Answer:
Volume of the cone is 1883.7 cm³
Step-by-step explanation:
The circumference of the full circle with radius 18 cm :
360 := 2*π*18 = 36π cm
125 := 125/360 * 36π
The new circumference is maller:
36π - 125/360 * 36π
36π * 0.652(7)
Calculate the new r based on the new circomference:
2*π * r = 36π * 0.652(7)
r = 36π/2π * 0.652(7)
r = 18 * 0.652(7)
r = 11.75 cm
Based on this radius you can calculate the area of the base of the cone.
area base = π*(11.75)²
The Volume V of this cone = 1/3 π r² * h
You can calculate the height h by using Pythagoras theorum.
The sector is the hypothenusa= 18 cm
The h is the height, which is the "unknown"
The r is the new radius = 11.75 cm
s² = r² + h²
h² = s² - r²
h = √(s² - r²)
h = √(18² - 11.75²)
h = 13.6358901432946 cm
h = 13.636 cm
V cone
V = 1/3 π 11.75² * h
V = 1/3 π 11.75² * √(18² - 11.75²)
V = 1/3 π 11.75² * 13.636
V = 1883.7 cm³
Number of multiples of 7 between 200 and 1000
Answer:
114
Step-by-step explanation:
Answer:
144Step-by-step explanation:
Any help would be great
Answer:
2/5
Step-by-step explanation:
What is the domain of the following set of ordered pairs (-2,-5),(-3,8),(12,6),(8,3),(4,0),(-5,7)
Answer:
domain = {-5, -3, -2, 4, 8, 12}
Step-by-step explanation:
The domain is the set containing the x-coordinates of all ordered pairs.
domain = {-2, -3, 12, 8, 4, -5}
If you'd like, you can put the numbers in ascending order:
domain = {-5, -3, -2, 4, 8, 12}
The mean number of hours of part-time work per week for a sample of 317 teenagers is 29. If the margin of error for the population mean with a 95% confidence interval is 2.1, construct a 95% confidence interval for the mean number of hours of part-time work per week for all teenagers.
Answer:
The degrees of freedom are given by:
[tex]df=n-1=317-1=316[/tex]
And replaicing we got:
[tex]29-2.1=26.9[/tex]
[tex]29+2.1=31.1[/tex]
The 95% confidence interval would be between 26.9 and 31.1
Step-by-step explanation:
Information given
[tex]\bar X= 29[/tex] represent the sample mean
[tex]\mu[/tex] population mean
s represent the sample standard deviation
[tex] ME= 2.1[/tex] represent the margin of error
n represent the sample size
Solution
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
And this formula is equivalent to:
[tex] \bar X \pm ME[/te]x
The degrees of freedom are given by:
[tex]df=n-1=317-1=316[/tex]
And replaicing we got:
[tex]29-2.1=26.9[/tex]
[tex]29+2.1=31.1[/tex]
The 95% confidence interval would be between 26.9 and 31.1
Pet Place sells pet food and supplies including a popular bailed hay for horses. When the stock of this hay drops to 20 bails, a replenishment order is placed. The store manager is concerned that sales are being lost due to stock outs while waiting for a replenishment order. It has been previously determined that demand during the lead-time is normally distributed with a mean of 15 bails and a standard deviation of 6 bails. The manager would like to know the probability of a stockout during replenishment lead-time. In other words, what is the probability that demand during lead-time will exceed 20 bails
Answer:
The probability that demand during lead-time will exceed 20 bails is 0.2033.
Step-by-step explanation:
We are given that it has been previously determined that demand during the lead-time is normally distributed with a mean of 15 bails and a standard deviation of 6 bails.
Let X = demand during the lead-time
So, X ~ Normal([tex]\mu=15, \sigma^{2} = 6^{2}[/tex])
The z-score probability distribution for the normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu=[/tex] population mean demand = 15 bails
[tex]\sigma[/tex] = standard deviation = 6 bails
Now, the probability that demand during lead-time will exceed 20 bails is given by = P(X > 20 bails)
P(X > 20 bails) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{20-15}{6}[/tex] ) = P(Z > 0.83) = 1 - P(Z [tex]\leq[/tex] 0.83)
= 1 - 0.7967 = 0.2033
The tensile strength of a certain metal component is normally distributed with a mean of 10,000 kilograms per square centimeter and a standard deviation of 100 kilograms per square centimeter. Measurements are recorded to the nearest 50 kilograms per square centimeter.
Required:
a. What proportion of these components exceed 10,150 kilograms per square centimeter in tensile strength?
b. If specifications require that all components have tensile strength between 9800 and 10,200 kilograms per square centimeter inclusive, what proportion of pieces would we expect to scrap?
Answer:
a. 0.0668
b. 0.9545
Step-by-step explanation:
We have the following information:
mean (m) = 10000
standard deviation (sd) = 100
(a)
We must calculate the proportion of the components exceed 10150 kilograms per square centimeter in tensile strength as follows:
P (x> 10150) = P [(x - m) / sd> (10150 - 1000 /) 100]
P (x> 10150) = P (z> 1.5)
P (x> 10150) = 1 - P (z <1.5)
P (x> 10150) = 1 - 0.9332 (attached table)
P (x> 10150) = 0.0668
Therefore the proportion of the components exceed 10150 kilograms per square centimeter in tensile strength is 0.0668
(b)
We must calculate the proportion of all components has tensile strength between 9800 and 10200, as follows:
P (9800 <x <10200) = P [(9800 - 1000 /) 100 <(x - m) / sd <(10200 - 1000 /) 100]
P (9800 <x <10200) = P (-2 <z <2)
P (9800 <x <10200) = P (z <2) - P (z <-2)
P (9800 <x <10200) = 0.9773 - 0.0228 (attached table)
P (9800 <x <10200) = 0.9545
the proportion of pieces that would expect to scrap is 0.9545
The amount of pollutants that are found in waterways near large cities is normally distributed with mean 8.5 ppm and standard deviation 1.4 ppm. 18 randomly selected large cities are studied. Round all answers to two decimal places.
A. xBar~ N( ____) (____)
B. For the 18 cities, find the probability that the average amount of pollutants is more than 9 ppm.
C. What is the probability that one randomly selected city's waterway will have more than 9 ppm pollutants?
D. Find the IQR for the average of 18 cities.Q1 =
Q3 =
IQR:
2. X ~ N(30,10). Suppose that you form random samples with sample size 4 from this distribution. Let xBar be the random variable of averages. Let ΣX be the random variable of sums. Round all answers to two decimal places.
A. xBar~ N(___) (____)
B. P(xBar<30) =
C. Find the 95th percentile for the xBar distribution.
D. P(xBar > 36)=
E. Q3 for the xBar distribution =
Answer:
1)
A) [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)
B) P([tex]\frac{}{X}[/tex] > 9)= 0.0552
C) P(X> 9)= 0.36317
D) IQR= 0.4422
2)
A) [tex]\frac{}{X}[/tex] ~ N(30;2.5)
B) P( [tex]\frac{}{X}[/tex]<30)= 0.50
C) P₉₅= 32.60
D) P( [tex]\frac{}{X}[/tex]>36)= 0
E) Q₃: 31.0586
Step-by-step explanation:
Hello!
1)
The variable of interest is
X: pollutants found in waterways near a large city. (ppm)
This variable has a normal distribution:
X~N(μ;σ²)
μ= 8.5 ppm
σ= 1.4 ppm
A sample of 18 large cities were studied.
A) The sample mean is also a random variable and it has the same distribution as the population of origin with exception that it's variance is affected by the sample size:
[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)
The population mean is the same as the mean of the variable
μ= 8.5 ppm
The standard deviation is
σ/√n= 1.4/√18= 0.329= 0.33 ⇒σ²/n= 0.33²= 0.108
So: [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)
B)
P([tex]\frac{}{X}[/tex] > 9)= 1 - P([tex]\frac{}{X}[/tex] ≤ 9)
To calculate this probability you have to standardize the value of the sample mean and then use the Z-tables to reach the corresponding value of probability.
Z= [tex]\frac{\frac{}{X} - Mu}{\frac{Sigma}{\sqrt{n} } } = \frac{9-8.5}{0.33}= 1.51[/tex]
Then using the Z table you'll find the probability of
P(Z≤1.51)= 0.93448
Then
1 - P([tex]\frac{}{X}[/tex] ≤ 9)= 1 - P(Z≤1.51)= 1 - 0.93448= 0.0552
C)
In this item, since only one city is chosen at random, instead of working with the distribution of the sample mean, you have to work with the distribution of the variable X:
P(X> 9)= 1 - P(X ≤ 9)
Z= (X-μ)/δ= (9-8.5)/1.44
Z= 0.347= 0.35
P(Z≤0.35)= 0.63683
Then
P(X> 9)= 1 - P(X ≤ 9)= 1 - P(Z≤0.35)= 1 - 0.63683= 0.36317
D)
The first quartile is the value of the distribution that separates the bottom 2% of the distribution from the top 75%, in this case it will be the value of the sample average that marks the bottom 25% symbolically:
Q₁: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₁)= 0.25
Which is equivalent to the first quartile of the standard normal distribution. So first you have to identify the first quartile for the Z dist:
P(Z≤z₁)= 0.25
Using the table you have to identify the value of Z that accumulates 0.25 of probability:
z₁= -0.67
Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:
z₁= ([tex]\frac{}{X}[/tex]₁-μ)/(σ/√n)
z₁*(σ/√n)= ([tex]\frac{}{X}[/tex]₁-μ)
[tex]\frac{}{X}[/tex]₁= z₁*(σ/√n)+μ
[tex]\frac{}{X}[/tex]₁= (-0.67*0.33)+8.5= 8.2789 ppm
The third quartile is the value that separates the bottom 75% of the distribution from the top 25%. For this distribution, it will be that value of the sample mean that accumulates 75%:
Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75
⇒ P(Z≤z₃)= 0.75
Using the table you have to identify the value of Z that accumulates 0.75 of probability:
z₃= 0.67
Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:
z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)
z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)
[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ
[tex]\frac{}{X}[/tex]₃= (0.67*0.33)+8.5= 8.7211 ppm
IQR= Q₃-Q₁= 8.7211-8.2789= 0.4422
2)
A)
X ~ N(30,10)
For n=4
[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)
Population mean μ= 30
Population variance σ²/n= 10/4= 2.5
Population standard deviation σ/√n= √2.5= 1.58
[tex]\frac{}{X}[/tex] ~ N(30;2.5)
B)
P( [tex]\frac{}{X}[/tex]<30)
First you have to standardize the value and then look for the probability:
Z= ([tex]\frac{}{X}[/tex]-μ)/(σ/√n)= (30-30)/1.58= 0
P(Z<0)= 0.50
Then
P( [tex]\frac{}{X}[/tex]<30)= 0.50
Which is no surprise since 30 y the value of the mean of the distribution.
C)
P( [tex]\frac{}{X}[/tex]≤ [tex]\frac{}{X}[/tex]₀)= 0.95
P( Z≤ z₀)= 0.95
z₀= 1.645
Now you have to reverse the standardization:
z₀= ([tex]\frac{}{X}[/tex]₀-μ)/(σ/√n)
z₀*(σ/√n)= ([tex]\frac{}{X}[/tex]₀-μ)
[tex]\frac{}{X}[/tex]₀= z₀*(σ/√n)+μ
[tex]\frac{}{X}[/tex]₀= (1.645*1.58)+30= 32.60
P₉₅= 32.60
D)
P( [tex]\frac{}{X}[/tex]>36)= 1 - P( [tex]\frac{}{X}[/tex]≤36)= 1 - P(Z≤(36-30)/1.58)= 1 - P(Z≤3.79)= 1 - 1 = 0
E)
Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75
⇒ P(Z≤z₃)= 0.75
z₃= 0.67
z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)
z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)
[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ
[tex]\frac{}{X}[/tex]₃= (0.67*1.58)+30= 31.0586
Q₃: 31.0586
Someone flips five coins, but you don’t see the outcome. The person reports that no tails are showing. What is the probability that the person flipped 5 heads?
Answer:
0.03125 = 3.125% probability that the person flipped 5 heads
Step-by-step explanation:
For each coin, there are only two possible outcomes. Either it was heads, or it was tails. The result of a coin toss is independent of other coin tosses. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
Five coins:
This means that n = 5.
Fair coin:
Equally as likely to be heads or tails, so p = 0.5.
What is the probability that the person flipped 5 heads?
This is P(X = 5).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 5) = C_{5,5}.(0.5)^{5}.(0.5)^{0} = 0.03125[/tex]
0.03125 = 3.125% probability that the person flipped 5 heads
identify the property being demonstrated
if x/5 = 7, then x=35
a. division
b. multiplication
c. reflexive
d. symmetric
Answer:
[tex] \: \: \: \: \: \: \: \: \: \: \dfrac{x}{5} = 7 \\ \implies \: x = 7 \times 5 \\ \implies \: x = 35[/tex]
So,b. multiplication
Answer:
A. division
Step-by-step explanation:
[tex]x/5=7[/tex]
[tex]x[/tex] is being divided by an integer.
[tex]x=35[/tex]
[tex]35/5=7[/tex]
35 divided by 5 is equal to 7.
What is the result of −18⋅16 2/3? Enter the result as an improper fraction and as a mixed number.
Answer:
-30000/100
300 0/1
Step-by-step explanation:
We have the following numbers -18 and 16 2/3, the first is an integer and the second is a mixed number, the first thing is to pass the mixed number to a decimal number.
16 2/3 = 16.67
We do the multiplication:
−18⋅16 2/3 = -300
We have an improper fraction is a fraction in which the numerator (top number) is greater than or equal to the denominator (bottom number), therefore it would be:
-30000/100
How mixed number would it be:
300 0/1
T-Mobile sells 6 different models of cell phones and have found that they sell an equal number of each model. The probability distribution that would describe this random variable is called:
Answer:
Option A is correct.
A uniform distribution.
Step-by-step explanation:
Complete Question
T-Mobile sells 6 different models of cell phones and have found that they sell an equal number of each model. The probability distribution that would describe this random variable is called:
A) Uniform Distribution
B) Continuous Distribution
C) Poisson Distribution
D) Relative Frequency Distribution
Solution
A uniform distribution is one in which all the variables have the same probability of occurring.
It is also known as a rectangular distribution, as every portion of the sample space has an equal chance of occurring, with equal length on the probability curve, leading to a rectangular probability curve.
And for this question, 6 different models of phones sell an equal number, hence, the probability of selling each model is equal to one another, hence, this is evidently a uniform distribution.
Hope this Helps!!!
Find the area of the smaller sector.
A
6 in
030°
Area = [? ]in?
B
Round your answer to the nearest hundredth.
Answer:
9.42 in²
Step-by-step explanation:
The area of whole circle S=pi*R² , where pi is appr. 3.14, R= 6 in
S= 3.14*6² =113.04 in²
The area of smaller sector is Ssec=S/360*30=113,04/12=9.42 in²
The area of the smaller sector with a central angle of 30 degrees and a radius of 6 inches is 9.42478 square inches.
To find the area of a sector, you can use the formula:
Area of sector = (θ/360) × π × r²
where θ is the central angle in degrees, r is the radius of the sector.
The central angle is 30 degrees and the radius is 6 inches.
Plugging these values into the formula:
Area of sector = (30/360) × π × 6²
= (1/12) × π × 36
= (1/12) × 3.14159 × 36
= 9.42478 square inches
To learn more on Area of sector click:
https://brainly.com/question/29055300
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Alan has reached 25% of his weekly exercise time goal so far this week. If he has exercised for a total of 42 minutes this week, what is his weekly exercise time goal in minutes
Answer:
His weekly exercise time goal is 168 minutes.
Step-by-step explanation:
This question can be solved using a rule of three.
42 minutes is 25% = 0.25 of the total
x minutes is 100% = 1 of the total.
Then
42 minutes - 0.25
x minutes - 1
[tex]0.25x = 42[/tex]
[tex]x = \frac{42}{0.25}[/tex]
[tex]x = 168[/tex]
His weekly exercise time goal is 168 minutes.
The manager of the Danvers-Hilton Resort Hotel stated that the mean guest bill for a weekend is $600 or less. A member of the hotel's accounting staff noticed that the total charges for guest bills have been increasing in recent months. The accountant will use a sample of future weekend guest bills to test the manager's claim. (a) Which form of the hypotheses should be used to test the manager's claim? H0: - Select your answer - Ha: - Select your answer - The member of the hotel's accounting staff suspects that the total charges for guest bills have Select in recent months. To test the manager’s claim, the staff member will conduct Select test of the population Select . (b) What conclusion is appropriate when H0 cannot be rejected? When H0 cannot be rejected, there Select enough evidence to conclude that the total charges for guest bills have Select in recent months. (c) What conclusion is appropriate when H0 can be rejected? When H0 can be rejected, there Select enough evidence to conclude that the total charges for guest bills have Select in recent m
Answer:
a) Null hypothesis (H0): the mean guest bill for a weekend is $600.
Alternative hypothesis (Ha): the mean guest bill for a weekend is significantly bigger than $600.
b) When H0 can not be rejected, the conclusion is that there is no enough evidence to claim that the mean guest bill had increased from $600.
c) When the H0 is rejected, they have enough evidence to claim that the mean guest bill is significantly bigger than $600.
Step-by-step explanation:
a) The accountant, as he wants to see if there is evidence to support the claim that the mean guest bill has increased significanty, should write the hypothesis like that:
Null hypothesis (H0): the mean guest bill for a weekend is $600.
Alternative hypothesis (Ha): the mean guest bill for a weekend is significantly bigger than $600.
A sample of bills of the period in study needs to be taken in order to have a representation of the actual population of bills and then perform a t-test, as the sample mean and standard deviation will be used to perform the test.
b) When H0 can not be rejected, the conclusion is that there is no enough evidence to claim that the mean guest bill had increased from $600. If the P-value was low but not enough, they may take another sample to perform the test again or leave it like that.
c) When the H0 is rejected, they have enough evidence to claim that the mean guest bill is significantly bigger than $600.
Find the area of a triangle that has the base of 5 inches and a height of 3 3/4 inches
Answer:
9.375 in^2
Step-by-step explanation:
Please help Solving linear and quadratic equations
Answer: B.
x ≈2.5
Step-by-step explanation:
[tex]-\left(u\right)^{-1}-6=-u+10[/tex]
[tex]u=8-\sqrt{65},\:u=8+\sqrt{65}[/tex]
[tex]x=\frac{\ln \left(8+\sqrt{65}\right)}{\ln \left(3\right)}[/tex]
x=2.52...
Answer:
x=2.5
Step-by-step explanation:
Determine whether the underlined value is a parameter or a statistic. In a national survey of high school students (grades 9 to 12), Modifying 25 % of the students who responded reported that someone had offered, sold, or given them an illegal drug on school property.
Answer:
I am assuming the underlined value is 25%. It is a parameter
Step-by-step explanation:
The value is is a parameter. This is because the parameter is a value that describes the population.
The survey carried out was a national survey of which there were 25% respondents who reported that someone had offered, sold, or given them an illegal drug on school property. It is not a statistics because a sample was not taken out of the population and a survey made on the sample.
The underlined 25% value is the value that summarizes the entire population of high school students
Two thousand dollars is deposited into a savings account at 8.5% interest compounded continuously. (a) What is the formula for A(t), the balance after t years? (b) What differential equation is satisfied by A(t), the balance after t years? (c) How much money will be in the account after 5 years? (d) When will the balance reach $3000? (e) How fast is the balance growing when it reaches $3000?
Answer:
a)[tex]A(t)=2000e^{0.085t}[/tex]
b)[tex]A'(t)=170e^{0.085t}[/tex]
c)$3059.1808
d)t=4.77 years
e) The balance growing is $254.99/year
Step-by-step explanation:
We are given that Two thousand dollars is deposited into a savings account at 8.5% interest compounded continuously.
Principal = $2000
Rate of interest = 8.5%
a) What is the formula for A(t), the balance after t years?
Formula [tex]A(t)=Pe^{rt}[/tex]
So,[tex]A(t)=2000e^{0.085t}[/tex]
B)What differential equation is satisfied by A(t), the balance after t years?
So, [tex]A'(t)=2000 \times 0.085 e^{0.085t}[/tex]
[tex]A'(t)=170e^{0.085t}[/tex]
c)How much money will be in the account after 5 years?
Substitute t = 5 in the formula "
[tex]A(t)=2000e^{0.085t}\\A(5)=2000e^{0.085(5)}\\A(5)=3059.1808[/tex]
d)When will the balance reach $3000?
Substitute A(t)=3000
So, [tex]3000=2000e^{0.085t}[/tex]
t=4.77
The balance reach $3000 in 4.77 years
e)How fast is the balance growing when it reaches $3000?
Substitute the value of t = 4.77 in derivative formula :
[tex]A'(t)=170e^{0.085t}\\A'(t)=170e^{0.085 \times 4.77}\\A'(t)=254.99[/tex]
Hence the balance growing is $254.99/year
6. Factor the expression.
9b2 + 48b + 64
A (3b + 8)2
B (-3b + 8)2
C (-3b - 82
D (3b - 8)2
70%
Answer:
A. [tex](3b+8)^2[/tex]
Step-by-step explanation:
[tex]9b^2+48b +64\\=(3b)^2 + 2\times 3b\times 8 +(8)^2\\=(3b+8)^2[/tex]
ali's typing rate between 8:00 am and noon is 48 words per minute . after lunch a lunch break, Ali's typing rate between 1:00 pm and 4:00 pm is 2,040 words per hour . what is Ali's average typing rate per minute for the whole time she works?
Answer:
41 word/min
Step-by-step explanation:
Before noon Ali works:
4 hours= 4*60 min= 240 minShe types:
240*48= 11520 wordsAfter lunch she works:
4 hoursShe types:
4*2040= 8160 wordsTotal Ali works= 4+4= 8 hours= 480 min
Total Ali types= 11520+8160= 19680 words
Average typing rate= 19680 words/480 min= 41 word/min
Consider three consecutive positive integers, such that the sum of the
squares of the two larger integers is 5 more than 40 times the smaller
one. Find the smaller integer.
Answer:
17
Step-by-step explanation:
Let x represent the smaller integer. Then we have ...
(x +1)² +(x +2)² = 40x +5
2x² +6x +5 = 40x +5
x² -17x = 0 . . . . . subtract (40x+5), divide by 2
x(x -17) = 0 . . . . . factor
The solution of interest is x = 17.
The smaller integer is 17.
Lara’s Inc. is currently an unlevered firm with 450,000 shares of stock outstanding, with a market price of $15 a share. The company has earnings before interest and taxes of $314,000. Lara's met with his bankers, Warne Incorporated and agreed to borrow $825,000, at 5 percent. You are an ardent investor and you currently own 20,000 shares of Lara's stock. If you seek to unlevered your position; how many shares of Lara's stock will you continue to own, if you can loan out funds at 5 percent interest? Ignore taxes in your deliberations. Kindly show all workings.
Answer:
tiStep-by-step explanaon:
The annual interest on a $14,000 investment exceeds the interest earned on a $7000 investment by $595 . The $14,000 is invested at a 0.5% higher rate of interest than the $7000 . What is the interest rate of each investment?
Answer:
x= 8.00 Interest rate on $14000
y= 7.50 Interest rate on $7000
Step-by-step explanation:
Let interest rate of $14000 be x%
and Interest rate for $7000 be y %
According to the first condition
14000 * x% - 7000 * y% = 595
multiply by 100
14000x-7000y = 59500
/700
20x-10y=85.................(1)
II condition
x%=y%+0.5%
x=y+0.5
x-y=0.5..................................(2)
solve (1) & (2)
20 x -10 y = 85 .............1
Total value
1 x -1 y = 0.50 .............2
Eliminate y
multiply (1)by 1
Multiply (2) by -10
20.00 x -10.00 y = 85.00
-10.00 x + 10.00 y = -5.00
Add the two equations
10.00 x = 80.00
/ 10.00
x = 8.00
plug value of x in (1)
20.00 x -10.00 y = 85.00
160.00 -10.00 y = 85.00
-10.00 y = 85.00 -160.00
-10.00 y = -75.00
y = 7.50
x= 8.00 Interest rate on $14000
y= 7.50 Interest rate on $7000
Consider random samples of size 900 from a population with proportion 0.75 . Find the standard error of the distribution of sample proportions. Round your answer for the standard error to three decimal places. standard error
Answer:
[tex] SE =\sqrt{\frac{p(1-p)}{n}}[/tex]
And replacing we got:
[tex] SE=\sqrt{\frac{0.75*(1-0.75)}{900}}= 0.014[/tex]
Step-by-step explanation:
For this case we have the following info given:
[tex] n=900[/tex] represent the sample size selected
[tex]p = 0.75[/tex] represent the population proportion
We want to find the standard error and we can use the distribution for the sample proportion and for this case since the sample size is large enough and we satisfy np>10 and n(1-p) >10 we have:
[tex] \hat p \sim N (p,\sqrt{\frac{p(1-p)}{n}})[/tex]
And the standard error is given;
[tex] SE =\sqrt{\frac{p(1-p)}{n}}[/tex]
And replacing we got:
[tex] SE= \sqrt{\frac{0.75* (1-0.75)}{900}}= 0.014[/tex]
Find the Laplace transform F(s)=L{f(t)} of the function f(t)=sin2(wt), defined on the interval t≥0. F(s)=L{sin2(wt)}= help (formulas) Hint: Use a double-angle trigonometric identity. For what values of s does the Laplace transform exist? help (inequalities)
The Laplace transform of the function [tex]\frac{1}{2} (\frac{1}{s} - \frac{s}{s^2 + 4w^2} )[/tex] .
The Laplace transform exist when s > 0 .
Here, the given function is f(t) = sin²(wt) .
The Laplace transform of the the function f(t),
F(s) = f(t) = { [tex]{\frac{1}{2} \times 2sin^2(wt) }[/tex] }
F(s) = { [tex]\frac{1}{2} \times (1- cos2wt)[/tex] }
F(s) = { [tex]\frac{1}{2} - \frac{1}{2} \times cos(2wt)\\[/tex] }
F(s) = [tex]\frac{1}{2} (\frac{1}{s} - \frac{s}{s^2 + 4w^2} )[/tex]
Next,
The above Laplace transform exist if s > 0 .
Know more about Laplace transform,
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The equation of a circle is x2 + y2 = 56.25. Find the radius of the circle?
Answer:
r = 7.5
Step-by-step explanation:
Circle equation: [tex](x - h)^2 + (y - k)^2 = r^2[/tex]
Since we are already give r², we simply just take the square root of 56.25, and we should get 7.5 as our final answer!
What is the slope of the line with the two
points A(-4, 8) and B(-9, 12)?
Answer:
slope = -4/5
Step-by-step explanation:
A line passes two points (x1, y1) and (x2, y2).
The slope of this line can be calculate by the formula:
s = (y2 - y1)/(x2 - x1)
=>The line that passes A(-4, 8) and B(-9, 12) has the slope:
s = (12 - 8)/(-9 - -4) = 4/(-5) = -4/5
Hope this helps!
The graph of the function f(x) =4 square root x is shown. what is the domain of the function?
Answer:
[0, positive infinity)
Step-by-step explanation:
The domain is all x values a graph inputs. In a square root function, you cannot have negative inputs as it will turn out imaginary numbers. Therefore, your domain is all values of x above and including 0.
Answer: d on Ed
Step-by-step explanation:
Just took the test
Black walnut trees contain chemicals that inhibit the growth of other plants. In a simple experiment to test whether this is true, you grow several tomato plants in soil with and without decomposing leaves from a black walnut tree. You collect data on plant height as a measure of growth. In this experiment, __________ is the independent variable, __________ is the dependent variable, and __________ is the control.
Answer:
Height of tomato plant is the dependent variable
Presence of walnut leaves in the soil is the independent variable
Tomato plants grown without walnut leaves is the control
Step-by-step explanation:
An independent variable is the variable in an experiment that can be altered to test for a certain result. It is independent, or does not change with change in other factors in the experiment. In this case, the presence or absence, or quantity of walnut available in the soil is the independent variable in the experiment.
A dependent variable varies, and depends on the independent variable. It is what is measured in the experiment. In this case, the height of the tomato plants is the dependent variable that depends on the presence, absence or quantity of walnut in the soil.
A control in an experiment, is a replicate experiment, that is manipulated in order to be able to test a single variable at a time. Controls are variables are held constant so as to minimize their effect on the system under study. In this case, some of the tomato plants are planted without walnut in the soil, to test the effect of the absence of the walnut in the soil.
A hotel manager believes that 27% of the hotel rooms are booked. If the manager is correct, what is the probability that the proportion of rooms booked in a sample of 423 rooms would differ from the population proportion by less than 6%
Answer:
The probability that the proportion of rooms booked in a sample of 423 rooms would differ from the population proportion by less than 6% is 0.9946.
Step-by-step explanation:
According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.
The mean of this sampling distribution of sample proportion is:
[tex]\mu_{\hat p}=p[/tex]
The standard deviation of this sampling distribution of sample proportion is:
[tex]\sigma_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
The information provided here is:
p = 0.27
n = 423
As n = 423 > 30, the sampling distribution of sample proportion can be approximated by the Normal distribution.
The mean and standard deviation of the sampling distribution of sample proportion are:
[tex]\mu_{\hat p}=p=0.27\\\\\sigma_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.27\times(1-0.27)}{423}}=0.0216[/tex]
Compute the probability that the proportion of rooms booked in a sample of 423 rooms would differ from the population proportion by less than 6% as follows:
[tex]P(|\hat p-p|<0.06)=P(p-0.06<\hat p<p+0.06)[/tex]
[tex]=P(0.27-0.06<\hat p<0.27+0.06)\\\\=P(0.21<\hat p<0.33)\\\\=P(\frac{0.21-0.27}{0.0216}<\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}}<\frac{0.33-0.27}{0.0216})\\\\=P(-2.78<Z<2.78)\\\\=P(Z<2.78)-P(Z<-2.78)\\\\=0.99728-0.00272\\\\=0.99456\\\\\approx 0.9946[/tex]
*Use a z-table.
Thus, the probability that the proportion of rooms booked in a sample of 423 rooms would differ from the population proportion by less than 6% is 0.9946.