Answer:
20 liters of the 40% solution
60-20=40 liters of the 70% solution
Step-by-step explanation:
40%x+70%(60-x)=60%(60)
0.4x+0.7(60-x)=0.6(60)
0.4x+42-0.7x=36
-0.3x+42=36
-0.3x=-42+36
-0.3x=-6
0.3x=6
3x=60
x=60/3
x=20 liters of the 40% solution
60-20=40 liters of the 70% solution
check: 0.4*20=8
0.7*40=28
8+28=36 and 0.6*60=36 also
What is the greatest common factor of 10x^2y^2 – 8xy?
(^=exponent)
Answer:
2xy
Step-by-step explanation:
10x²y² - 8xy
2xy(5xy - 4)
The greatest common factor (GCF) of the expressions 10x^2y^2 and 8xy is 2xy. The GCF for the numbers is 2, and for the variables x and y it's x and y, respectively.
Explanation:The subject is the greatest common factor (GCF), applied to algebraic expressions. The GCF is the largest expression that can be multiplied by another expression to get the original expression. For the expressions 10x^2y^2 and 8xy, the numbers are 10 and 8. Their GCF is 2. The term x appears in both expressions, and the lowest power of x is 1. The term y also appears in both expressions, with the lowest power being 1. So, 2x and y are the GCFs for the numbers and the variables respectively. Combining them, we get 2xy, which is the GCF of 10x^2y^2 and 8xy.
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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."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.
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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PLEASE HELP 69 POINTS, BRAINLIEST, 5 STARS, AND THANKS.
Answer:
a) 125 scoops b) 31 scoops
Step-by-step explanation:
The sink = 4000/3 pi
a) The volume of a Cone is 1/3 (pi)(h)(r)^2, meaning...
(1/3)(pi)(8)(2^2)=32/3 pi, and...
(4000pi/3)/(32pi/3) =
4000/32 =
125
b) The volume of a Cone is 1/3 (pi)(h)(r)^2, meaning...
(1/3)(pi)(8)(4^2) = 128pi/3, and...
(4000pi/3)/(128pi/3) =
4000/128 =
31.25, which is approx. 31
Answer:
a) 125
b) 31
Step-by-step explanation:
a) radius = 2
Volume of cup = ⅓×pi×r²×h
= 32pi/3
No. of scoops = 4000pi/3 ÷ 32pi/3
= 125
b) radius = 4
Volume of cup = ⅓×pi×r²×h
= 128pi/3
No. of scoops = 4000pi/3 ÷ 128pi/3
= 31.25
Nearest whole number: 31
A shipping company handles containers in three different sizes: (1) 27 ft3 (3 × 3 × 3), (2) 125 ft3, and (3) 512 ft3. Let Xi (i = 1, 2, 3) denote the number of type i containers shipped during a given week. With μi = E(Xi) and σi2 = V(Xi), suppose that the mean values and standard deviations are as follows: μ1 = 220 μ2 = 250 μ3 = 120 σ1 = 9 σ2 = 13 σ3 = 8 (a) Assuming that X1, X2, X3 are independent, calculate the expected value and variance of the total volume shipped. [Hint: Volume = 27X1 + 125X2 + 512X3.] expected value ft3 variance ft6 (b) Would your calculations necessarily be correct if the Xi's were not independent? Explain. Both the expected value and the variance would be correct. The expected value would be correct, but the variance would not be correct. Neither the expected value nor the variance would be correct. The expected value would not be correct, but the variance would be correct.
a. Expectation is linear, so that
[tex]E[27X_1+125X_2+512X_3]=27E[X_1]+125E[X_2]+512E[X_3]=98,630[/tex]
The variance of a sum of independent random variables is equal to a weighted sum of the variances, with weights equal to the squares of the coefficients of the [tex]X_i[/tex]:
[tex]V[27X_1+125X_2+512X_3]=27^2V[X_1]+125^2V[X_2]+512^2V[X_3]=2,306,838[/tex]
b. If the [tex]X_i[/tex] were dependent on one another, we would have the same expectation, but now the variance of a sum of random variables becomes the sum of their covariances:
[tex]\displaystyle\sum_{i=1}^3V[\alpha_iX_i]=\sum_{i=1}^3\sum_{j=1}^3\mathrm{Cov}[X_i,X_j]=\sum_{i=1}^3{\alpha_i}^2V[X_i]+2\sum_{i\neq j}\alpha_i\alpha_j\mathrm{Cov}[X_i,X_j][/tex]
where
[tex]\mathrm{Cov}[X_i,X_j]=E[(X_i-E[X_i])(X_j-E[X_j])]=E[X_iX_j]-E[X_i]E[X_j][/tex]
When we assumed independence, we were granted [tex]E[X_iX_j]=E[X_i]E[X_j][/tex], but this may not be the case if the variables are dependent.
The expected value of the total volume shipped is 98630 ft3. The variance of the total volume shipped is 17423640 ft6. If the variables were not independent, only the expected value would remain correct.
Explanation:To calculate the expected value of the total volume shipped, we use the linearity of the expectation:
E(Total Volume) = E(27X1 + 125X2 + 512X3)
E(Total Volume) = 27E(X1) + 125E(X2) + 512E(X3)
E(Total Volume) = 27(μ1) + 125(μ2) + 512(μ3)
E(Total Volume) = 27(220) + 125(250) + 512(120)
E(Total Volume) = 5940 + 31250 + 61440
E(Total Volume) = 98630 ft3
To calculate the variance of the total volume shipped, given that X1, X2, X3 are independent:
V(Total Volume) = V(27X1 + 125X2 + 512X3)
V(Total Volume) = 272V(X1) + 1252V(X2) + 5122V(X3)
V(Total Volume) = 729(σ12) + 15625(σ22) + 262144(σ32)
V(Total Volume) = 729(92) + 15625(132) + 262144(82)
V(Total Volume) = 729(81) + 15625(169) + 262144(64)
V(Total Volume) = 59049 + 2639375 + 16777216
V(Total Volume) = 17423640 ft6
If the Xi's were not independent, the expected value would still be correct but the variance would not be correct because the calculation of variance for non-independent variables includes covariance terms that are not present for independent variables.
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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).
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%
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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Baseball's World Series is a maximum of seven games, with the winner being the first team to win four games. Assume that the Atlanta Braves and the Minnesota Twins are playing in the World Series and that the first two games are to be played in Atlanta, the next three games at the Twins' ballpark, and the last two games, if necessary, back in Atlanta. Taking into account the projected starting pitchers for each game and the home field advantage, the probabilities of Atlanta winning each game are as follows: Game 1 2 3 4 5 6 7 Probability of Win 0.4 0.55 0.42 0.56 0.55 0.39 0.52 Set up a spreadsheet simulation model for which whether Atlanta wins or loses each game is a random variable. What is the probability that the Atlanta Braves win the World Series? If required, round your answer to two decimal places. What is the average number of games played regardless of winner? If required, round your answer to one decimal place.
Answer:
For part ;a)
probability that the Atlanta Braves win the World Series=0.40
For part b ;b)
average number of games played regardless of winner =5.8
Step-by-step explanation:
using the rand() function for the simulation of the probabilities and compare with given probabilites if less than above then Atlanta wins otherwise loses
a)
probability that the Atlanta Braves win the World Series=0.40
b)
average number of games played regardless of winner =5.8
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
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
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]
In a sample of 100 steel canisters, the mean wall thickness was 8.1 mm with a standard deviation of 0.5 mm. Someone says that the mean thickness is less than 8.2 mm. With what level of confidence can this statement be made? (Express the final answer as a percent and round to two decimal places.)
Answer:
This statement can be made with a level of confidence of 97.72%.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 8.1 mm
Standard Deviation, σ = 0.5 mm
Sample size, n = 100
We are given that the distribution of thickness is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
Standard error due to sampling:
[tex]=\dfrac{\sigma}{\sqrt{n}} = \dfrac{0.5}{\sqrt{100}} = 0.05[/tex]
P(mean thickness is less than 8.2 mm)
P(x < 8.2)
[tex]P( x < 8.2)\\\\ = P( z < \displaystyle\frac{8.2 - 8.1}{0.05})\\\\ = P(z < 2)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x < 8.2) =0.9772 = 97.72\%[/tex]
This statement can be made with a level of confidence of 97.72%.
The z-score calculation reveals that the mean thickness is less than 8.2 mm with approximately 97.50% confidence.
Explanation:To determine the level of confidence that the mean thickness is less than 8.2 mm, we can calculate the z-score of the sample mean and then find the corresponding percentile in the standard normal distribution. The z-score is calculated by the formula:
z = (x - μ) / (σ / √n)
where x is the sample mean, μ is the population mean, σ is the standard deviation, and n is the sample size.
Using the given information, x = 8.1 mm, μ = 8.2 mm (the mean to test against), σ = 0.5 mm, and n = 100, we get:
z = (8.1 mm - 8.2 mm) / (0.5 mm / √100) = -0.1 / 0.05 = -2
The z-score corresponds to a percentile in the standard normal distribution, which can be found using a z-table or statistical software. A z-score of -2 corresponds to approximately the 2.5th percentile, which means that the sample mean is less than 8.2 mm with about 97.5% confidence.
Therefore, the statement that the mean thickness is less than 8.2 mm can be made with approximately 97.50% confidence.
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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.
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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In an experiment, A,B, C, andD are events with probabilitiesP[A UB] = 5/8,P[A] =3/8,
P[C ∩D] = 1/3, andP[C] =1/2. Furthermore, Aand B are disjoint, whileC and D areindependent.
(a) Find P[A∩ B],P[B],P[A ∩Bc], andP[A UBc].
(b) Are A andB independent?
(c) FindP[D],P[C ∩Dc],P[Cc∩ Dc],andP[C|D].
(d) Find P[CU D] andP[C UDc].
(e) Are C andDcindependent?
Answer:
Step-by-step explanation:
Hello!
You have 4 events A, B, C and D
With probabilities:
P(A∪B)= 5/8
P(A)= 3/8
P(C∩D)= 1/3
P(C)= 1/2
A and B are disjoint events, this means that there are no shared elements between then and their intersection is void, symbolically A∩B= ∅, in consequence, these events are mutually exclusive.
C and D are independent events, this means that the occurrence of one of them does not affect the probability of occurrence of the other one in two consecutive repetitions.
a.
i. P(A∩B)= 0
⇒ Since A and B are disjoint events, the probability of their intersection is zero.
ii. A and B are mutually exclusive events, this means that P(A∪B)= P(A)+P(B)
⇒ From this expression, you can clear the probability of b as P(B)= P(A∪B)-P(A)= 5/8-3/8= 1/4
iii. If Bc is the complementary event of B, its probability would be P(Bc)= 1 - P(B)= 1 - 1/4= 3/4. If the events A and B are mutually exclusive and disjoint, it is logical to believe that so will be the events A and Bc, so their intersection will also be void:
P(A∩Bc)= 0
vi.P(A∪Bc)= P(A) + P(Bc)= 3/8+3/4= 9/8
b.
If A and B are independent then the probability of A is equal to the probability of A given B, symbolically:
P(A)= P(A/B)
[tex]P(A/B)= \frac{P(AnB)}{P(B)}= \frac{0}{1/4}= 0[/tex]
P(A)= 3/8
P(A) ≠ P(A/B) ⇒ A and B are not independent.
c.
i. P(D) ⇒ Considering C and D are two independent events, then we know that P(C∩D)= P(C)*P(D)
Then you can clear the probability of D as:
P(D)= P(C∩D)/P(C)= (1/3)/(1/2)= 2/3
ii. If Dc is the complementary event of D, then its probability is P(Dc)= 1 - P(D) = 1 - 2/3= 1/3
P(C∩Dc)= P(C)*P(Dc)= (1/2)*(1/3)= 1/6
iii. Now Cc is the complementary event of C, its probability is P(Cc)= 1 - P(C)= 1 - 1/2= 1/2
P(Cc∩Dc)= P(Cc)*P(Dc)= (1/2)*(1/3)= 1/6
vi. and e.
[tex]P(C/D)= \frac{P(CnD)}{P(D)} = \frac{1/3}{2/3} = 1/2[/tex]
P(C)=1/2
As you can see the P(C)=P(C/D) ⇒ This fact proves that the events C and D are independent.
d.
i. P(C∪D)= P(C) + P(D) - P(C∩D)= 1/2 + 2/3 - 1/3= 5/6
ii. P(C∪Dc)= P(C) + P(Dc) - P(C∩Dc)= 1/2 + 1/3 - 1/6= 2/3
I hope it helps!
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 .
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]
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%.
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:
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.
A student needs 11 more classes to graduate. If she has met the prerequisites for all the classes, how many possible schedules for next semester could she make if she plans to take 3 classes?
Using the combination formula, the student can make 165 different possible schedules for next semester if she is planning to take three classes from the eleven available that she has met the prerequisites for.
Explanation:This problem is a combination problem where students need to select 3 classes from 11 available classes to plan her schedule. The combination formula is used when the order of election does not matter. Here, the formula to find a combination is written as C(n, r) = n! / [r!(n - r)!].
In this particular case, the student has 11 classes (n=11) and plans to take 3 classes (r=3). So, the combination becomes C(11,3) = 11! / [3!(11-3)!] = (11*10*9) / (3*2*1) = 165.
This means, the student can make 165 different possible schedules for next semester assuming she is planning to take three classes from the eleven available classes.
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The student could make 165 different schedules.
The formula for combinations is given by:
[tex]\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \][/tex]
where n! (n factorial) is the product of all positive integers up to n, and r is the number of items to choose.
Given that the student has 11 classes left and she plans to take 3 classes next semester, we substitute n = 11 and r = 3 into the combination formula:
[tex]\[ \binom{11}{3} = \frac{11!}{3!(11-3)!} \][/tex]
Now, we simplify the factorial expressions:
[tex]\[ \binom{11}{3} = \frac{11!}{3! \times 8!} \][/tex]
We can cancel out the common terms in the numerator and the denominator:
[tex]\[ \binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} \][/tex]
Now, we perform the arithmetic:
[tex]\[ \binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = \frac{990}{6} = 165 \][/tex]
Thus, the student has 165 different possible schedules for next semester if she is to take 3 classes out of the remaining 11.
Stay in school: In a recent school year in the state of Washington, there were 315000 high school students. Of these, were girls and were boys. Among the girls, dropped out of school, and among the boys, dropped out. A student is chosen at random. Round the answers to four decimal places. (a) What is the probability that the student is female? (b) What is the probability that the student dropped out? (c) What is the probability that the student is female and dropped out? (d) Given that the student is female, what is the probability that she dropped out? (e) Given that the student dropped out, what is the probability that the student is female? Part: 0 / 50 of 5 Parts Complete
Answer:
Step-by-step explanation:
Hello!
Since the data on the text is missing, I've found a similar exercise with the same questions so I'm going to use the data of that one to explain how to calculate the probabilities.
We have a sample of 323000 high school students
154000 are girls, of those 47700 dropped out
16900 are boys, of those 10300 dropped out
a) To calculate this probability you have to divide the total number of girls by the total number of students:
P(G)= 154000/323000= 0.476 ≅ 0.48
b) In this item you have to calculate the probability of dropouts, the total of students that dropped out are the girls + boys that dropped out:
Dropouts: 47700+10300= 58000
Now you divide it by the total of students to reach the probability:
P(D)= 58000/323000= 0.179≅ 0.18
c) Now you have to calculate the probability of the intersection between the events "female" and "dropped out", symbolically:
P(G∩D)= P(G)*P(D)= 0.48*0.18= 0.0864
d) The probability of the student dropping out given that it is female is a conditional probability and you can calculate using the following formula:
P(D/G)= P(G∩D) = 0.0864 = 0.18
P(G) 0.48
e) Now you have to calculate the probability of the student being a femal, given that she dropped out of school, symbolically:
P(G/D)= P(G∩D) = 0.0864 = 0.48
P(D) 0.88
Note:
As you can see in d) P(D/G)=P(D)=0.18 and in e) P(G/D)=P(G)=0.48, this means that both events "the student is a girl" and "the student dropped out" are independent.
Remember two events are not independent when the occurrence of one modifies the probability of occurrence of the other.
I hope it helps!
Consider the following. (If an answer does not exist, enter DNE.) f '(x) = x2 + x − 20 (a) Find the open intervals on which f '(x) is increasing or decreasing.
Answer:
f'(x) is increasing in the open interval (-1/2, +∞), and it is decreasing in the interval (-∞, -1/2).
Step-by-step explanation:
f'(x) is a quadratic function with positive main coefficient, then it will be decreasing until the x-coordinate of its vertex and it will be increasing from there onwards. The x-coordinate of the vertex is given by the equation -b/2a = -1/2. Hence
- f'(x) is incresing in the interval (-1/2, +∞)
- f'(x) is decreasing in the interval (-∞, -1/2)
Final answer:
To find intervals where f'(x) is increasing or decreasing, locate the critical points by solving f'(x) = 0 and test the signs of f'(x) around these points. This process determines where the original function f(x) is increasing or decreasing.
Explanation:
To determine the open intervals in which the derivative of a function f(x), denoted as f'(x), is increasing or decreasing, one must:
Find all solutions of f'(x) = 0 within the interval [a, b]. These points are known as critical or stationary points.Examine the sign of f'(x) at points other than the critical points to determine the intervals where f'(x) is > 0 (indicating increasing behavior of f(x)) and where f'(x) is < 0 (indicating decreasing behavior).Identify the changes in the sign of f'(x) at the critical points to confirm if those points are local maxima or minima. The function f(x) will be evaluated at each of these stations to determine local extrema.In the scenario one would solve the equation for f'(x) = 0 to find the critical points. Once these are found, the signs of f'(x) to the left and right of these points can be tested to see where the function is increasing or always decreasing.
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).
A similar problem is given at https://brainly.com/question/15872108
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
(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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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.
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! :)