The length of the angle bisector of angle ∠A is approximately 4.62 feet.
To find the length of the angle bisector of angle ∠A in triangle ΔABC, we can use the Angle Bisector Theorem, which states that in a triangle, the angle bisector of a vertex divides the opposite side into segments proportional to the adjacent sides.
In triangle ΔABC, let AD be the angle bisector of ∠A, where D lies on BC. According to the Angle Bisector Theorem:
AC/CD = AB/BD
Given AC = 5 ft and AB = 13 ft, we can plug in these values:
5/CD = 13/BD
To find BD, we use the Pythagorean theorem:
BD = √(AB² - AD²) = √(13² - 5²) = √(169 - 25) = √144 = 12 ft
Now, using the Angle Bisector Theorem:
5/CD = 13/12
Cross-multiply:
5 × 12 = 13 × CD
CD = (5 × 12) / 13 = 60 / 13 ≈ 4.62 ft
Find the solution of the system of equations 3x+4y=10 and x−y=1. Give the x value followed by the y value, separated by a comma
Answer:
Solution of the system is (2,1).
Step-by-step explanation:
We are given the following system of equation:
[tex]3x+4y=10\\x - y = 1[/tex]
We would use the elimination method to solve the following system of equation.
Multiplying the second equation by 4 and adding the two equation we gwt:
[tex]3x + 4y = 10\\4\times (x-y = 1)\\\Rightarrow 4x - 4y = 4\\\text{Adding equations}\\3x + 4y + (4x-4y) = 10 + 4\\7x = 14\\\Rightarrow x = 2\\\text{Substituting value of x in second equation}\\2 - y = 1\\\Rightarrow y = 1[/tex]
Solution of the system is (2,1).
Answer: the solution is (2, 1)
Step-by-step explanation:
The given system of simultaneous equations is given as
3x+4y=10 - - - - - - - - - - - - -1
x−y=1 - - - - - - - - - - - - 2
We would eliminate x by multiplying equation 1 by 1 an equation 2 by 3. It becomes
3x + 4y = 10
3x - 3y = 3
Subtracting, it becomes
7y = 7
Dividing the left hand side and the right hand side of the equation by 7, it becomes
7y/7 = 7/7
y = 1
Substituting y = 1 into equation 2, it becomes
x - 1 = 1
Adding 1 to the left hand side and the right hand side of the equation, it becomes
x - 1 + 1= 1 + 1
x = 2
Let S = {2,4,6} and T = {1,3,5}. Use the set-roster notation to write each of the following sets, and indicate the number of elements that are in each set?
a. S X T
b. T X S
c. S X S
d. T X T
Answer:
a. [tex]S\times T=\{(2,1),(2,3),(2,5),(4,1),(4,3),(4,5),(6,1),(6,3),(6,5)\}[/tex] . Set S × T has 9 elements.
b. [tex]T\times S=\{(1,2),(1,4),(1,6),(3,2),(3,4),(3,6),(5,2),(5,4),(5,6)\}[/tex]. Set T × S has 9 elements.
c. [tex]S\times S=\{(2,2),(2,4),(2,6),(4,2),(4,4),(4,6),(6,2),(6,4),(6,6)\}[/tex] . Set S × S has 9 elements.
d. [tex]T\times T=\{(1,1),(1,3),(1,5),(3,1),(3,3),(3,5),(5,1),(5,3),(5,5)\}[/tex] . Set T × T has 9 elements.
Step-by-step explanation:
The given sets are S={2,4,6} and T={1,3,5}.
We need to find the set-roster notation to write each of the following sets, and the number of elements that are in each set.
a.
[tex]S\times T=\{(s,t)|s\in S and t\in T\}[/tex]
[tex]S\times T=\{(2,1),(2,3),(2,5),(4,1),(4,3),(4,5),(6,1),(6,3),(6,5)\}[/tex]
Set S × T has 9 elements.
b.
[tex]T\times S=\{(t,s)|s\in S and t\in T\}[/tex]
[tex]T\times S=\{(1,2),(1,4),(1,6),(3,2),(3,4),(3,6),(5,2),(5,4),(5,6)\}[/tex]
Set T × S has 9 elements.
c.
[tex]S\times S=\{(s,s)|s\in S \}[/tex]
[tex]S\times S=\{(2,2),(2,4),(2,6),(4,2),(4,4),(4,6),(6,2),(6,4),(6,6)\}[/tex]
Set S × S has 9 elements.
d.
[tex]T\times T=\{(t,t)|t\in T\}[/tex]
[tex]T\times T=\{(1,1),(1,3),(1,5),(3,1),(3,3),(3,5),(5,1),(5,3),(5,5)\}[/tex]
Set T × T has 9 elements.
Using the set roster notation, the number of elements in each of the sets would be the product of the number of elements in each individual set, which is 9.
S = {2, 4, 6}T = {1, 3, 5}A.) S × T :
{2, 4, 6} × {1, 3, 5}
{(2, 1), (2, 3), (2, 5), (4, 1), (4, 3), (4, 5), (6, 1), (6, 3), (6, 5)}
Number is of elements = 9
B.) T × S :
{1, 3, 5} × {2, 4, 6}
{(1,2), (1,4), (1,6), (3,2), (3,4), (3, 6), (5, 2), (5,4), (5,6)}
Number is of elements = 9
C.) S × S :
{1, 3, 5} × {1, 3, 5}
{(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 3), (5, 1), (5, 3), (5, 5)}
Number is of elements = 9
D.) T × T :
{2, 4,6} × {2, 4, 6}
{(2,2), (2, 4), (2, 6), (4,2), (4, 4), (4, 6), (6,2), (6, 4), (6,6)}
Number is of elements = 9
Hence, the distribution of values of each set.
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Without solving the equation 3x2−12x+7=0 find the sum of the squares of its roots.
Answer:
Step-by-step explanation:
3x²-12x+7=0
x = (-b ± √ b²-4ac) / 2a
from these equation: a = 3 , b = 12 , c = 7
Accordingly, b² - 4ac = 12²- 4 (3 x 7) = 144 - 4(21) = 144 - 84 = 60
2a= 2 x 3 = 6
Applying the quadratic formula :
x = (-12 ± √ 60 )/6
√ 60 rounded to 2 decimal digit using calculator 7.75
x = (-12 ± 7.75 )/6
(-12 ± 7.75 ) = (-12 + 7.75 ) = 5.75
(-12 - 7.75 ) = -19.75
x = 5.75/6 = 0.96
-19.75/6 = 3.29
sum of the square of roots = 0.96² + 3.29²
0.9261 + 10.8241 = 11.75
The sum of the squares of its roots of the quadratic equation
3x² - 12x + 7 = 0 is p² + q² = 34/3.
What is a quadratic equaton?A quadratic equation is an algebraic expression in the form of variables and constants.
A quadratic equation has two roots as its degree is two.
We have a quadratic equation 3x² - 12x + 7 = 0.
We know a quadratic equation ax² - bx + c = 0 can be written as,
x² - (b/a)x + (c/a) = 0 ⇒ x² - (p + q)x + pq = 0 where p and q are the roots.
∴ 3x² - 12x + 7 = 0.
x² - (12/3)x + 7/3 = 0.
Hence (p + q) = 4 and pq = 7/3.
Now, (p + q)² = p² + q² + 2pq.
16 = p² + q² + 14/3.
p² + q² = 16 - 14/3.
p² + q² = (48 - 14)/3.
p² + q² = 34/3.
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Let X and Y be independent random variables representing the lifetime (in 100 hours) of Type A and Type B light bulbs, respectively.
Both variables have exponential distributions, and the mean of X is 2 and the mean of Y is 3.
Find the joint pdf f(x|y) of X and Y.
Find the conditional pdf f_2(y|x) of Y given X = x.
Find the probability that a Type A bulb lasts at least 300 hours and a Type B bulb lasts at least 400 hours.
Given that a Type B bulb fails at 300 hours, find the probability that a Type A bulb lasts longer than 300 hours.
What is the expected total lifetime of two Type A bulbs and one Type B bulb?
What is the variance of the total lifetime of two Type A bulbs and one Type B bulb?
Answer:
a. is below in the explanation
b. 1/3e^(-y/3)
c.0.05882
d. 0.2231
e. 7 f.25
Step-by-step explanation:
Let and be independent random variables representing the lifetime (in 100 hours) of Type A and Type B light bulbs, respectively. Both variables have exponential distributions, and the mean of X is 2 and the mean of Y is 3.
a can be solved as follows [tex]\frac{1}{\alpha\beta } e^{\frac{-x-y}{\alpha\beta } }[/tex]
[tex]\frac{1}{6} e^{\frac{-\alpha }{2} -\beta/3 }[/tex]
1/6[tex]e^{\frac{-(3x+2y}{6 }[/tex]
b.
f(y/x)=f(x).f(y)/{f(y)}=f(y)
1/3e^(-y/3)
c. Find the probability that a Type A bulb lasts at least 300 hours and a Type B bulb lasts at least 400 hours.
[tex]\int\limits^\alpha _4 {1/6e^{-(3x+2y)/6} } \, dy[/tex]
1/2e^-(3x+8)/6
[tex]\int\limits^\alpha _3 {e^{-(3x+800)} } \, dx \\[/tex]
0.05882
Given that a type B fails at 300 hours . we find the probability that type A bulb lasts longer than 300hr
f(x>y|y=3)=f(x>3)
[tex]\int\limits^a_3 {1/2e^{-x/2} } \, dx[/tex]
0.2231
e.e) What is the expected total lifetime of two Type A bulbs and one Type B bulb?
E(2A+B)
(2E(A)+E(y)
2*2+3
=7
f. variance of the total lifetime of two types A bulbs and one type B bulb
V(2x+y)=
4*4+9
=25
The joint pdf of X and Y is (1/6)e^(-x/2-y/3). The conditional pdf of Y given X = x is (1/2)e^(-y/3). The probability that a Type A bulb lasts at least 300 hours and a Type B bulb lasts at least 400 hours is 9/250. The probability that a Type A bulb lasts longer than 300 hours, given that a Type B bulb fails at 300 hours, is approximately 0.9999. The expected total lifetime of two Type A bulbs and one Type B bulb is 7 100 hours. The variance of the total lifetime of two Type A bulbs and one Type B bulb is 10 100 hours.
Explanation:To find the joint pdf f(x|y) of X and Y, we need to find the product of the individual pdfs of X and Y since they are independent. Since X and Y follow exponential distributions with means of 2 and 3 respectively, their pdfs can be written as f(x) = (1/2)e^(-x/2) and f(y) = (1/3)e^(-y/3). Therefore, the joint pdf f(x|y) is given by:
f(x|y) = f(x)*f(y) = (1/2)e^(-x/2) * (1/3)e^(-y/3) = (1/6)e^(-x/2-y/3).
To find the conditional pdf f_2(y|x) of Y given X = x, we use Bayes' theorem:
f_2(y|x) = (f(x|y)*f(y))/f(x) = [(1/6)e^(-x/2-y/3) * (1/3)e^(-y/3)] / [(1/2)e^(-x/2)] = (1/2)e^(-y/3).
To find the probability that a Type A bulb lasts at least 300 hours and a Type B bulb lasts at least 400 hours, we integrate the joint pdf over the given range:
P(X >= 300, Y >= 400) = ∫∫(1/6)e^(-x/2-y/3)dxdy = ∫(3/2)e^(-x/2)dx * ∫e^(-y/3)dy. Solving the integrals, we get P(X >= 300, Y >= 400) = 9/250.
Given that a Type B bulb fails at 300 hours, we need to find the probability that a Type A bulb lasts longer than 300 hours. This is equivalent to finding P(X > 300 | Y = 300). Using the conditional pdf f_2(y|x) = (1/2)e^(-y/3), we integrate from 300 to infinity:
P(X > 300 | Y = 300) = ∫(1/2)e^(-y/3)dy = 1 - ∫(1/2)e^(-y/3)dy = 1 - (1/2)e^(-y/3) = 1 - (1/2)e^(-300/3) = 1 - e^(-100) ≈ 0.9999.
The expected total lifetime of two Type A bulbs and one Type B bulb is obtained by summing the means of X and Y twice and the mean of Y once:
E[Total Lifetime] = 2*E[X] + E[Y] = 2*2 + 3 = 7.
The variance of the total lifetime of two Type A bulbs and one Type B bulb is obtained by summing the variances of X and Y twice and the variance of Y once:
Var[Total Lifetime] = 2*Var[X] + Var[Y] = 2*(2^2) + 3^2 = 10.
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If 400 mL of a 20% w/v solution were diluted to 2 L, what would be the final percentage strength?
Answer:
4%
Step-by-step explanation:
A 20% solution with a total volume of 400 mL has 20% * 400 mL of solute.
20% * 400 mL = 80 mL
When you dilute the solution to 2 L, you introduce additional water, but no additional solute, so now you have the same 80 mL of solute in 2 L of total solution.
The concentration is:
(80 mL)/(2 L) = (80 mL)/(2000 mL) = 0.04
As a percent it is:
0.04 * 100% = 4%
"4" will be the final percentage strength.
According to the question,
20% solution with 400 mL has 20% × 400 mL of solute
then,
→ [tex]20 \ percent\times 400=80[/tex]
hence,
The concentration will be:
→ [tex]\frac{80 \ mL}{2 \ L} = \frac{80 \ mL}{2000 \ mL}[/tex]
[tex]= 0.04[/tex]
or,
[tex]= 0.04\times 100[/tex]
[tex]= 4[/tex] (%)
Thus the answer above is right.
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(Anderson, 1.14) Assume that P(A) = 0.4 and P(B) = 0.7. Making no further assumptions on A and B, show that P(A ∩ B) satisfies 0.1 ≤ P(A ∩ B) ≤ 0.
Answer with Step-by-step explanation:
We are given that
P(A)=0.4 and P(B)=0.7
We know that
[tex]P(A)+P(B)+P(A\cap B)=P(A\cup B)[/tex]
We know that
Maximum value of [tex]P(A\cup B)[/tex]=1 and minimum value of [tex]P(A\cup B)[/tex]=0
[tex]0\leq P(A\cup B )\leq 1[/tex]
[tex]0\leq P(A)+P(B)-P(A\cap B)\leq 1[/tex]
[tex]0\leq 0.4+0.7-P(A\cap B)\leq 1[/tex]
[tex]0\leq 1.1-P(A\cap B)\leq 1[/tex]
[tex]0\leq 1.1-P(A\cap B)[/tex]
[tex]P(A\cap B)\leq 1.1[/tex]
It is not possible that [tex]P(A\cap B)[/tex] is equal to 1.1
[tex]1.1-P(A\cap B)\leq 1[/tex]
[tex]-P(A\cap B)\leq 1-1.1=-0.1[/tex]
Multiply by (-1) on both sides
[tex]P(A\cap B)\geq 0.1[/tex]
Again, [tex]P(A\cup B)\geq P(B)[/tex]
[tex]0.4+0.7-P(A\cap B)\geq 0.7[/tex]
[tex]1.1-P(A\cap B)\geq 0.7[/tex]
[tex]-P(A\cap B)\geq -1.1+0.7=-0.4[/tex]
Multiply by (-1) on both sides
[tex]P(A\cap B)\leq 0.4[/tex]
Hence, [tex]0.1\leq P(A\cap B)\leq 0.4[/tex]
Solve the right triangle
SR =sin (angle) = opposite leg/ hypotenuse
sin(52) = SR/ 7.6
SR = 7.6 x sin(52)
SR = 5.988 ( round answer as needed.)
QR = cos(angle) = adjacent leg / hypotenuse
Cos(52) = QR/ 7.6
QR = 7.6 x cos(52)
QR = 4.679. ( round answer as needed).
Angle S = 180 - 90 -52 = 38 degrees
What is the simple interest on a loan of $16 500 for 2 years at 8 per year ?
Answer:
$2 640
Step-by-step explanation:
Principal = $16 500; Time = 2 years Rate = 8%
Simple Interest = P × R × T/100
= $16 500 × 8 × 2/100
=$264 000/100
Simple Interest = $2 640
An article reported the following data on oxygen consumption (mL/kg/min) for a sample of ten firefighters performing a fire-suppression simulation: 29.6 49.3 30.1 28.3 28.7 26.3 33.5 29.8 23.7 31.0 Compute the following. (Round your answers to four decimal places.)
(a) the sample range(b) the sample variance s2 from the definition(c) the sample standard deviation(d) s2 using the shortcut method
Answer:
a) Range = 25.6
b) Variance = 48.1446
c) s = 6.9386
d) See below
Step-by-step explanation:
(a) the sample range
To obtain the sample range, we must sort the data increasingly :
23.7, 26.3, 28.3, 28.7, 29.6, 29.8, 30.1, 31, 33.5, 49.3
Then, find the distance between the largest and the lowest
Range = 49.3 - 23.7 = 25.6
(b) the sample variance [tex]s^2[/tex] from the definition
In order to find the variance from the definition, we need first the mean. The mean is defined as the average
[tex]\bar x=\displaystyle\frac{\displaystyle\sum_{i=1}^{n}x_i}{n}[/tex]
where the [tex]x_i[/tex] are the values of the data collected and n=10 the size of the sample.
So, the mean is
[tex]\bar x=31.03[/tex]
Now, the variance of the sample is defined as
[tex]s^2=\displaystyle\frac{\displaystyle\sum_{i=1}^n(x_i-\bar x)^2}{n-1}[/tex]
and we have that the variance is
[tex]s^2=48.1446[/tex]
(c) the sample standard deviation
The sample standard deviation is nothing but the square root of the variance
[tex]s=\sqrt{48.1446}=6.9386[/tex]
d) [tex]s^2[/tex] using the shortcut method
The shortcut method figures out the variance without having to compute the mean. The formula is
[tex]s^2=\frac{n\sum x_i^2-(\sum x_i)^2}{n(n-1)}[/tex]
where n=10 is the sample size .
So, using the shortcut method,
[tex]s^2=\frac{10*10,061.91-96,286.09}{10*9}=48.1446[/tex]
A prize was awarded to 56 women and 642 men.
a. What fraction of the prize winners were women?
b. What fraction were men?
SIMPLIFY EACH plz
New drug strategies have centered on cholesterol ______ statins. Select one:
a. lowering b. raising c. destroying d. improving
Answer: lowering
Step-by-step explanation:
Statins are drugs that are administered in order to help reduce cholesterol level in the body. Statins achieves this by making it impossible for the body to have access to substances that helps to produce cholesterol. This drug reduces cholesterol up to 50% and are taken once every day. Examples includes rosuvastatin, pitavastatin, simvastatin.
Sarah inherited a sum of 1,539 dollars with an interest rate of 1% per year what is the accrued value after 5 years, 10 years, and 15 years. Show all work.
Answer:
The accrued value after 5 years is $1,605.95.
The accrued value after 10 years is $1,672.9.
The accrued value after 15 years is $1,739.85.
Step-by-step explanation:
This is a simple interest problem.
The simple interest formula is given by:
[tex]E = P*I*t[/tex]
In which E are the earnings, P is the principal(the initial amount of money), I is the interest rate(yearly, as a decimal) and t is the time.
After t years, the total amount of money is:
[tex]T = E + P[/tex].
In this problem, we have that:
[tex]P = 1539, I = 0.01[/tex]
Accrued value after 5 years
This T when t = 5. So
[tex]E = P*I*t[/tex]
[tex]E = 1339*0.01*5 = 66.95[/tex]
The total is
[tex]T = E + P = 66.95 + 1539 = 1605.95[/tex]
The accrued value after 5 years is $1,605.95.
Accrued value after 10 years
This T when t = 10. So
[tex]E = P*I*t[/tex]
[tex]E = 1339*0.01*10 = 133.9[/tex]
The total is
[tex]T = E + P = 133.9 + 1539 = 1672.9[/tex]
The accrued value after 10 years is $1,672.9.
Accrued value after 15 years
This T when t = 15. So
[tex]E = P*I*t[/tex]
[tex]E = 1339*0.01*15 = 200.85[/tex]
The total is
[tex]T = E + P = 200.85 + 1539 = 1739.85[/tex]
The accrued value after 15 years is $1,739.85.
What is 4^150 divided by 2^150?
Find the Surface area
Answer:
27m²
Step-by-step explanation:
surface area is area of all sides
these is two triangles and a rectangle
area of triangle = (base x height) /2 = (3 x 4)/2= 12/2 =6
area of rectangle = length x breadth = 5x3 = 15
surface area = area of triangle + area of triangle + area of rectangle
6 + 6 + 15 = 27m²
Find values for θ that make each statement true (show steps)
1) sinθ = cos(25)
θ =
2) sin(θ/3 + 10 ) = cos θ
θ =
Answer:
Step-by-step explanation:
1. sinФ = cos 25
25 ° is in the between 0 and 90°
therefore it can simply represent
cosФ= sin (90-Ф) = sin (90 - 25) = sin 65
2. sin(Ф/3 + 10) = cos Ф
cos Ф = sin (90 -Ф)
sin(Ф/3 + 10) = cos Ф
sin(Ф/3 + 10) = cos Ф = sin(90-Ф)
Ф/3+10=90-Ф
10Ф+3/30 = 90-Ф
10Ф+3 = 30(90-Ф)
10Ф+3 = 2700-30Ф
10Ф+30Ф=2700-3
40Ф = 2697
Ф = 2697 / 40 = 67.425 ≅ 67.4°
uppose that a certain fortunate person has a net worth of $71.0 billion ($7.10×1010). If her stock has a good year and gains $3.20 billion (3.20×109) in value, what is her new net worth
The new net worth after a gain in stock value is $74.2 billion, or in the scientific notation it is $7.42×10¹0.
Explanation:In this problem, we have a person with a starting net worth of $71.0 billion, or $7.10×1010 in scientific notation. This person's stock has increased by $3.20 billion, or 3.20×109 in scientific notation. To find the new net worth, these two amounts should be added together.
Step 1: Write the starting net worth and stock increase in standard form: $71.0×10⁹ + $3.2×10⁹. Step 2: Add these two amounts together to find the new net worth. $74.2×10⁹ This is the new net worth in standard form.
To express this amount in scientific notation, it becomes $7.42×10¹0, which is your final answer.
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The cost, c, of a ham sandwich at a deli varies directly with the number of sandwiches, n. If c = $54 when n is 9, what is the
cost of the sandwiches when n is 3?
$18
$21
$27
$48
Mark this and return
Next
Submit
Answer:
$18
Step-by-step explanation:
The cost, c, of a ham sandwich at a deli varies directly with the number of sandwiches, n. c = $54 when n is 9
This means that to buy 9 sandwiches, it costs $54. So one sandwich costs 54/9 = $6.
What is the cost of the sandwiches when n is 3?
To buy three sandwiches, it costs $6*3 = $18.
So the correct answer is:
$18
Answer:A
Step-by-step explanation:
$18
At a certain college, 30% of the students major in engineering, 20% play club sports, and 10% both major in engineering and play club sports. A student is selected at random. What is the probability that the student is majoring in engineering?
The probability that a student selected at random majors in engineering is 30% which is 0.3.
The probability that the student both majors in engineering and play club sports is 10% which is 0.1.
For a student who is selected at random to be one who majors in engineering, there are two possible ways.
The student majors in engineering OR the Student both majors in engineering and plays club
The Probabilty that the student majors in Enginnering =
The probability that the student majors in engineering plus the probability that Student both majors in engineering and plays club sport
= 0.3+0.1= 0.4
There is a mound of g pounds of gravel in a quarry. Throughout the day, 300 pounds of gravel are added to the mound. Two orders of 800 pounds are sold and the gravel is removed from the mound. At the end of the day, the mound has 1,500 pounds of gravel. Write the equation that describes the situation. Then solve for g
Final answer:
The equation that describes the situation is g + 300 - 1600 = 1500. Simplifying this equation gives g = 2800.
Explanation:
To write the equation that describes the situation, we start with the initial weight of the mound as g pounds. Throughout the day, 300 pounds are added to the mound, so the new weight is g + 300 pounds. Two orders of 800 pounds each are sold, so 1600 pounds are removed from the mound. At the end of the day, the weight of the mound is 1500 pounds. The equation that describes the situation is:
g + 300 - 1600 = 1500
Simplifying this equation, we have:
g - 1300 = 1500
Adding 1300 to both sides, the equation becomes:
g = 2800
The table below shows selected points from a function.
The rate of change for the interval shown in the table is
A. constant
B. not constant
part b
A. linear
B. not linear
The rate of change is A. constant, B. linear
Step-by-step explanation:
Rate of change is the ratio of change in value of y with corresponding value of x
Rate of change=Δy/Δx
Rate of change=5-2/4-1 =3/3=1
The rate of change is constant in the given interval and linear with a positive slope.
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math answers- the length of segment AB is 3.6. the length of segment AC is 11.8. what is the length of segment BC? write and solve a decimal addition equation and a decimal subtraction equation.
Answer:
answer is AC^2
Step-by-step explanation:
Charlie is a car salesman. He earns $800 plus a 9% commission on all of his sales for the week. If Charlie received a paycheck for 1,916 this week, how much were his total sales?
Answer:
$12,400
Step-by-step explanation:
Given:
Week earning + 9% commission = $1,916
Weekly Earning = $800
Commission for the Week = $1,916 - $800 = $1,116
Let total sales for the week = x
Therefore, x = ($1,116 * 100)/9
x = $12,400
Answer:
$12400
Step-by-step explanation:
Wages = Basic pay + Commission
1916 = 800 + 9%A
Let A be the total sales
9%A = 1916 - 800
9%A = 1116
A = $12400
What is the difference between relative frequency and cumulative frequency?
A. Relative frequency of a class is the percentage of the data that falls in that class, while cumulative frequency of a class is the sum of the frequencies of that class and all previous classes.
B. Relative frequency of a class is the proportion of the data in that class, while cumulative frequency of a dass is the number of observations in that class.
C. Relative frequency of a class is the number of observations in that dass, while cumulative frequency of a dass is the sum of all the frequencies.
D. There is no difference between the two.
Answer:
A
Step-by-step explanation:
The relative frequency is calculated by dividing the frequency of that class to the sum of frequencies. It can be represented as
[tex]Relative frequency=\frac{f}{sum(f)}[/tex]
Hence, the relative frequency of class is the percentage or proportion of data lies in that class.
The cumulative frequency of a class is computed by adding the frequency of the respective class to the frequencies of all previous classes. The cumulative frequency of first class will always be equal to the frequency of first class.
Hence, the cumulative frequency for a class is the sum of frequency for that class and the frequencies of all previous classes.
Relative frequency is the proportion of total data that falls into a class, while cumulative frequency is the sum of frequencies of that class and all previous classes. Therefore, relative frequency illustrates the percentage of data in a certain class, while cumulative frequency shows the accumulation of data up to that point.
Explanation:The terms relative frequency and cumulative frequency both pertain to statistics; however, they represent different concepts. Relative frequency of a class is the percentage or proportion of the whole set of data that falls in that class. It's calculated by dividing the frequency of that class by the total number of data points.
On the other hand, cumulative frequency of a class is the sum of the frequencies of that class and all previous classes in a dataset. Essentially, it accumulates counts as you proceed through the dataset.
For instance, if you have five classes with frequencies of 1, 2, 3, 4 and 5, the relative frequencies would be 0.0667, 0.1333, 0.2, 0.2667, and 0.3333 respectively (assuming we divide each frequency by the total data points, which is 15 in this case). In contrast, the cumulative frequencies would be 1, 3, 6, 10 and 15, indicating the total count up to each class.
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Wich values from the given replacement set make up the solution set of the inequality
Answer:
ljgiuipuiopu
Step-by-step explanation:
When 258 college students are randomly selected and surveyed, it is found that 106 own a car. Find a 99% confidence interval for the true proportion of all college students who own a car. 0.351 < p < 0.471 0.332 < p < 0.490 0.339 < p < 0.482 0.360 < p < 0.461
Answer: 0.332 < p < 0.490
Step-by-step explanation:
We know that the confidence interval for population proportion is given by :-
[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
, where n= sample size
[tex]\hat{p}[/tex] = sample proportion
z* = critical z-value.
As per given , we have
n= 258
Sample proportion of college students who own a car = [tex]\hat{p}=\dfrac{106}{258}\approx0.411[/tex]
Critical z-value for 99% confidence interval is 2.576. (By z-table)
Therefore , the 99% confidence interval for the true proportion(p) of all college students who own a car will be :[tex]0.411\pm (2.576)\sqrt{\dfrac{0.411(1-0.411)}{258}}\\\\=0.411\pm (2.576)\sqrt{0.00093829}\\\\= 0.411\pm (2.576)(0.0306315197142)\\\\=0.411\pm 0.0789=(0.411-0.0789,\ 0.411+0.0789)\\\\=(0.3321,\ 0.4899)\approx(0.332,\ 0.490)[/tex]
Hence, a 99% confidence interval for the true proportion of all college students who own a car : 0.332 < p < 0.490
The sales of digital cameras (in millions of units) in year t is given by the function
f(t) = 3.06t + 6.84 (0 ≤ t ≤ 3)
where t = 0 corresponds to the end of 2001. Over that same period, the sales of film cameras (in millions of units) is given by
g(t) = −1.85t + 16.48 (0 ≤ t ≤ 3).
(a) Show that more film cameras than digital cameras were sold in 2001.
digital cameras
million
film cameras
million
(b) When did the sales of digital cameras first exceed those of film cameras?
Explanation on determining which type of cameras sold more in 2001 and when digital camera sales surpassed film camera sales.
Explanation:(a) To show that more film cameras than digital cameras were sold in 2001, we need to substitute t = 0 into both functions:
For digital cameras: f(0) = 3.06(0) + 6.84 = 6.84 million units
For film cameras: g(0) = -1.85(0) + 16.48 = 16.48 million units
Therefore, more film cameras (16.48 million units) were sold in 2001 than digital cameras (6.84 million units).
(b) To find when digital camera sales first exceeded film camera sales, we need to set the two functions equal to each other and solve for t:
3.06t + 6.84 = -1.85t + 16.48
4.91t = 9.64
t ≈ 1.96 years, which is approximately at the end of 2003.
A sample is a subset of all possible data values for a given subject under consideration.
(A) True
(B) False
Answer: True
Step-by-step explanation:
A subset is a term used in the sets topic In mathematics to describe a Part of A larger values being considered.
Sample In experimental procedures is known as Part of what is to be used to determine the reaction of a larger mass of specimen.
Taking these two definitions by side, it is quite evident that both of them represent taking part of a larger picture to form a smaller picture which does not totally deviate from the larger picture being considered.
Cooley et al. (2009) randomly assigned either of two treatments, naturopathic care (diet counseling, breathing techniques, vitamins, and a herbal medicine) or stan- dardized psychotherapy (psychotherapy with breathing techniques and a placebo added), to 81 individuals having moderate to severe anxiety. Anxiety scores decreased an average of 57% in the naturopathic group and 31% in the psychotherapy group. Is this an experimental or observational study?
Answer: Experimental study
Step-by-step explanation:
This is an experimental study because it involves determination of the effect of each treatment on separate individuals. It involves selecting the 81 individuals randomly to be treated with a particular treatment (either of the two treatment). In experimental study the researcher apply separate treatments on a different groups and measure its effect on each of the groups.
This is an experimental study where participants were randomly assigned to either naturopathic care or standardized psychotherapy to assess their effects on anxiety scores.
Explanation:This is an example of an experimental study. In an experimental study, researchers assign participants to different groups and manipulate variables to determine cause-and-effect relationships. In this case, Cooley et al. randomly assigned individuals to either naturopathic care or standardized psychotherapy, which are the two treatments being compared. They then measured the effects of these treatments on anxiety scores.
The fact that participants were randomly assigned to the two treatments suggests that this study was experimental rather than observational.
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The list of individuals from which a sample is actually selected is called the sampling frame. Ideally, the frame should list every individual in the population, but in practice this is often difficult. A frame that leaves out part of the population is a common source of undercoverage.
Suppose that a sample of households in a community is selected at random from the telephone directory.
What types of people do you think are likely to live in these households?These people will probably be underrepresented in the sample. (Select all that apply.)1.People with extra time on their hands.2.People without the extra income to keep a phone line will largely comprise the population without a phone line.3.Very busy people or those interested in maintaining privacy will largely comprise those with unlisted numbers.4.People who do not like to take surveys are likely to live in these households.
Answer:
Step-by-step explanation:
answer to Step 1
Households with unlisted numbers or without telephones.
answer to Step 2
2.People without the extra income to keep a phone line will largely comprise the population without a phone line.
3.Very busy people or those interested in maintaining privacy will largely comprise those with unlisted numbers.
USA Today reported that about 20% of all people in the United States are illiterate. Suppose you take eleven people at random off a city street. (a) Make a histogram showing the probability distribution of the number of illiterate people out of the eleven people in the sample.
Final answer:
To create a histogram showing the probability distribution of the number of illiterate people out of eleven people in a sample, you can use the binomial probability formula. Calculate the probability for each possible number of illiterate people and represent it in a histogram.
Explanation:
This question is asking for a histogram showing the probability distribution of the number of illiterate people out of eleven people in a sample, based on the information that about 20% of all people in the United States are illiterate.
To create the histogram, we can use the binomial probability formula. The formula is P(X = k) = C(n, k) * p^k * (1 - p)^(n - k), where C(n, k) represents combinations, n is the number of trials, k is the number of successes, and p is the probability of success.
We can calculate the probability for each possible number of illiterate people out of the eleven, ranging from 0 to 11, and represent it in a histogram.