. What is the P—value that should be used to test the claim that the population mean is less than 45? a. 0.0608 b. 0.1216 c. 0.4696 d. 0.9392 e. The P—value cannot be determined from the given information.
Answer:
b. 0.1216
Step-by-step explanation:
Given that a sample of 15 from a normal population yields a sample mean of 43 and a sample standard deviation of 4.7.
We have to check the p value for the claim that mean <45
[tex]H_0: \mu =45\\H_a: \mu <45[/tex]
(Left tailed test for population mean)
Sample size n = 15
Sample mean = 45
Sample std dev s = 4.7
Since sample std deviation is being used, we use t test only
Std error of mean = [tex]\frac{s}{\sqrt{n} } \\=1.214[/tex]
Mean difference = 43-45 = -2
t statistic = mean difference/std error
= -1.176
df = n-1 = 14
p value = 0.1216
Exercise 3.23 introduces a husband and wife with brown eyes who have 0.75 probability of having children with brown eyes, 0.125 probability of having children with blue eyes, and 0.125 probability of having children with green eyes. (a) What is the probability that their first child will have green eyes and the second will not?
Answer:
There is a 10.9375% probability that their first child will have green eyes and the second will not.
Step-by-step explanation:
We have these following probabilities:
0.75 probability of having children with brown eyes, 0.125 probability of having children with blue eyes, and 0.125 probability of having children with green eyes.
(a) What is the probability that their first child will have green eyes and the second will not?
There is a 0.125 probability a child will have green eyes and an 1-0.125 = 0.875 probability a child will not have green eyes.
So
0.125*0.875 = 0.109375
There is a 10.9375% probability that their first child will have green eyes and the second will not.
What are the rectangular coordinates of the point whose cylindrical coordinates are (r=9, θ=2π3, z=3)(r=9, θ=2π3, z=3) ?
Answer:
The point is [tex](-\frac{9}{2},\frac{9\sqrt{3}}{2},3)[/tex] in rectangular coordinates.
Step-by-step explanation:
To convert from cylindrical to rectangular coordinates we use the relations
[tex]x=r \cdot cos(\theta)\\y=r\cdot sin(\theta)\\z=z[/tex]
To convert the point [tex](9,\frac{2}{3}\pi ,3)[/tex] from cylindrical to rectangular coordinates we use the above relations
Since [tex]r=9[/tex], [tex]\theta=\frac{2}{3} \pi[/tex], and [tex]z=3[/tex],
[tex]x=r \cdot cos(\theta)=9\cdot cos(\frac{2}{3}\pi )=-\frac{9}{2}[/tex]
[tex]y=r\cdot sin(\theta)=9\cdot sin(\frac{2}{3} \pi )=\frac{9\sqrt{3}}{2}[/tex]
[tex]z=z=3[/tex]
Thus, the point is [tex](-\frac{9}{2},\frac{9\sqrt{3}}{2},3)[/tex] in rectangular coordinates.
The powerful survival impulse that leads infants to seek closeness to their caregivers is called:A)attachment.B)imprinting.C)habituation.D)assimilation.E)the rooting reflex
Answer:
A. Attachment
Step-by-step explanation:
The powerful survival impulse that leads infants to seek closeness to their caregivers is called Attachment. The infant can count on the caregiver possibly parent(s) for care which gives the infant a solid foundation for dependence and survival.
Final answer:
The instinctual behavior that drives infants to seek closeness with their caregivers is known as A) attachment. It's fostered by reflexes that ensure physical contact and is crucial for an infant's survival, ensuring they receive the necessary care, protection, and opportunity to develop securely.
Explanation:
The powerful survival impulse that leads infants to seek closeness to their caregivers is called A) attachment. This is an intrinsic part of human development and is crucial for the infant's survival. Infants have a set of innate behaviors and reflexes that promote closeness and contact with their caregivers, such as the Moro reflex and the grasping reflex, which help the infant to hold onto the caregiver and thus reduce the risk of falling.
Additionally, behaviors such as crying and the sucking reflex are instinctive methods for infants to express needs and receive care. Furthermore, the rooting reflex is an instinctive behavior that helps the infant find the nipple to feed by touching. John Bowlby's evolutionary theory underscores the importance of attachment by suggesting that the ability to maintain proximity to an attachment figure would have increased the chances of an infant surviving to reproductive age.
Attachments are not just reactions to the provision of food and warmth by the caregivers but are biological imperatives that ensure an infant remains close to those who provide security, learning, and protection, thereby enhancing their chance of survival.
Let D be the region bounded by the paraboloids; z = 6 - x² - y² and z = x² + y².
Write six different triple iterated integrals for the volume of D. Evaluate one of the integrals.
Answer:
∫∫∫1 dV=4\sqrt{3}π
Step-by-step explanation:
From Exercise we have
z=6-x^{2}-y^{2}
z=x^{2}+y^{2}
we get
2z=6
z=3
x^{2}+y^{2}=3
We use the polar coordinates, we get
x=r cosθ
y=r sinθ
x^{2}+y^{2}&=r^{2}
r^{2}=3
We get at the limits of the variables that well need for our integral
x^{2}+y^{2}≤z≤3
0≤r ≤\sqrt{3}
0≤θ≤2π
Therefore, we get a triple integral
\int \int \int 1\, dV&=\int \int \left(\int_{x^2+y^2}^{3} 1\, dz\right) dA
=\int \int \left(z|_{x^2+y^2}^{3} \right) dA
=\int \int\ \left(3-(x^2+y^2) \right) dA
=\int \int\ \left(3-r^2 \right) dA
=\int_{0}^{2\pi}\int_{0}^{\sqrt{3}} (3-r^2) dr dθ
=3\int_{0}^{2\pi}\int_{0}^{\sqrt{3}} 1 dr dθ-\int_{0}^{2\pi}\int_{0}^{\sqrt{3}} r^2 dr dθ
=3\int_{0}^{2\pi} r|_{0}^{\sqrt{3}} dθ-\int_{0}^{2\pi} \frac{r^3}{3}|_{0}^{\sqrt{3}}dθ
=3\sqrt{3}\int_{0}^{2\pi} 1 dθ-\sqrt{3}\int_{0}^{2\pi} 1 dθ
=3\sqrt{3} ·2π-\sqrt{3}·2π
=4\sqrt{3}π
We get
∫∫∫1 dV=4\sqrt{3}π
We find the volume of the region D bounded by the paraboloids z = 6 - x² - y² and z = x² + y² by setting up triple iterated integrals. Six different integrals are presented, and one is evaluated using cylindrical coordinates. The volume is determined to be 9π.
To find the volume of the region D bounded by the paraboloids z = 6 - x² - y² and z = x² + y², we need to set up triple iterated integrals.
The intersection of the two surfaces occurs when 6 - x² - y² = x² + y²,
which simplifies to 6 = 2(x² + y²) or x² + y² = 3, defining a circle of radius √3 in the xy-plane.
Possible Triple Iterated Integrals
Here are six different triple iterated integrals to find the volume of the region D:
[tex]\int_{-\sqrt{3}}^{\sqrt{3}} \int_{-\sqrt{3-x^2}}^{\sqrt{3-x^2}} \int_{x^2+y^2}^{6-x^2-y^2} dz \, dy \, dx = dx \, dy \, dz.\end{equation}[/tex][tex]\begin{equation}\int_{0}^{2\pi} \int_{0}^{\sqrt{3}} \int_{r^2}^{6-r^2} r \, dz \, dr \, d\theta \, dv = dx \, dy \, dz.\end{equation}[/tex][tex]\begin{equation}\int_{0}^{2\pi} \int_{-\sqrt{3}\cos\theta}^{\sqrt{3}\cos\theta} \int_{r^2}^{6-r^2} r \, dz \, dr \, d\theta \, dv = dx \, dy \, dz.\end{equation}[/tex][tex]\begin{equation}\int_{-\sqrt{3}}^{\sqrt{3}} \int_{-\sqrt{3-y^2}}^{\sqrt{3-y^2}} \int_{y^2+x^2}^{6-y^2-x^2} dz \, dx \, dy \, dv = dx \, dy \, dz.\end{equation}[/tex][tex]\begin{equation}\int_{0}^{2\pi} \int_{-\sqrt{3}\cos\theta}^{\sqrt{3}\cos\theta} \int_{x^2}^{6-x^2-\theta} dz \, dx \, d\theta \, dv = dx \, dy \, dz.\end{equation}[/tex][tex]\begin{equation}\int_{-\sqrt{3}}^{\sqrt{3}} \int_{y-x}^{y+x} \int_{r^2}^{6-r^2} r \, dz \, dr \, d\theta \, dv = dx \, dy \, dz.\end{equation}[/tex]Evaluating One of the Integrals
Let's evaluate the triple iterated integral in cylindrical coordinates:[tex]\int_{0}^{2\pi} \int_{0}^{\sqrt{3}} \int_{r^2}^{6 - r^2} r \, dz \, dr \, d\theta[/tex]
First, integrate with respect to z:[tex]\int_{r^2}^{6 - r^2}\, dz = \left[ z \right]_{z=r^2}^{z=6-r^2}[/tex]
[tex]= (6-r^2) - (r^2)[/tex]
[tex]= 6-2r^2[/tex]
Next, integrate with respect to r:[tex]int_{0}^{\sqrt{3}} r(6 - 2r^2) dr = \int_{0}^{\sqrt{3}} (6r - 2r^3) dr[/tex]
[tex]= \left[ 3r^2 - \frac{1}{2}r^4 \right]_{r=0}^{r= \sqrt{3}}[/tex]
[tex]= \left[ 3(3) - \frac{1}{2}(9) \right][/tex]
= 9 - 4.5
= 4.5
Finally, integrate with respect to θ:[tex]\int_{0}^{2\pi} 4.5 \, d\theta = 4.5 \cdot 2\pi = 9\pi[/tex]
So the volume of the region D is 9π.
Use the Pythagorean theorem to determine which of the following give the measures of the legs and hypotenuse of a right triangle. Which apply. 3,4,5. Or. 4,11,14. Or. 9,14,17. Or 8,14,16. Or. 8,15,17
Answer: 3, 4,5 and 17, 15,8
give the measures of the legs and hypotenuse of a right triangle.
Step-by-step explanation:
In order for the measures of the legs and hypotenuse given to form a right angle triangle, they must be Pythagorean triples. A Pythagoras triple is a set of numbers that perfectly satisfy the Pythagorean theorem. The Pythagorean theorem is expressed as
Hypotenuse² = opposite side² + adjacent side². We will apply the theorem to each set of numbers given.
1) 3, 4, 5
5² = 3² + 4² = 9 + 16
25 = 25
It is a Pythagorean triple
2) 4, 11, 14
14² = 11² + 4² = 121 + 16
196 = 137
It is a Pythagorean triple
3) 9, 14, 17
17² = 14² + 9² = 196 + 81
289 = 277
It is not a Pythagorean triple
4) 8, 14, 16
16² = 14² + 8² = 196 + 64
256 = 260
It is not a Pythagorean triple
5) 8, 15 , 17
17² = 15² + 8²
289 = 225 + 64
289 = 289
It is a Pythagorean triple
Therefore, 3, 4,5 and 17, 15,8
give the measures of the legs and hypotenuse of a right triangle.
Walleye is a common game fish. Adult walleye have a length with a mean of 44 cm and a standard deviation of 4 cm, and the distribution of lengths is approximately Normal. What fraction of fish are greater than 41 cm in length?
Select one O a. -0.75 O b. 0.22 ? ?.077 O d. 0.75
Answer:b. 0.22
Step-by-step explanation:
Since the lengths of adult walleye fishes are normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = lengths of walleye fishes.
µ = mean length
σ = standard deviation
From the information given,
µ = 44 cm
σ = 4 cm
We want to find the probability or fraction of fishes that are greater than 41 cm in length. It is expressed as
P(x > 41) = 1 - P(x ≤ 41)
For x = 41,
z = (41 - 44)/4 = - 0.75
Looking at the normal distribution table, the probability corresponding to the z score is 0.22
To find the fraction of fish that are greater than 41 cm in length, calculate the z-score with the mean and standard deviation.
Explanation:To find the fraction of fish that are greater than 41 cm in length, we need to calculate the z-score of 41 cm using the mean and standard deviation. The z-score formula is z = (x - μ) / σ. Plugging in the values, we have z = (41 - 44) / 4 = -0.75. We can then look up the corresponding value on the z-table to find the fraction of fish with a length greater than 41 cm, which is approximately 0.7734. Therefore, the answer is option d, 0.75.
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Write the equation in vertex form for the parabola with focus (0,5) and directrix y=
–
5.
Simplify any fractions.
Answer: [tex]x^{2} = 20y[/tex]
Step-by-step explanation:
The directrix given is vertical , so we will use the formula :
[tex](x-h)^{2}=4p(y-k)[/tex]
P is the distance between the focus , that is 5 - 0 = 5
Therefore : p = 5
(h,k) is the mid point between the focus and the directrix , that is
(h,k) = [tex](\frac{x_{1}+x_{2} }{2},\frac{y_{2}+y_{1}}{2})[/tex] = [tex](\frac{0+0}{2} , \frac{5-5}{2})[/tex] = [tex](0,0)[/tex]
Therefore:
h =0
k = 0
substituting into the formula : we have
[tex](x-h)^{2}=4p(y-k)[/tex]
[tex](x-0)^{2}[/tex] = 4(5)([tex]y-0)[/tex]
[tex]x^{2} = 20y[/tex]
Therefore : the equation in vertex form is [tex]x^{2} = 20y[/tex]
A chain lying on the ground is 10 m long and its mass is 70 kg. How much work (in J) is required to raise one end of the chain to a height of 4 m?
Answer:
[tex] W= 34.3 \frac{kg}{s^2} (4^2-0^2)m^2 =548.8 \frac{kg m^2}{s^2} =548.8 J[/tex]
Step-by-step explanation:
Data Given: m = 70 kg , g = 9.8 ms^-2, h =10m.
For this case we can use the following formula:
[tex] W = \int_{x_i}^{x_f} F(x) dx[/tex]
For this case we need to find an expression for the force in terms of the distance. And since on this case the total distance is 10 m long we can write the expression like this:
[tex] F(x) = \frac{ma}{10m}= \frac{mg}{10m} x[/tex]
The only acceleration on this case is the gravity and if we replace the values given we got:
[tex] \frac{70 kg *9.8 m/s^2}{10m} x=68.6 x\frac{kg}{s^2}[/tex]
Now we can find the required work with the following integral:
[tex] W= 68.6 \frac{kg}{s^2} \int_{0}^4 x dx[/tex]
[tex] W= 34.3 \frac{kg}{s^2} x^2 \Big|_0^4[/tex]
[tex] W= 34.3 \frac{kg}{s^2} (4^2-0^2)m^2 =548.8 \frac{kg m^2}{s^2} =548.8 J[/tex]
The amount of work that is required to raise one end of the chain is 548.8 Joules.
Given the following data:
Length of chain = 10 meters.Mass of chain = 70 kg.Height = 4 meters.To calculate the amount of work that is required to raise one end of the chain:
How to calculate the work done.We would solve for the magnitude of the force acting on the chain with respect to the distance and this is given by this expression:
[tex]Force = \frac{mgx}{10} \\\\Force = \frac{70 \times 9.8 \times x}{10}[/tex]
Force = 68.6x Newton.
Now, we can calculate the amount of work by using this formula:
[tex]W=\int\limits^{x_2}_{x_1} F({x}) \, dx \\\\W= 68.6 \int\limits^{4}_{0} x \, dx\\\\W= 34.3 x^2 |^4_0\\\\W=34.3 [4^2 -0^]\\\\W=34.3 \times 16[/tex]
W = 548.8 Joules.
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Which of the following is independent variable?
I think the answer is D. hours because its variation does not depend on another variable
Step1: find the Laplace transform of the solution Y(s).Y(s). Write the solution as a single fraction in s
Complete Question :
The question is shown on the first uploaded image
Answer:
The solution is the second uploaded image
Step-by-step explanation:
The step by step explanation is on the third, fourth and fifth uploaded image
What is the variable x and y equal in the equation 0.4x+0.6y=6.2?
Answer:
The solution is (6.2,6.2)
Step-by-step explanation:
we have
[tex]0.4x+0.6y=6.2[/tex] ----> equation A
For variable x and y equal
[tex]x=y[/tex] ----> equation B
Solve the system by substitution
substitute equation B in equation A
[tex]0.4y+0.6y=6.2[/tex]
solve for y
combine like terms
[tex]y=6.2[/tex]
so
[tex]x=6.2[/tex]
therefore
The solution is (6.2,6.2)
State all possible names for each figure.
Answer:
Step-by-step explanation:
square
quadrilateral - all of these
polygon - squares
trapezoid - others - non square
rhombus - squares
hope this helps.
A statistician controls ____________ by establishing the risk he or she is willing to take in terms of rejecting a true null hypothesis.
a) Alpha
b) beta
c) mean
d) standard deviation
Answer:
a) Alpha
Step-by-step explanation:
The correct option is alpha because alpha known as type I error is the probability of reject the null hypothesis when null hypothesis is true. If we take alpha 5%, it means that we are taking 5 out of 100 chance of rejecting the null hypothesis when it is true.
So, statistician controls alpha by establishing the risk of rejecting a null hypothesis when its true.
Mary wants to get her boyfriend a nice birthday present. She has to decide on a choice of one of five books that he would like, one of four ties or one of four X-box games. What is the total number of different outcomes available to her?
Answer: 13 outcomes
Step-by-step explanation:
Given:
She wants to get a gift from the following choices;
- one of five books
- one of four ties
- one of four X-box
Since, the three groups of choices are joined with "OR" but not "AND" that means she is getting just one gift from any of the 3 groups.
Total number of gift she needed = 1
Total number of choices = 3 groups with total of 13 options
N = 13P1 = 13!/(13-1)! = 13!/12! =13
N = 13 outcomes.
Two random variables X and Y are independent. Each has a binomial distribution with success probability 0.4 and 2 trials.
(a) Find the joint probability distribution function f(x,y).
(b) Give the joint probabilities using a table. Hint, the size of the tables is 3 by 3.
Answer:
Step-by-step explanation:
Given that two random variables X and Y are independent. Each has a binomial distribution with success probability 0.4 and 2 trials.
When x and y are independent joint probability would be product of individual probabilities
pdf of X
X is Binom (2,0.4)
and Y is Binomi (2,0.4)
Hence joint distribution of XY would be
P(X=x, Y=y) =[tex]2Cx (0.4)^x (0.6)^{2-x} *2Cy (0.4)^y (0.6)^{2-y}[/tex]
for x=0,1,2 and y =0,1,2
b) Joint probability using table
PDF of X is
X 0 1 2
p 0.36 0.48 0.16
and same for Y also
Joint prob would be
X Y 0 1 2
0 0.1296 0.1728 0.0576
1 0.1728 0.2304 0.0768
2 0.0576 0.0768 0.0256
Joint probability distribution function are used to represent the probability of multiply variables
The joint probability distribution function is [tex]f(x,y) = ^2C_x *0.4^x * 0.6^{2- x} *^2C_y * 0.4^y * 0.6^{2- y}[/tex]
The given parameters are:
[tex]p = 0.4[/tex] --- the probability of success
[tex]n = 2[/tex] ----the number of trials
The joint probability distribution function f(x,y) is calculated as:
[tex]f(x,y) = ^nC_x * p^x * (1 -p)^{n- x} *^nC_y * p^y * (1 -p)^{n- y}\\[/tex]
So, we have:
[tex]f(x,y) = ^2C_x *0.4^x * (1 -0.4)^{2- x} *^2C_y * 0.4^y * (1 -0.4)^{2- y}[/tex]
Evaluate the differences
[tex]f(x,y) = ^2C_x *0.4^x * 0.6^{2- x} *^2C_y * 0.4^y * 0.6^{2- y}[/tex]
The above represents the joint probability distribution function f(x,y)
When x = 0, y = 0;
We have:
[tex]f(0,0) = 0.130[/tex]
When x = 0, y = 1;
We have:
[tex]f(0,1) = 0.173[/tex]
When x = 0, y = 2;
We have:
[tex]f(0,2) = 0.058[/tex]
When x = 1, y = 0;
We have:
[tex]f(1,0) = 0.173[/tex]
When x = 1, y = 1;
We have:
[tex]f(1,1) = 0.230[/tex]
When x = 1, y = 2;
We have:
[tex]f(1,2) = 0.077[/tex]
When x = 2, y = 0;
We have:
[tex]f(2,0) = 0.058[/tex]
When x = 2, y = 1;
We have:
[tex]f(2,1) = 0.077[/tex]
When x = 2, y = 2;
We have:
[tex]f(2,2) = 0.026[/tex]
So, the joint probability as a table is:
X /Y 0 1 2
0 0.1296 0.1728 0.0576
1 0.1728 0.2304 0.0768
2 0.0576 0.0768 0.0256
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what is the answer for 6+3×2
Data obtained from a nominal scale: A. must be alphabetic B. can be either numeric or nonnumeric C. must be numeric D. must rank order the data
Answer:
Option B
Step-by-step explanation:
The data constitutes nominal scale of measurement when the observations can be classified into groups. For example, students are classified into groups on the basis of eye color. The numerical values can also be use in nominal scale for grouping. For example, the students can be categorize into 1,2 and 3 if they have brown, black and green eye color. But they have no numerical significance. Thus, data obtained from a nominal scale can be either numeric or non-numeric.
If the interest rate is 7%, how many years will it take for your bank balance to double from $1,000 to $2,000?Enter the following data into your calculator:
It takes 10.3 years for your bank balance to double from $1,000 to $2,000.
Given that,
The interest rate is 7%.
Principal amount = $1000
Final amount = $2000
Used the formula for the time,
A = P (1 + r)ⁿ
Where, A = Final amount
P = Principal amount
r = interest rate
n = number of years
Substitute all the values,
2000 = 1000 (1 + 0.07)ⁿ
2000/1000 = (1.07)ⁿ
2 = (1.07)ⁿ
Take natural logs on both sides,
ln 2 = n ln (1.07)
0.69 = n × 0.067
n = 0.69/0.067
n = 10.3 years
Therefore, the time for your bank balance to double from $1,000 to $2,000 is 10.3 years.
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It will take approximately 10.1351 years for your bank balance to double from $1,000 to $2,000 at an interest rate of 7%.
To calculate the number of years it will take for your bank balance to double from $1,000 to $2,000 at an interest rate of 7%, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A is the final amount (in this case, $2,000),
P is the initial principal (in this case, $1,000),
r is the annual interest rate (7% or 0.07 as a decimal),
n is the number of times interest is compounded per year (we'll assume it's compounded annually),
and t is the number of years.
We can rearrange the formula to solve for t:
t = (log(A/P) / log(1 + r/n)) / n
Plugging in the values:
A = $2,000
P = $1,000
r = 0.07
n = 1 (since it's compounded annually)
t = (log(2,000/1,000) / log(1 + 0.07/1)) / 1
Simplifying the expression:
t = (log(2) / log(1.07)) / 1
Using a calculator to evaluate the logarithms:
t ≈ (0.3010 / 0.0296) / 1
t ≈ 10.1351 / 1
t ≈ 10.1351
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Consider the vector b⃗ b→b_vec with length 4.00 mm at an angle 23.5∘∘ north of east. What is the y component bybyb_y of this vector?
Answer:
[tex]\large\boxed {1.59 mm}[/tex]Explanation:
1. Given vector:
length: 4.00 mm = magnitude of the vectorangle: 23.5º north of east = 23.5º from the x-axys (counterclockwise)2. y-component
The y-component may be determined using the sine ratio, the angle from the x-axys (counterclockwise direction), and the magnitude of the vector.
sine (23.5º) = y-component / magnitudey-component = magnitude × sine (23.5º) = 4.00 mm × sine (23.5º) = 1.59 mm.[tex]\large\boxed{y-component = 1.59 mm}[/tex]Harry Potter approaches with a strange bag full of balls, numbered 1 to k. As you reach in to pick one, he notes that they are not all equally likely because of magic: ball 1 is least likely to be chosen, with probability c, where c is some constant. Ball 2 has probability 2c, Ball 3 has probability 3c, . . . , Ball k − 1 has probability (k − 1)c, and Ball k has probability kc.
1. What is the expected value of the ball number you pick? Your answer can’t use the constant c, but will use k.
Answer:
[k*(k+1)*(2*k+1)] / 6
Step-by-step explanation:
We have balls numbered as: 1, 2, 3, ... , k with probabilities as: c, 2*c, 3*c, ... , k*c
Let Y be the discrete random variable defined as: Y = ball number
We know that Expected value of discrete Random Variable is:
E[X] = Σ₁ⁿ xₐ*f(xₐ) ,where f(xₐ) is probability of xₐ
then,
E[Y] = 1*c + 2*2*c + 3*3*c + ... + k*k*c
E[Y] = c*(1 + 2*2 + 3*3 + ... + k*k)
E[Y] = c*(1^2 + 2^2 + 3^2 + ... + k^2)
consider c = 1 (because it's constant so you can suppose any you wish)
E[Y] = 1^2 + 2^2 + 3^2 + ... + k^2
using formula of first n squares natural numbers (as attached picture)
E[Y] = [k*(k+1)*(2*k+1)] / 6
In a survey of 447 registered voters, 157 of them wished to see Mayor Waffleskate lose her next election. The Waffleskate campaign claims that no more than 27% of registered voters wish to see her defeated. Does the 98% confidence interval for the proportion support this claim? (Hint: you should first construct the 98% confidence interval for the proportion of registered voters who wish to see Waffleskate defeated.) (0.299, 0.404)
A. No
B. Yes
Let p be the true proportion of registered voters wish to see Mayor Waffleskate defeated.
As per given , we have
[tex]H_0: p\leq0.27\\\\ H_a: p >0.27[/tex]
Sample size : n= 447
Number of of registered voters wish to see Mayor Waffleskate defeated = 157
I.e. sample proportion : [tex]\hat{p}=\dfrac{157}{447}\approx0.3512[/tex]
Confidence interval for population proportion is given by :-
[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
, where n= sample size
[tex]\hat{p}[/tex] = sample proportion
z* = critical z-value.
Critical z-value for 98% confidence interval is 2.33. (By z-table)
Then, the 98% confidence interval for the proportion of registered voters who wish to see Waffleskate defeated will be :
[tex]0.3512\pm2.33\sqrt{\dfrac{0.3512(1-0.3512)}{447}}\\\\=0.3512\pm (2.33)(0.022577656)\\\\=0.3512\pm 0.05260593848\\\\=(0.3512-0.05260593848,\ 0.3512+0.05260593848)\\\\=(0.29859406152,\ 0.40380593848)\approx(0.299,\ 0.404)[/tex]
Since the 0.27 < 0.299 , it means 0.27 does not belong to the above confidence interval.
So , we reject the null hypothesis ([tex]H_0[/tex]).
So , 98% confidence interval does not support the claim.
Let Upper A equals left bracket Start 2 By 2 Matrix 1st Row 1st Column negative 2 2nd Column 4 2nd Row 1st Column 1 2nd Column 3 EndMatrix right bracketA=
−2 4
1 3
, and Upper B equals left bracket Start 2 By 2 Matrix 1st Row 1st Column negative 2 2nd Column 1 2nd Row 1st Column 3 2nd Column 7 EndMatrix right bracketB=
−2 1
3 7
.a. Find
ABAB,
if possible. b. Find
BABA,
if possible.
c. Are the answers in parts a and b the same?
d. In general, for matrices A and B such that AB and BA both exist, does AB always equal BA?
a. Find
ABAB,
if possible.
Answer:
not
Step-by-step explanation:
[tex]\left[\begin{array}{ccc}-2&4\\1&3\end{array}\right] *\left[\begin{array}{ccc}-2&1\\3&7\end{array}\right]=[/tex]
First is A and Second is B
Let's find A*B
[tex]\left[\begin{array}{ccc}-2(-2)+4*3&-2*1+4*7\\1(-2)+3*3&1*1+3*7\end{array}\right] =\left[\begin{array}{ccc}16&26\\7&22\end{array}\right][/tex]
b)
[tex]\left[\begin{array}{ccc}-2&1\\3&7\end{array}\right] \left[\begin{array}{ccc}-2&4\\1&3\end{array}\right] =[/tex]
Now let's find B*A
[tex]\left[\begin{array}{ccc}-2(-2)+1*1&-2*4+1*3\\3(-2)+7*1&3*4+7*3\end{array}\right] =\left[\begin{array}{ccc}5&-5\\1&23\end{array}\right][/tex]
c) They are not
Michael Beasley is shooting free throws. Making or missing free throws doesn't change the probability that he will make his next one, and he makes his free throws 75%, percent of the time. What is the probability of Michael Beasley making all of his next 4 free throw attempts?
A. .75^8
B. .375^4
C. .75^4
D. 1.50^2
Answer: C. [tex]0.75^4[/tex]
Step-by-step explanation:
Let x be the binomial variable that denotes the number of makes.
Since each throw is independent from the other throw , so we can say it follows Binomial distribution .
So [tex]X\sim Bin(n=4 , p=0.75)[/tex]
Binomial distribution formula: The probability of getting x success in n trials :
[tex]P(X=x)=^nC_xp^n(1-p)^{n-x}[/tex] , where p = probability of getting success in each trial.
Then, the probability of Michael Beasley making all of his next 4 free throw attempts will be :
[tex]P(X=4)=^4C_4(0.75)^4(1-0.75)^{0}[/tex]
[tex]=(1)(0.75)^4(1)\ \ [\because\ ^nC_n=1]\\\\=(0.75)^4[/tex]
Thus, the probability of Michael Beasley making all of his next 4 free throw attempts is [tex]=0.75^4[/tex]
Hence, the correct answer is C. [tex]0.75^4[/tex].
The family of functions y=ce−2x+e−x is solution of the equation y+2y=e−x
Find the constant c which defines the solution which also satisfies the initial condition y(−5)=6. c=
Answer:
c = 6*e^(-10) - e^(-5) ( ≈ -e⁻⁵ = -6.74*10⁻³)
Step-by-step explanation:
for the function
y=c*e^(−2x)+e^(−x)
as a solution of y'+2y=e^(−x)
then for y(x=−5)=6
6 =c*e^(−2(-5))+e^(−(-5)) = c*e^10 + e^5
6 = c*e^10 + e^5
c = (6 - e^5)/*e^10 = 6*e^(-10) - e^(-5)
c = 6*e^(-10) - e^(-5) ( ≈ -e⁻⁵ = -6.74*10⁻³)
Find an equation of the largest sphere that is centered at (5,4,9) and has interior contained in the first octant.
Answer:
[tex](x - 5)^{2} + (y - 4)^{2} + (z - 9)^{2} = 16[/tex]
Step-by-step explanation:
The general equation of a sphere is as follows:
[tex](x - x_{c})^{2} + (y - y_{c})^{2} + (z - z_{c})^{2} = r^{2}[/tex]
In which the center is [tex](x_{c}, y_{c}, z_{c})[/tex], and r is the radius.
In this problem, we have that:
[tex]x_{c} = 5, y_{c} = 4, z_{c} = 1[/tex]
So
[tex](x - 5)^{2} + (y - 4)^{2} + (z - 9)^{2} = r^{2}[/tex]
Interior contained in the first octant:
The first octant is bounded by:
The xy plane, in which z is 0. The distance from the center of the sphere to the xy plane is 9.
The xz plane, in which y is 0. The distance from the center of the sphere to the xz plane is 4.
The yz plane, in which x is 0. The distance from the center of the sphere to the yz plane is 5.
This means that if the radius is higher than four, the sphere will cross into a different octant.
So the radius for the largest sphere is 4.
The equation is
[tex](x - 5)^{2} + (y - 4)^{2} + (z - 9)^{2} = 4^{2}[/tex]
[tex](x - 5)^{2} + (y - 4)^{2} + (z - 9)^{2} = 16[/tex]