Answer:
(a) Probability that a single randomly selected element X of the population is between 57,000 and 58,000 = 0.46411
(b) Probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = 0.99621
Step-by-step explanation:
We are given that a normally distributed population has mean 57,800 and standard deviation 75, i.e.; [tex]\mu[/tex] = 57,800 and [tex]\sigma[/tex] = 750.
Let X = randomly selected element of the population
The z probability is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
(a) So, P(57,000 <= X <= 58,000) = P(X <= 58,000) - P(X < 57,000)
P(X <= 58,000) = P( [tex]\frac{X-\mu}{\sigma}[/tex] <= [tex]\frac{58000-57800}{750}[/tex] ) = P(Z <= 0.27) = 0.60642
P(X < 57000) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{57000-57800}{750}[/tex] ) = P(Z < -1.07) = 1 - P(Z <= 1.07)
= 1 - 0.85769 = 0.14231
Therefore, P(31 < X < 40) = 0.60642 - 0.14231 = 0.46411 .
(b) Now, we are given sample of size, n = 100
So, Mean of X, X bar = 57,800 same as before
But standard deviation of X, s = [tex]\frac{\sigma}{\sqrt{n} }[/tex] = [tex]\frac{750}{\sqrt{100} }[/tex] = 75
The z probability is given by;
Z = [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
Now, probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = P(57,000 < X bar < 58,000)
P(57,000 <= X bar <= 58,000) = P(X bar <= 58,000) - P(X bar < 57,000)
P(X bar <= 58,000) = P( [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] <= [tex]\frac{58000-57800}{\frac{750}{\sqrt{100} } }[/tex] ) = P(Z <= 2.67) = 0.99621
P(X < 57000) = P( [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{57000-57800}{\frac{750}{\sqrt{100} } }[/tex] ) = P(Z < -10.67) = P(Z > 10.67)
This probability is that much small that it is very close to 0
Therefore, P(57,000 < X bar < 58,000) = 0.99621 - 0 = 0.99621 .
Using the normal distribution and the central limit theorem, it is found that:
There is a 0.4641 = 46.41% probability that a single randomly selected element X of the population is between 57,000 and 58,000.For samples of size 100, the mean is of 57800 and the standard deviation is 75.There is a 0.9965 = 99.65% probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000.In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
It measures how many standard deviations the measure is from the mean. After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.By the Central Limit Theorem, for sampling distributions of samples of size n, the mean is [tex]\mu[/tex] and the standard deviation is [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].In this problem:
Mean of 57800, thus [tex]\mu = 57800[/tex].Standard deviation of 750, thus [tex]\sigma = 750[/tex].The probability that a single randomly selected element X of the population is between 57,000 and 58,000 is the p-value of Z when X = 58000 subtracted by the p-value of Z when X = 57000, thus:
X = 58000:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{58000 - 57800}{750}[/tex]
[tex]Z = 0.27[/tex]
[tex]Z = 0.27[/tex] has a p-value of 0.6064.
X = 57000:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{57000 - 57800}{750}[/tex]
[tex]Z = -1.07[/tex]
[tex]Z = -1.07[/tex] has a p-value of 0.1423.
0.6064 - 0.1423 = 0.4641.
0.4641 = 46.41% probability that a single randomly selected element X of the population is between 57,000 and 58,000.
For samples of size 100, [tex]n = 100[/tex], and then:
[tex]s = \frac{750}{\sqrt{100}} = 75[/tex]
For samples of size 100, the mean is of 57800 and the standard deviation is 75.
Then, the probability is:
X = 58000:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{58000 - 57800}{75}[/tex]
[tex]Z = 2.7[/tex]
[tex]Z = 2.7[/tex] has a p-value of 0.9965.
X = 57000:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{57000 - 57800}{75}[/tex]
[tex]Z = -10.7[/tex]
[tex]Z = -10.7[/tex] has a p-value of 0.
0.9965 - 0 = 0.9965.
0.9965 = 99.65% probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000.
A similar problem is given at https://brainly.com/question/24663213
what is the domain and range on this graph?
Answer:
The answer to your question is Domain (-∞, ∞) Range [-4, ∞)
Step-by-step explanation:
The Domain is the set of all possible values of the independent variable (x).
The Range is the set of all the possible values of the dependent variable when substitute the domain in the function.
On the graph, we find the domain looking at the x-axis
On a graph, we find the range, looking at all the y-axis
In this graph, x has values from -infinite to infinite, then, the domain is (-∞, ∞).
In this graph, y has values from -4 to infinite, then, the range is [-4, ∞)
Find the indicated complement. A certain group of women has a 0.55% rate of red/green color blindness. If a woman is randomly selected, what is the probability that she does not have red/green color blindness? What is the probability that the woman selected does not have red/green color blindness? nothing (Type an integer or a decimal. Do not round.)
Answer:
The probability that the woman selected does not have red/green color blindness is 0.9945.
Step-by-step explanation:
Let X = a woman has red/green color blindness.
It is provided that, in a certain group of women the rate of red/green color blindness is P (X) = 0.0055.
A complement of an event E, is defined as the event of not E.
The probability of complement of an event E is:
[tex]P(E^{c})=1-P(E)[/tex]
Compute the probability that the woman selected does not have red/green color blindness as follows:
[tex]P (X^{c})=1-P(X)\\=1-0.0055\\=0.9945[/tex]
Thus, the probability of the complement of the event of a woman having red/green color blindness is 0.9945.
The probability that the woman selected does not have red/green color blindness is 0.9945.
Calculation of probability:Since A certain group of women has a 0.55% rate of red/green color blindness.
Here we assume a woman has red/green color blindness be X
So, here the probability should be
= 1-0.55%
= 0.9945
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If every student is independently late with probability 10%, find the probability that in a class of 30 students: a) nobody is late, 4.2% 8.0% 17.4% 33.3% unanswered b) exactly 1 student is late. 3.33% 5.25% 7.75% 14.1%
Answer:
a) 4.2%
b) 14.1%
Step-by-step explanation:
a) 0.9³⁰ = 0.0423911583
b) 30C1 × 0.1 × 0.9²⁹ = 0.1413038609
Answer:
a.) 4.2%
b.) 14.1%
Step-by-step explanation:
We solve using the probability distribution formula for selection and this formula uses the combination formula for estimation.
When choosing a random selection of "r" items from a sample of "n" items, The formula is generally denoted by:
P(X=r) = nCr × p^r × q^n-r.
Where p = probability of success
q= probability of failure.
From the given question,
n = number of samples =30,
p = Probability that a student is late = 10% = 0.1,
q=0.9
a.) when no student is late, that is when r = 0, then
P(X=0) = 30C0 × 0.1^0 × 0.9^30
P(X=0) = 0.0424 = 4.24 ≈ 4.2%
b.) when exactly one student is late, that is when r=1, then
P(X=1) = 30C1 × 0.1¹ × 0.9^29
P(X=1) = 0.1413 = 14.13 ≈ 14.1%
Particles are a major component of air pollution in many areas. It is of interest to study the sizes of contaminating particles. Let X represent the diameter, in micrometers, of a randomly chosen particle. Assume that in a certain area, the probability density function of X is inversely proportional to the volume of the particle ; that is, assume that fX(x) = c x 3 , x > 1, where c is a constant. (a) Find the value of c so that fX is a probability density function. (b) The term PM10 refers to particles 10 µ m or less in diameter. What proportion of contaminating particles are PM10 ? (c) The term PM2.5 refers to particles 2.5 µ m or less in diameter. What proportion of contaminating particles are PM2.5 ? (d) What proportion of the PM10 particles are PM2.5 ?
Answer:
a) c = 2
b) 0.99
c) 0.84
d) 0.8485
Step-by-step explanation:
We are given the following in the question:
[tex]f(x) = \dfrac{c}{x^3}, x > 1[/tex]
a) Value of c
Property of probability density function
[tex]\displaystyle\int^{\infty}_{-\infty} f(x) = 1[/tex]
Putting values, we get,
[tex]\displaystyle\int^{\infty}_{1} \frac{c}{x^3} = 1\\\\\Rightarrow -\frac{c}{2}\bigg[\frac{1}{x^2}\bigg]^{\infty}_{1} = 1\\\\\Rightarrow \frac{c}{2} = 1\\\\\Rightarrow c = 2[/tex]
Thus, the value of c is 2.
[tex]f(x) = \dfrac{2}{x^3}, x > 1[/tex]
b) proportion of contaminating particles are PM10
We have to evaluate
[tex]P( x \leq 10) =\displaystyle\int ^{10}_{1}\frac{2}{x^3}dx\\\\=\bigg(\frac{2}{-2x^2}\bigg)^{10}_{1}\\\\=-(\frac{1}{100}-1)\\\\=0.99[/tex]
c) proportion of contaminating particles are PM2.5
[tex]P( x \leq 2.5) =\displaystyle\int ^{2.5}_{1}\frac{2}{x^3}dx\\\\=\bigg(\frac{2}{-2x^2}\bigg)^{2.5}_{1}\\\\=-(\frac{1}{6.25}-1)\\\\=0.84[/tex]
d) proportion of the PM10 particles are PM2.5
[tex]P(PM ~2.5|PM~10) = \dfrac{0.84}{0.99} = 0.8485[/tex]
Neil is a drummer who purchases his drumsticks online. When practicing with the newest pair, he notices they feel heavier than usual. When he weighs one of the sticks, he finds that it is 2.33 oz. The manufacturer's website states that the average weight of each stick is 1.75 oz with a standard deviation of 0.22 oz. Assume that the weight of the drumsticks is normally distributed. What is the probability of the stick's weight being 2.33 oz or greater
Answer:
The probability of the stick's weight being 2.33 oz or greater is 0.0041 or 0.41%.
Step-by-step explanation:
Given:
Weight of a given sample (x) = 2.33 oz
Mean weight (μ) = 1.75 oz
Standard deviation (σ) = 0.22 oz
The distribution is normal distribution.
So, first, we will find the z-score of the distribution using the formula:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Plug in the values and solve for 'z'. This gives,
[tex]z=\frac{2.33-1.75}{0.22}=2.64[/tex]
So, the z-score of the distribution is 2.64.
Now, we need the probability [tex]P(x\geq 2.33 )=P(z\geq 2.64)[/tex].
From the normal distribution table for z-score equal to 2.64, the value of the probability is 0.9959. This is the area to the left of the curve or less than z-score value.
But, we need area more than the z-score value. So, the area is:
[tex]P(z\geq 2.64)=1-0.9959=0.0041=0.41\%[/tex]
Therefore, the probability of the stick's weight being 2.33 oz or greater is 0.0041 or 0.41%.
A local department store is going out of business and is selling every item for 40% off the original price.The discount will be taken at the register. Gene buys 2 pairs of shorts for $18.77 each,one polo shirt for $21.87, and a pair of shoes for $34.24. Gene wants to know approximately how much he will pay. A) use your own words to describe the big-picture ideas in the scenario. B) use your own words to list out each important detail in the scenario. C) Estimate (round first) what Gene will pay for the items after the discount is taken.
Answer:
Gene would pay approximately $56.19 for his total purchase of two pairs of shorts, one polo shirt, and one pair of shoes.
Step-by-step explanation:
First identify what you know:
1) Every item in the store is 40% off the ticketed price (original price)
2) Gene buys two pairs of shorts for $18.77 each.
3) Gene buys one polo shirt for $21.87.
4) Gene buys one pair of shoes for $34.24.
5) Discount is taken at the register.
So, assuming that each price given, IS the original price, we need to figure out exactly how much Gene paid with the 40% discount. We can solve this in two different ways.
1) Individually calculate 40% of each price of each item, and then add them up to find the total discounted price of all the items.
2) Add all the original items and calculate 40% off the original price, and subtract that amount from the original price.
I find it simplest to add up the original prices, and find the total, and then calculate the discounted (40%) total.
Original total = (2 x $18.77) + ($21.87) + ($34.24)
Original Price = ($37.54) + ($21.87) + ($34.24)
Original Total Price: $93.65
Now, we know that without the discount, the total for ALL the items would be $93.65. Now, we just need to find out what 40% of 93.65 is! TO do this, simply multiply 93.65 by 40% OR multiply 93.65 by 0.40.
93.65 x 0.40 = 37.46
So, 40% of $93.65 is $37.46. Now we just need to subtract that percentage to find the discounted price!
$93.65 - $37.46 = $56.19
Gene paid approximately $56.19 for two pairs of short, one polo shirt, and one pair of shoes, at a discount of 40% off.
Hope this helps! :)
A recently published study shows that 50% of Americans adults take multivitamins regularly. Another recent study showed that 20.6% of American adults work out regularly. Suppose that these two variables are independent. The probability that a randomly selected American adult takes multivitamins regularly and works out regularly is _______.
Answer:
The probability that a randomly selected American adult takes multivitamins regularly and works out regularly is 0.103
Step-by-step explanation:
p(Americans adults take multivitamins regularly) = 50% = 0.50
p(American adults work out regularly) = 20.6 = 0.206
p(Americans adults take multivitamins regularly and American adults work out regularly) = 0.50* 0.206 = 0.103
Thus, p(Americans adults take multivitamins regularly and American adults work out regularly) = 0.103
If f(x, y) = x(x2 + y2)−3/2 esin(x2y), find fx(1, 0). [Hint: Instead of finding fx(x, y) first, note that it's easier to use the following equations.] fx(a, b) = g'(a) where g(x) = f(x, b) fx(a, b) = lim h→0 f(a + h, b) − f(a, b) h
By first reducing the following function f(x, y) to g(x)=x3 with y=0 and then determining its derivative at x=1, we can obtain fx(1, 0) equals 3.
Explanation:You are required to determine the value of fx(1, 0) in the given function f(x, y) = x(x(x2 + y2)3/2 e)(sin(x2y)). The tip suggests that you can address this problem by computing g'(1) where g(x) = f(x, 0). When y = 0, the function changes to f(x,0) = x * x2 * e0, which is then expressed as x3. G(x)=x3 has a derivative, g(x)=3x2. Consequently, fx(1, 0) equals g'(1), which equals 3.
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Let u solve cu = 0. Show that any derivative, say w = uxt, also solves cw = 0. In cu = 0, u = u(t, x) do the change of variables (ξ, η) specified below, to find the equation for v(ξ, η). Is it vξξ − c 2vηη = 0? (a) Translation ξ = t − T, η = x − y where y, T are fixed. (b) Dilation ξ = at, η = ax for any constant a. (c) Find the change of variables (ξ, η) = (?, ?) such that v(ξ, η) satisfies 1v = vξξ −vηη = 0
Answer:
See the pictures attached
Step-by-step explanation:
"A researcher asks participants to estimate the height (in inches) of a statue that was in a waiting area. The researcher records the following estimates: 40, 46, 30, 50, and 34. If the researcher removes the estimate of 40 (say, due to an experimenter error), then the value of the mean will"
Answer:
We conclude that the value of the mean is 40.
Step-by-step explanation:
We know that the researcher records the following estimates: 40, 46, 30, 50, and 34. If the researcher removes the estimate of 40 (say, due to an experimenter error).
Now we have the following estimates: 46, 30, 50, and 34.
We calculate the value of the mean. We get:
[tex]x=\frac{46+30+50+34}{4}\\\\x=\frac{160}{4}\\\\x=40\\[/tex]
We conclude that the value of the mean is 40.
In Exercises 29 and 30, describe the possible echelon forms of the standard matrix for a linear transformation T . Use the notation of Example 1 in Section 1.2. 29. T W R3 ! R4 is one-to-one. 30. T W R4 ! R3 is onto.
Answer:
note:
solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment
Credit sales are collected as follows:65 percent in the month of the sale.25 percent in the month after the sale.10 percent in the second month after the sale. The accounts receivable balance at the end of the previous quarter was $94,000 ($64,000 of which were uncollected December sales). a.Calculate the sale
Answer:
Step-by-step explanation:
a) November sales = (Total Uncollected Sales - Uncollected Sales from December) / Collection rate after two months
= ($121,100 - $87,300) / 0.20
November sales = $169, 000.00
b) December sales = Uncollected sales from December / Collection rate of the previous month sales,
Therefore: December sales = $87,300 / 0.45 = $194,000
December sales = $194,000.00
c) Each month's collection for the company are:
Collections for ech month = 0.20(Sales from 2 months ago) + 0.25(Last month's sales) + 0.55 (Current sales)
January collections = 0.20($169000.00) + 0.25($194,000.00) + 0.55($132,000)
January collections = $154,900.00
February collections = 0.20($194,000.00) + 0.25($132,000) + 0.55($149,000)
February collections = $153,750.00
March collections = 0.20($132,000) + 0.25($149,000) + 0.55($164,000)
March collections = $153,850.00
The Maclaurin series for sin−1(x) is given by sin−1(x) = x + [infinity] n = 1 1 · 3 · 5 (2n − 1) 2 · 4 · 6 (2n) x2n+1 2n + 1 . Use the first five terms of the Maclaurin series above to approximate sin−1 3 7 . (Round your answer to eight decimal places.)
Answer:
0.44290869
Step-by-step explanation:
The Maclaurin series for sin⁻¹(x) is given by
sin⁻¹(x) = x + [tex]x^{\alpha } _{n=1}[/tex] [tex]\frac{1.3.5...(2n - 1)}{2.4.6...(2n)} * \frac{x^{2n + 1} }{2n + 1}[/tex]
Use the first five terms of the Maclaurin series above to approximate sin⁻¹ [tex]\frac{3}{7}[/tex]. (Round your answer to eight decimal places.)
Answer
sin⁻¹(x) = x + [tex]x^{\alpha } _{n=1}[/tex] [tex]\frac{1.3.5...(2n - 1)}{2.4.6...(2n)} * \frac{x^{2n + 1} }{2n + 1}[/tex]
in the above equation [tex]x^{\alpha } _{n=1}[/tex] summation from n=1 to ∞
we are estimating this for the first 5 terms as follows
sin⁻¹(x) = x + [tex]\frac{1}{2} * \frac{x^{3} }{3}[/tex] + [tex]\frac{1*3}{2*4} * \frac{x^{5} }{5}[/tex] + [tex]\frac{1*3*5}{2*4*6} * \frac{x^{7} }{7}[/tex] + [tex]\frac{1*3*5*7}{2*4*6*8} * \frac{x^{9} }{9}[/tex]
sin⁻¹(x) = x + [tex]\frac{x^{3} }{6}[/tex] + [tex]\frac{3x^{5} }{40}[/tex] +[tex]\frac{15x^{7} }{336}[/tex] + [tex]\frac{105x^{9} }{3456}[/tex]
now to get
sin⁻¹([tex]\frac{3}{7}[/tex]) substitute
hence,
sin⁻¹([tex]\frac{3}{7}[/tex]) = [tex]\frac{3}{7} + \frac{\frac{3}{7} ^{3} }{6} + \frac{3 * \frac{3}{7} ^{5} }{40} + \frac{15* \frac{3}{7} ^{7} }{336} + \frac{105* \frac{3}{7} ^{9} }{3456}[/tex]
sin⁻¹([tex]\frac{3}{7}[/tex]) = 0.42857142 + 0.01311953 + 0.00108437 + 0.00011855 + 0.00001482
= 0.44290869
8–8 If the derivative property of phasors is multiplication of the phasor by jω, the integral property of phasors is division of the phasor by jω. Use phasors and these properties to find the sinusoids in each of the following: υ 2 ( t ) = 1 100 d υ 1 ( t ) d t + 20 υ 1 ( t ) and υ1(t) = 10 cos(100t + 90°) V i 2 ( t ) = 10 ∫ i 1 ( t ) d t − 3 i 1 ( t ) and i1(t) = − 4 cos(5t) A
Answer:
solution attached below
Step-by-step explanation:
To find the sinusoids in the given equations, we can use the derivative and integral properties of phasors. For the first equation, υ2(t) = 1/100 dυ1(t)/dt + 20υ1(t), you can find the phasor representation of υ1(t) using the given equation υ1(t) = 10cos(100t + 90°) V and apply the derivative property of phasors. For the second equation, i2(t) = 10∫i1(t)dt - 3i1(t), you can find the phasor representation of i1(t) using the given equation i1(t) = -4cos(5t) A and apply the integral property of phasors. Therefore, the sinusoids in the given equations are υ2(t) ≈ 10cos(100t + 90°) + 2000cos(100t + 90°) V and i2(t) ≈ (4/5)cos(5t + 180°) + 12cos(5t + 180°) A·s.
Explanation:To find the sinusoids in the given equations, we can use the derivative and integral properties of phasors.
For the first equation, υ2(t) = 1/100 dυ1(t)/dt + 20υ1(t), we can start by finding the phasor representation of υ1(t) using the given equation υ1(t) = 10cos(100t + 90°) V:
υ1(t) = 10e^(j(100t + 90°))
Next, we differentiate υ1(t) with respect to t to find dυ1(t)/dt:
dυ1(t)/dt = -1000sin(100t + 90°) V/s
Using the derivative property of phasors (multiplying by jω), we have:
jωυ1(t) = j(100)(10)e^(j(100t + 90°)) = 1000ej(100t + 90°) V/s
Now, we can substitute these phasor representations back into the original equation:
υ2(t) = 1/100 (1000ej(100t + 90°) V/s) + 20(10e^(j(100t + 90°)) V
Simplifying, we get:
υ2(t) ≈ 10ej(100t + 90°) + 2000ej(100t + 90°) V
For the second equation, i2(t) = 10∫i1(t)dt - 3i1(t), we can find the phasor representation of i1(t) using the given equation i1(t) = -4cos(5t) A:
i1(t) = -4e^(j(5t + 180°))
Next, we can integrate i1(t) with respect to t to find ∫i1(t)dt:
∫i1(t)dt = (-4/5)e^(j(5t + 180°)) A·s
Using the integral property of phasors (dividing by jω), we have:
(1/jω)i1(t) = (-1/(j5))(4)e^(j(5t + 180°)) = (4/5)ej(5t + 180°) A·s
Substituting these phasor representations back into the original equation, we get:
i2(t) ≈ (4/5)ej(5t + 180°) - 3(-4)e^(j(5t + 180°)) A·s
Therefore, the sinusoids in the given equations are υ2(t) ≈ 10cos(100t + 90°) + 2000cos(100t + 90°) V and i2(t) ≈ (4/5)cos(5t + 180°) + 12cos(5t + 180°) A·s.
An arc with a measure of 190° has an arc length of 40[tex]\pi[/tex]centimeters. What is the radius of the circle on which the arc sits?
Answer:
37.9 cm
Step-by-step explanation:
Arc length is:
s = 2πr (θ/360°)
where r is the radius and θ is the arc angle.
40π cm = 2πr (190°/360°)
20 cm = r (190°/360°)
r = 37.9 cm
Answer: the radius of the circle on which the arc sits is 3.8 cm
Step-by-step explanation:
The formula for determining the length of an arc is expressed as
Length of arc = θ/360 × 2πr
Where
θ represents the central angle.
r represents the radius of the circle.
π is a constant whose value is 3.14
From the information given,
θ = 190 degrees
Length of arc = 40π centimeters
Therefore,
40π = 190/360 × 2 × π × r
Dividing both sides of the equation by π, it becomes
40 = 380r/360
380r = 40 × 360 = 1440
r = 1440/380
r = 3.8 to the nearest tenth
Verify that y1(t) =t2 and y2(t) =t−1 are two solutions of the differential equation t2y−2y=0 for t > 0.Then show that y=c1t2 +c2t−1 is also a solution of this equation for any c1 and c2.
Answer:
Step-by-step explanation:
Consider first
[tex]y_1(t) = t^2\\[/tex]
We differentiate this two times to get
[tex]y_1'(t) = 2t\\y_1"(t) =2[/tex]
Substitute in the given equation
[tex]t^2 (2) -2(t^2 =0[/tex]
Hence satisfied
Consider II equation
[tex]y_2(t) = t^{-1} \\y_2'(t) = - t^{-2}\\y_2"(t) = -2 t^{-3}[/tex]
Substitute in the given equation to get
[tex]t^2 (-2 t^{-3})+2 t^{-1} = 0[/tex]
Hence satisfied
Together if we have
[tex]y = c_1 t^2 +c_2 t^{-1}[/tex]
being linear combination of two solutions
automatically this also will satisfiy the DE Or
this is a solution to the given DE
Since this expression is identically zero for any values of c1 and c2, we can conclude that y=c1t2 +c2t−1 is also a solution of the differential equation for t > 0.
Verifying that y1(t) =t2 is a solution of the differential equation t2y−2y=0:
Substituting y1(t) =t2 into the differential equation, we get:
t2(t2) - 2(t2) = 0
t4 - 2t2 = 0
2t2(t2 - 1) = 0
Since t > 0, t2 - 1 = 0
t2 = 1
t = 1 or t = -1
However, t > 0, so the only valid solution is t = 1. Therefore, y1(t) =t2 is indeed a solution of the differential equation for t > 0.
Verifying that y2(t) =t−1 is a solution of the differential equation t2y−2y=0:
Substituting y2(t) =t−1 into the differential equation, we get:
t2(t−1) - 2(t−1) = 0
t3 - t - 2t + 2 = 0
t3 - 3t + 2 = 0
(t - 1)(t2 - 2t + 2) = 0
Since t > 0, t2 - 2t + 2 = 0
(t - 1)(t - 1) = 0
t = 1 or t = 1
However, t > 0, so the only valid solution is t = 1. Therefore, y2(t) =t−1 is indeed a solution of the differential equation for t > 0.
Showing that y=c1t2 +c2t−1 is also a solution of this equation for any c1 and c2:
Substituting y=c1t2 +c2t−1 into the differential equation, we get:
t2(c1t2 +c2t−1) - 2(c1t2 +c2t−1) = 0
c1t4 +c2t3 - 2c1t2 - 2c2t + 2 = 0
Yvonne bought a new computer and printer for college. The total cost was 2500, which she put on her new credit card that has an interest rate of 13.5%. She makes a $75.00 monthly payment. How many months will it take to pay off the credit card balance? Enter your answer as a whole number such as: 23.
Answer:
43 months
Step-by-step explanation:
The amortization formula is ...
A = P(r/n)/(1 -(1 +r/n)^(-nt))
We want to find the value of t for monthly payment A, financed amount P, interest rate r, and compounding monthly (n=12). Filling in the values, we have ...
75 = 2500(0.135/12)/(1 -(1 +0.135/12)^(-12t))
1 -(1.01125^(-12t) = 2500·0.135/(12·75) = 0.375
1 -0.375 = 1.01125^(-12t) . . . . . add 1.01125^(-12t) -0.375
Next, take logarithms and divide by the coefficient of t.
log(0.625)/(-12log(1.01125)) = t ≈ 3.50106 . . . . years
In months, that is ...
3.50106×12 ≈ 42.01
The balance will be not quite zero after 42 payments. (It will be about $0.94.) It will take 43 payments to pay off the credit card balance.
Answer:
42
I did the test and got it right hope this helps
Assume that you have $60 a week to spend on bottled water and chips. A bottle of water costs $2 and a bag of chips costs $3. If you buy 15 bottles of water, how many bags of chips can you purchase
Answer:
10 bags of chips
Step-by-step explanation:
15(2)+3x=60
30+3x=60
3x=30
x=10
Therefore, 10 bags of chips
There are many applications of exponentials and logarithms, including exponential growth and decay, half life, doubling time, Carbon dating, compound interest. Here are a couple of examples.
You find out that in the year 1800 an ancestor of yours invested 100 dollars at 6 percent annual interest, compounded yearly. You happen to be her sole known descendant and in the year 2005 you collect the accumulated tidy sum of _______________ dollars. You retire and devote the next 10 years of your life to writing a detailed biography of your remarkable ancestor.
Strontium-90 is a biologically important radioactive isotope that is created in nuclear explosions. It has a half life of 28 years. To reduce the amount created in a particular explosion by a factor 1,000 you would have to wait______________ years. Round your answer to the nearest integer.
Seeds found in a grave in Egypt proved to have only 53% of the Carbon-14 of living tissue. Those seeds were harvested ________________ years ago. The half life of Carbon-14 is 5,730 years.
Answer:
a) $15,406,443
b) 279.04 years = 279 years.
c) 5246.9 years = 5247 years.
Step-by-step explanation:
Compound interest
The final amount obtainable, A, from saving an initial amount, P, compounded at a rate of r in t number of years is given as
A = P (1 + r)ᵗ
A = ?
P = $100
r = 6% = 0.06
t = 2005 - 1800 = 205
A = 100 (1 + 0.06)²⁰⁵ = 100 × 154064.43 = $15406443
b) Radioactivity
Let the initial amount of Strontium be A
After 1 half life,
Amount remaining is A/2
After two half lives,
Amount remaining = A/2²
After 3 half lives,
Amount remaining = A/2³
After n half lives,
Amount remaining = A/2ⁿ
So, for this question,
(A)/(A/2ⁿ) = 1000
2ⁿ = 1000
In 2ⁿ = In 1000
n = (In 1000)/(In 2)
n = 9.966
1 half life = 28 years
n half lives = n × 28 = 9.966 × 28 = 279.04 years.
c) Carbon dating
The general relation of amount left to amount of Carbon-14 that is started with follows a first order rate of decay kinetics like every radioactive decay
A = A₀ e⁻ᵏᵗ
A = amount of Carbon-14 left at any time
A₀ = initial amount of Carbon-14
k = rate constant = (In 2)/(half life) = 0.693/5730 = 0.000121 /year.
(A/A₀) = 53% = 0.53
e⁻ᵏᵗ = 0.53
-kt = In 0.53 = -0.6349
t = 0.6349/k = 0.6349/0.000121 = 5246.9 years = 5247 years
Using the principles of compound interest, radioactive decay and carbon dating, we calculate that the investment made in 1800 would equal about $8,680,204 in 2005, that it would take approximately 280 years to reduce the amount of Strontium-90 by a factor 1000, and that the seeds were harvested around 8463 years ago.
Explanation:The sum you would've collected in the year 2005 from a 100 dollars investment at 6 percent annual interest, compounded yearly since the year 1800 depends on the principle of compound interest. The formula to calculate compound interest is A = P(1 + r/n)^(nt), where A is the total sum, P is the principal amount, r is the annual rate of interest, n is the number of compounding periods a year, and t is the time in years. In your case, we have P = 100, r = 0.06, n = 1 (compounded yearly), and t = 2005 - 1800 = 205 years. So, A = 100(1 + 0.06/1)^(1*205) which equals approximately $8,680,204.
Strontium-90, with a half life of 28 years, would require waiting for a certain amount of years to reduce by a factor of 1000. For half life calculations, use the formula N = N0*(1/2)^(t/h), where N is the final amount, N0 is the initial amount, t is time and h is the half life. We want N0/N = 1000, so log2(1000) = t/28, which gives t = 28*log2(1000) = approximately 280 years.
The carbon-14 dating principle lets us calculate the age of the seeds. Using log2(1/0.53) = t/5730 to get t = 5730 * log2(1/0.53), we find the seeds were harvested about 8463 years ago.
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Mrs. Jones recorded the time, in minutes, she spends reading each day for two weeks. The results are shown. What is the IQR for each week? Week 1 Week 2 81 50 63 58 39 72 104 62 54 110 72 68 34 79 A. The IQR for Week 1 is 65, and the IQR for Week 2 is 76. B. The IQR for Week 1 is 63, and the IQR for Week 2 is 68. C. The IQR for Week 1 is 50, and the IQR for Week 2 is 54. D. The IQR for Week 1 is 31, and the IQR for Week 2 is 25.
Answer:
D.
The IQR for Week 1 is 31, and the IQR for Week 2 is 25.
Step-by-step explanation:
Answer: D.The IQR for week 1 is 31 and the IQR for week 2 is 25
Step-by-step explanation: When you divid a data set in groups of 4 and measure the bulk of the values, this is called Interquatile Range (IQR).
It's calculated as follows:
1) Put the data in order;
week 1 : 39 50 58 63 72 81 104
week 2 : 34 54 62 68 72 79 110
2) Find the median of each data set:
*Note: Median is the middle value of a set.
week 1 : Median 63
week 2 : Median 68
3) Find Q1, which is the median in the lower half of the set:
week 1: the lower set is 39 50 58.
The middle value is 50.
So Q1 = 50.
week 2: the lower set is 34 54 62
The middle value is 54.
So Q1 = 54.
4) Find Q3, which is the middle value of the upper half:
week 1 : the upper half is 72 81 104
Q2 = 81
week 2 : the upper half is 72 79 110
Q2 = 79
5) To determine IQR, subtract Q1 and Q3:
week 1: 81 - 50 = 31
week 2 : 79 - 54 = 25
In conclusion, the IQR for week 1 is 31 and for week 2 is 25.
QUESTION 1 (0.5 POINTS) In a large restaurant an average of 3 out of every 5 customers ask for water with their meal. A random sample of 10 customers is selected. What is the probability that less than 3 customers ask for water with their meal? a). 0.012 b). 0.055 c). 0.042 d). 0.011
Answer:
Option A) 0.012
Step-by-step explanation:
We are given the following information:
We treat customers asking for water as a success.
P(customers ask for water) = [tex]\dfrac{3}{5}[/tex] = 0.6
Then the number of customers follows a binomial distribution, where
[tex]P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}[/tex]
where n is the total number of observations, x is the number of success, p is the probability of success.
Now, we are given n = 10
We have to evaluate:
[tex]P(x < 3) = P(x = 0) + P(x = 1) + P(x = 2) \\= \binom{10}{0}(0.6)^0(1-0.6)^{10} +\binom{10}{1}(0.6)^1(1-0.6)^{9} + \binom{10}{2}(0.6)^2(1-0.6)^{8}\\= 0.0001 +0.0015 + 0.0106\\=0.012[/tex]
0.012 is the probability that less than 3 customers ask for water with their meal.
The correct answer is option a) 0.012
The probability of success in each trial is 3 out of 5 (or 0.6), and the probability of failure is 2 out of 5 (or 0.4). Denoting the number of customers who ask for water as a binomial random variable X with n = 10 and p = 0.6.
The binomial probability formula is given by:
[tex]P(X=k) ={n \choose k}p^k(1-p)^{n-k}[/tex]
where:
n is the number of trials (10 in this case),
k is the number of successes (in this case, less than 3),
p is the probability of success (0.6), and
1−p is the probability of failure (0.4).
Probability that fewer than 3 customers ask for water needs to be found. This means we need to find:
P(X<3)=P(X=0)+P(X=1)+P(X=2)
Calculating each of these probabilities using the binomial formula:
For k=0:
[tex]P(X=0) ={10 \choose 0}(0.6)^0(0.4)^{10-0} = 1\times1\times0.4^{10} = 0.0001048576[/tex]
For k=1:
[tex]P(X=1) ={10 \choose 1}(0.6)^1(0.4)^{10-1} = 10\times0.6\times0.4^{9} = 0.001572864[/tex]
For k=2:
[tex]P(X=2) ={10 \choose 2}(0.6)^2(0.4)^{10-2} = 45\times0.36\times0.4^{8} = 0.010616832[/tex]
Now, we sum these probabilities to get the total probability that fewer than 3 customers ask for water:
P(X<3) = P(X=0) + P(X=1) + P(X=2)
P(X<3) = 0.0001048576 + 0.001572864 + 0.010616832
P(X<3) = 0.0122945536
Therefore rounding this value results in a probability of approximately 0.012.
Suppose you are interested in the effect of skipping lectures (in days missed) on college grades. You also have ACT scores and high school GPA (HSGPA). You run the following regression model numbers in parentheses below each coefficient represent standard errors of each coefficient) colGPA =2.52+0.38H SGPA+0.015 ACT-0.5skip (0.2) (0.3) (0.0001)
(a) Interpret the intercept in this model.
(b) Interpret BACT from this model.
(c) What is the predicted college GPA for someone who scored a 25 on the ACT, had a 3.2 high school GPA and missed 4 lectures. Show your work.
(d) Is the estimate of skipping class statistically significant? How do you know? Is the estimate of skipping class economically significant? How do you know? (Hint: Suppose there are 45 lectures in a typical semester long class).
Answer:
a) For this case the intercept of 2.52 represent a common effect of measure for any student without taking in count the other variables analyzed, and we know that if HSGPA=0, ACT= 0 and skip =0 we got [tex] colGPA=2.52[/tex]
b) This value represent the effect into the ACT scores in the GPA, we know that:
[tex]\hat \beta_{ACT} = 0.015[/tex]
So then for every unit increase in the ACT score we expect and increase of 0.015 in the GPA or the predicted variable
c) If we are interested in analyze if we have a significant relationship between the dependent and the independent variable we can use the following system of hypothesis:
Null Hypothesis: [tex]\beta_i = 0[/tex]
Alternative hypothesis: [tex]\beta_i \neq 0[/tex]
Or in other wouds we want to check if an specific slope is significant.
The significance level assumed for this case is [tex]\alpha=0.05[/tex]
Th degrees of freedom for a linear regression is given by [tex]df=n-p-1 = 45-3-1 = 41[/tex], where p =3 the number of variables used to estimate the dependent variable.
In order to test the hypothesis the statistic is given by:
[tex]t=\frac{\hat \beta_i}{SE_{\beta_i}}[/tex]
And replacing we got:
[tex] t = \frac{-0.5}{0.0001}=-5000[/tex]
And for this case we see that if we find the p value for this case we will get a value very near to 0, so then we can conclude that this coefficient would be significant for the regression model .
Step-by-step explanation:
For this case we have the following multiple regression model calculated:
colGPA =2.52+0.38*HSGPA+0.015*ACT-0.5*skip
Part a
(a) Interpret the intercept in this model.
For this case the intercept of 2.52 represent a common effect of measure for any student without taking in count the other variables analyzed, and we know that if HSGPA=0, ACT= 0 and skip =0 we got [tex] colGPA=2.52[/tex]
(b) Interpret [tex]\hat \beta_{ACT}[/tex] from this model.
This value represent the effect into the ACT scores in the GPA, we know that:
[tex]\hat \beta_{ACT} = 0.015[/tex]
So then for every unit increase in the ACT score we expect and increase of 0.015 in the GPA or the predicted variable
(c) What is the predicted college GPA for someone who scored a 25 on the ACT, had a 3.2 high school GPA and missed 4 lectures. Show your work.
For this case we can use the regression model and we got:
[tex] colGPA =2.52 +0.38*3.2 +0.015*25 - 0.5*4 = 26.751[/tex]
(d) Is the estimate of skipping class statistically significant? How do you know? Is the estimate of skipping class economically significant? How do you know? (Hint: Suppose there are 45 lectures in a typical semester long class).
If we are interested in analyze if we have a significant relationship between the dependent and the independent variable we can use the following system of hypothesis:
Null Hypothesis: [tex]\beta_i = 0[/tex]
Alternative hypothesis: [tex]\beta_i \neq 0[/tex]
Or in other wouds we want to check if an specific slope is significant.
The significance level assumed for this case is [tex]\alpha=0.05[/tex]
Th degrees of freedom for a linear regression is given by [tex]df=n-p-1 = 45-3-1 = 41[/tex], where p =3 the number of variables used to estimate the dependent variable.
In order to test the hypothesis the statistic is given by:
[tex]t=\frac{\hat \beta_i}{SE_{\beta_i}}[/tex]
And replacing we got:
[tex] t = \frac{-0.5}{0.0001}=-5000[/tex]
And for this case we see that if we find the p value for this case we will get a value very near to 0, so then we can conclude that this coefficient would be significant for the regression model .
A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below:
City Price ($) Sales
River Falls 1.30 100
Hudson 1.60 90
Ellsworth 1.80 90
Prescott 2.00 40
Rock Elm 2.40 38
Stillwater 2.90 32
Referring to the above listed table, what is the estimated slope parameter for the candy bar price and sales data?
(A) 161.386
(B) 0.784
(C) -3.810
(D) -48.193
Answer:
(D) -48.193
Step-by-step explanation:
We know that regression equation is
y=a+bx where a is intercept and b is slope of the regression equation.
[tex]Slope=b=\frac{sum(x-xbar)(y-ybar)}{sum(x-xbar)^2}[/tex]
City Price ($)(x) Sales(Y)
River Falls 1.30 100
Hudson 1.60 90
Ellsworth 1.80 90
Prescott 2.00 40
Rock Elm 2.40 38
Stillwater 2.90 32
xbar=sumx/n=12/6=2
ybar=sumy/n=390/6=65
x y x-xbar y-ybar (x-xbar)(y-ybar) (x-xbar)²
1.3 100 -0.7 35 -24.5 0.49
1.6 90 -0.4 25 -10 0.16
1.8 90 -0.2 25 -5 0.04
2 40 0 -25 0 0
2.4 38 0.4 -27 -10.8 0.16
2.9 32 0.9 -33 -29.7 0.81
Total -80 1.66
[tex]Slope=b=\frac{sum(x-xbar)(y-ybar)}{sum(x-xbar)^2}[/tex]
b=-80/1.66
b=-48.193
Final answer:
The estimated slope parameter, which represents the rate of change in candy bar sales for every dollar change in price, is -3.810, indicating that there is a decrease in sales as the price increases. So the correct option is C.
Explanation:
To estimate the slope parameter of the regression line representing the relationship between candy bar price and sales, we can apply the formula for the slope (m) of the linear regression line:
m = Σ((Xi - μx) × (Yi - μy)) / Σ(Xi - μx)²
where Xi and Yi are the individual sample points and μx and μy are the mean values for the independent (X) and dependent (Y) variables, respectively.
Using the data provided, we can calculate the means (μx and μy) and then use the points to calculate the sum of the products of the deviations and the sum of the squared deviations. We substitute these into the slope formula to find the estimated slope parameter, which would provide the rate of change in candy bar sales concerning the price change.
After performing the calculations, we find that the estimated slope parameter is -3.810. Thus, the correct answer is (C).
(c) Assume this is a simple random sample of U.S. women. Use the Empirical Method to estimate the probability that a woman has more than six children. Round your answer to four decimal places.
Complete Question
The complete question is shown on the first uploaded image
Answer:
p(X> 6 ) = 0.01132 + 0.01672 = 0.0280
Step-by-step explanation:
The probability that the US woman would have more than six children is equal to the probability that she would have children that are equal to 7 or that she would that she would have children equal to 8 or more as shown on the table in the question
Now representing this mathematically we have
p(X > 6) = p(x =7 ) + p(x = 8 or more)
From the table the p(x = 7 ) [tex]= \frac{523}{46165} = 0.01132[/tex]
And the p(x= 8 or more ) [tex]= \frac{772}{46165} = 0.01672[/tex]
Hence p(X> 6 ) = 0.01132 + 0.01672 = 0.0280
An estimation of the probability that a U.S. woman has more than six children using the Empirical Method involves counting how many women in the sample data have more than six children divided by the total number of women in the sample. The specific calculation can't be provided without the concrete data.
Explanation:To estimate the probability that a U.S. woman has more than six children using the Empirical Method, you would need specific data from the sample. However, since this data hasn't been provided, it's important to explain the process.
First, list the number of women in your sample data who have more than six children. Let's call this number 'X'. Then, calculate the total number of women in your sample data. Let's call this number 'Y'. The empirical probability is then calculated by dividing the number of successful outcomes (X) by the total number of outcomes (Y).
So, if we had data, the equation would be: Probability = X/ Y. Remember to round your answer to four decimal places.
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A random sample of 6 homes in Gainesville, Florida between 1800 and 2200 square feet had a mean of 212990 and a standard deviation of 14500. Construct a 95% confidence interval for the average price of a home in Gainesville of this size. Group of answer choices (201387, 224592) (197773, 228207) (196318, 229662) (196557, 229422)
Answer:
Option B) (197773, 228207)
Step-by-step explanation:
We are given the following in the question:
Sample mean, [tex]\bar{x}[/tex] = 212990
Sample size, n = 6
Alpha, α = 0.05
Sample standard deviation = 14500
95% Confidence interval:
[tex]\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}[/tex]
Putting the values, we get,
[tex]t_{critical}\text{ at degree of freedom 5 and}~\alpha_{0.05} = \pm 2.5705[/tex]
[tex]212990 \pm 2.5705(\dfrac{14500}{\sqrt{6}} ) \\\\= 212990 \pm 15216.33 \\= (197773.67 ,228206.33)\\\approx (197773, 228207)[/tex]
Option B) (197773, 228207)
Using the t-distribution, it is found that the 95% confidence interval for the average price of a home in Gainesville of this size is (197773, 228207).
The first step is finding the number of degrees of freedom, which is the sample size subtracted by 1, so df = 6 - 1 = 5.
Now, we look at the t-table for the critical value for a 95% confidence interval with 5 df, which is t = 2.5706.
The margin of error is:
[tex]M = t\frac{s}{\sqrt{n}}[/tex]
For this problem, [tex]s = 14500, n = 6[/tex], thus:
[tex]M = 2.5706\frac{14500}{\sqrt{6}}[/tex]
[tex]M = 15217[/tex]
The confidence interval is:
[tex]\overline{x} \pm M[/tex]
For this problem, [tex]\overline{x} = 212990[/tex], then:
[tex]\overline{x} - M = 212990 - 15217 = 197773[/tex]
[tex]\overline{x} + M = 212990 + 15217 = 228207[/tex]
The correct option is (197773, 228207).
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If X=5 and 5=y, then x=y
Ar algebraic equality property justifies the above statement?
Answer:
Yes.
Step-by-step explanation:
According to the Transitive property of the Algebraic properties of equality, two values are said to be equal is they are differently equal to a corresponding third party. This is, If a=b and b=c then a=c.
Hence, if x=5 and 5=y, then following the transitive property of Algebraic properties of equality, x=y, hence the Algebraic property of equality Justifies the statement.
The time rate of change of a rabbit population PP is proportional to the square root of PP. At time t=0t=0 (months) the population numbers 100100 rabbits and is increasing at the rate of 1010 rabbits per month. Let P′=kP12P′=kP12 describe the growth of the rabbit population, where kk is a positive constant to be found. Find the formulas for kk and for the rabbit population P(t)P(t) after tt months.
Answer:
[tex] \frac{dP}{\sqrt{P}} = k dt[/tex]
And if we integrate both sides we got:
[tex] 2 \sqrt{P} = kt +C[/tex]
Where C is a constant., we can rewrite the expression like this:
[tex] \sqrt{P} = \frac{1}{2} (kt +C)[/tex]
If we square both sides we got:
[tex] P = \frac{1}{4} (kt +C)^2 [/tex]
If we use the initial condition we have that:
[tex] P(0) = 100 = \frac{1}{4} (k*0 +C)^2 [/tex]
And we can solve for C like this:
[tex] 400 = C^2[/tex]
[tex] C = 20[/tex]
And now we can find the derivate of the function and we got:
[tex] P'(t) = 2* \frac{1}{4} (kt + 20) * k[/tex]
Using the condition [tex] P'(0) = 10 [/tex] we got:
[tex] 10 = \frac{1}{2} k (k*0 +20)[/tex]
[tex] 20 = 20 k[/tex]
k= 1
And then the model is defined as:
[tex] P = \frac{1}{4} (t +20)^2 [/tex]
And for t =12 months we have:
[tex] P(12) = \frac{1}{4} (12 +20)^2 = 256 [/tex]
Step-by-step explanation:
For this case we cna use the proportional model given by:
[tex] \frac{dP}{dt} = k \sqrt{P}[/tex]
Where k is a proportional constant, P the population and the represent the number of months
For this case we know the following initial condition [tex] P(0) =100[/tex] and [tex] P'(0) = 10 [/tex]
we can rewrite the differential equation like this:
[tex] \frac{dP}{\sqrt{P}} = k dt[/tex]
And if we integrate both sides we got:
[tex] 2 \sqrt{P} = kt +C[/tex]
Where C is a constant., we can rewrite the expression like this:
[tex] \sqrt{P} = \frac{1}{2} (kt +C)[/tex]
If we square both sides we got:
[tex] P = \frac{1}{4} (kt +C)^2 [/tex]
If we use the initial condition we have that:
[tex] P(0) = 100 = \frac{1}{4} (k*0 +C)^2 [/tex]
And we can solve for C like this:
[tex] 400 = C^2[/tex]
[tex] C = 20[/tex]
And now we can find the derivate of the function and we got:
[tex] P'(t) = 2* \frac{1}{4} (kt + 20) * k[/tex]
Using the condition [tex] P'(0) = 10 [/tex] we got:
[tex] 10 = \frac{1}{2} k (k*0 +20)[/tex]
[tex] 20 = 20 k[/tex]
k= 1
And then the model is defined as:
[tex] P = \frac{1}{4} (t +20)^2 [/tex]
And for t =12 months we have:
[tex] P(12) = \frac{1}{4} (12 +20)^2 = 256 [/tex]
to compound a custom solution, you are asked to add 10mg of drug to a 500ml bag of 5% dextrose injection. the drug is available in pre mixed bottles of 0.5% solution. how many ml should you add
Answer:
2 mL
Step-by-step explanation:
If the drug is mixed in a solution of 0.5%, it means that for each ml of the solution, there is 0.5% or 0.005 grams of the drug. In order to get 10 mg of the drug, the volume required is:
[tex]0.005\frac{g}{mL}*V = 10*10^{-3} g\\ V= 2\ mL[/tex]
You should add 2 mL of the solution.
*Note that the volume of the bag should not be used since the amount of drug needed was specified in weight and not in concentration.
(6 pts) The average age of CEOs is 56 years. Assume the variable is normally distributed. If the SD is four years, find the probability that the age of randomly selected CEO will be between 50 and 55 years old.
Answer:
The probability that the age of a randomly selected CEO will be between 50 and 55 years old is 0.334.
Step-by-step explanation:
We have a normal distribution with mean=56 years and s.d.=4 years.
We have to calculate the probability that a randomly selected CEO have an age between 50 and 55.
We have to calculate the z-value for 50 and 55.
For x=50:
[tex]z=\frac{x-\mu}{\sigma}=\frac{50-56}{4}=\frac{-6}{4}= -1.5[/tex]
For x=55:
[tex]z=\frac{x-\mu}{\sigma}=\frac{55-56}{4}=\frac{-1}{4}=-0.25[/tex]
The probability of being between 50 and 55 years is equal to the difference between the probability of being under 55 years and the probability of being under 50 years:
[tex]P(50<x<55)=P(x<55)-P(x<50)\\\\P(50<x<55)=P(z<-0.25)-P(z<-1.5)\\\\P(50<x<55)=0.40129-0.06681\\\\P(50<x<55)=0.33448[/tex]
Final answer:
The probability that the age of a randomly selected CEO will be between 50 and 55 years old is approximately 0.3345.
Explanation:
To find the probability that the age of a randomly selected CEO will be between 50 and 55 years old, we need to calculate the z-scores for both values and then use the standard normal distribution table.
Step 1: Calculate the z-score for 50 years old:
z = (50 - 56) / 4 = -1.5
Step 2: Calculate the z-score for 55 years old:
z = (55 - 56) / 4 = -0.25
Step 3: Use the standard normal distribution table to find the area to the left of each z-score:
P(z < -1.5) = 0.0668
P(z < -0.25) = 0.4013
Step 4: Calculate the probability between the two z-scores:
P(-1.5 < z < -0.25) = P(z < -0.25) - P(z < -1.5) = 0.4013 - 0.0668 = 0.3345
Therefore, the probability that the age of a randomly selected CEO will be between 50 and 55 years old is approximately 0.3345.
For each of the following pairs of Treasury securities (each with $ 1 comma 000 par value), identify which will have the higher price: a. A three-year zero-coupon bond or a five-year zero-coupon bond? b. A three-year zero-coupon bond or a three-year 4 % coupon bond? c. A two-year 5 % coupon bond or a two-year 6 % coupon bond?
Answer:
Step-by-step explanation:
A. The three-year zero coupon bond, because the future value is receiver sooner, thus the present value is higher.
B. The three year 4% coupon bond because it pays interest payments, whereas the zero coupon bond is pure discount bond.
C. The 6% coupon bond because the coupon (interest) payments are higher.
Final answer:
In summary, a three-year zero-coupon bond has a higher price than a five-year zero-coupon bond due to its shorter maturity. A three-year 4% coupon bond will have a higher price than a three-year zero-coupon bond because it provides interest payments. A two-year 6% coupon bond will have a higher price than a two-year 5% coupon bond due to its higher interest payments.
Explanation:
The Treasury security prices are determined by several factors, including maturity and coupon rates. In answering the questions presented by the student, we compare the prices of different types of Treasury securities based on these attributes.
A three-year zero-coupon bond will have a higher price than a five-year zero-coupon bond because it has a shorter time to maturity, meaning less time for interest rates to affect its price.Between a three-year zero-coupon bond and a three-year 4% coupon bond, the coupon bond will generally have a higher price since it provides periodic interest payments in addition to the repayment of par at maturity.Comparing a two-year 5% coupon bond with a two-year 6% coupon bond, the bond with the higher coupon rate (6%) will typically have a higher price, as it offers more substantial periodic payments.