Answer:
(a)[tex]D(t)=\dfrac{50000}{1.9t+9}[/tex]
(b)[tex]D'(115)=-1.8355[/tex]
Step-by-step explanation:
The demand function for a product is given by :
[tex]D(p)=\dfrac{50000}{p}[/tex]
Price, p is a function of time given by [tex]p=1.9t+9[/tex], where t is in days.
(a)We want to find the demand as a function of time t.
[tex]\text{If } D(p)=\dfrac{50000}{p},$ and p=1.9t+9\\Then:\\D(t)=\dfrac{50000}{1.9t+9}[/tex]
(b)Rate of change of the quantity demanded when t=115 days.
[tex]\text{If } D(t)=\dfrac{50000}{1.9t+9}[/tex]
[tex]\dfrac{\mathrm{d}}{\mathrm{d}t}\left[\dfrac{50000}{\frac{19t}{10}+9}\right]}}=50000\cdot \dfrac{\mathrm{d}}{\mathrm{d}t}\left[\dfrac{1}{\frac{19t}{10}+9}\right]}[/tex]
[tex]=-50000\cdot\dfrac{d}{dt} \dfrac{\left[\frac{19t}{10}+9\right]}{\left(\frac{19t}{10}+9\right)^2}}}[/tex]
[tex]=\dfrac{-50000(1.9\frac{d}{dt}t+\frac{d}{dt}9)}{\left(\frac{19t}{10}+9\right)^2}}}[/tex]
[tex]=-\dfrac{95000}{\left(\frac{19t}{10}+9\right)^2}\\$Simplify/rewrite to obtain:$\\\\D'(t)=-\dfrac{9500000}{\left(19t+90\right)^2}[/tex]
Therefore, when t=115 days
[tex]D'(115)=-\dfrac{9500000}{\left(19(115)+90\right)^2}\\D'(115)=-1.8355[/tex]
If a tank holds 4500 gallons of water, which drains from the bottom of the tank in 50 minutes, then Toricelli's Law gives the volume V of water remaining in the tank after t minutes as V = 4500 (1 − 1 /50 t )^2. 0≤ t ≤ 50. At what time is the water flowing out the fastest?
Answer:
t = 0
Before it starts rushing that's when it will be fastest
Step-by-step explanation:
For the water ib the tank to flow very fast it means that there is a big volume of water present.
And for volume of water to be present that much it means that the water must
have not leaked much or at all.
And for that it signifies large volume of water.
If we do the calculation we'd see that time will be actually equal to zero for the pressure and the volume of the water to be biggest.
V = 4500 (1 − 1 /50 t )^2
V = 4500
4500 = 4500(1- 1/50t)²
1 = 1- 1/50t
0 = -1/50t
t = 0
Given that 9 x − 4 y = 20 Find y when x = − 2 Give your answer as an improper fraction in its simplest form
Answer:
[tex]\boxed{\df\ \dfrac{-19}{2}}[/tex]
Step-by-step explanation:
Hi,
x=-2
it gives
9*(-2)-4y=20
<=> -18-4y=20
<=> 18-18-4y=20+18=38
<=> -4y=38
<=> y = -38/4=-19/2
hope this helps
B
Round your answer to the nearest hundredth.
A
9
B
5
Answer:
56.25°
Step-by-step explanation:
The definition of the cosine function tells you that
cos(B) = BC/BA
B = arccos(BC/BA) = arccos(5/9)
B ≈ 56.25°
State the size of angle 'n' in the triangle illustrated below.
Answer:
Option B
Step-by-step explanation:
<r = 32 degrees (alternate angles )
<r = <n = 32 degrees (vertical angles)
Justin spent $23 on fruit at grocery store. He spent a total of $25 at the store. What percentage of the total did he spend on fruit?
Step-by-step explanation:
In my opinion maybe he has spent 98%
Please help with this problem
Answer:
The length of the short side is 14.5 units, the length of the other short side is 18.5 units, and the length of the longest side is 23.5 units.
Step-by-step explanation:
The Pythagorean Theorem
If a and b are the lengths of the legs of a right triangle and c is the length of the hypotenuse, then the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse.
This relationship is represented by the formula:
[tex]a^2+b^2=c^2[/tex]
Applying the Pythagorean Theorem to find the lengths of the three sides we get:
[tex](x)^2+(x+4)^2=(x+9)^2\\\\2x^2+8x+16=x^2+18x+81\\\\2x^2+8x-65=x^2+18x\\\\2x^2-10x-65=x^2\\\\x^2-10x-65=0[/tex]
Solve with the quadratic formula
[tex]\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}[/tex]
[tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]\mathrm{For\:}\quad a=1,\:b=-10,\:c=-65:\quad x_{1,\:2}=\frac{-\left(-10\right)\pm \sqrt{\left(-10\right)^2-4\cdot \:1\left(-65\right)}}{2\cdot \:1}\\\\x_{1}=\frac{-\left(-10\right)+ \sqrt{\left(-10\right)^2-4\cdot \:1\left(-65\right)}}{2\cdot \:1}=5+3\sqrt{10}\\\\x_{2}=\frac{-\left(-10\right)- \sqrt{\left(-10\right)^2-4\cdot \:1\left(-65\right)}}{2\cdot \:1}=5-3\sqrt{10}[/tex]
Because a length can only be positive, the only solution is
[tex]x=5+3\sqrt{10}\approx 14.5[/tex]
The length of the short side is 14.5, the length of the other short side is [tex]14.5+4=18.5[/tex], and the length of the longest side is [tex]14.5+9=23.5[/tex].
100 pts You have a bag of 15 marbles: 5 blue, 3 red, 4 green, and 3 yellow. You draw 3 marbles without replacement. Which action, performed before the draws, increases the probability of drawing 3 green marbles in a row?
Answer:
see below
Step-by-step explanation:
You can remove one or more of the other color marbles to increase the probability of drawing a green marble
or
You can add one or more green marbles to have more green marbles in the bag
According to the National Association of Theater Owners, the average price for a movie in the United States in 2012 was $7.96. Assume the population standard deviation is $0.50 and that a sample of 30 theaters was randomly selected.
Required:
a. Calculate the standard error of the mean.
b. What is the probability that the sample mean will be less than $7.75?
c. What is the probability that the sample mean will be less than $8.10?
d. What is the probability that the sample mean will be more than $8.20?
Answer:
(a) The standard error of the mean is 0.091.
(b) The probability that the sample mean will be less than $7.75 is 0.0107.
(c) The probability that the sample mean will be less than $8.10 is 0.9369.
(d) The probability that the sample mean will be more than $8.20 is 0.0043.
Step-by-step explanation:
We are given that the average price for a movie in the United States in 2012 was $7.96.
Assume the population standard deviation is $0.50 and that a sample of 30 theaters was randomly selected.
Let [tex]\bar X[/tex] = sample mean price for a movie in the United States
The z-score probability distribution for the sample mean is given by;
Z = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean price for a movie = $7.96
[tex]\sigma[/tex] = population standard deviation = $0.50
n = sample of theaters = 30
(a) The standard error of the mean is given by;
Standard error = [tex]\frac{\sigma}{\sqrt{n} }[/tex] = [tex]\frac{0.50}{\sqrt{30} }[/tex]
= 0.091
(b) The probability that the sample mean will be less than $7.75 is given by = P([tex]\bar X[/tex] < $7.75)
P([tex]\bar X[/tex] < $7.75) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{7.75-7.96}{\frac{0.50}\sqrt{30} } }[/tex] ) = P(Z < -2.30) = 1 - P(Z [tex]\leq[/tex] 2.30)
= 1 - 0.9893 = 0.0107
The above probability is calculated by looking at the value of x = 2.30 in the z table which has an area of 0.9893.
(c) The probability that the sample mean will be less than $8.10 is given by = P([tex]\bar X[/tex] < $8.10)
P([tex]\bar X[/tex] < $8.10) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{8.10-7.96}{\frac{0.50}\sqrt{30} } }[/tex] ) = P(Z < 1.53) = 0.9369
The above probability is calculated by looking at the value of x = 1.53 in the z table which has an area of 0.9369.
(d) The probability that the sample mean will be more than $8.20 is given by = P([tex]\bar X[/tex] > $8.20)
P([tex]\bar X[/tex] > $8.20) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{8.20-7.96}{\frac{0.50}\sqrt{30} } }[/tex] ) = P(Z > 2.63) = 1 - P(Z [tex]\leq[/tex] 2.63)
= 1 - 0.9957 = 0.0043
The above probability is calculated by looking at the value of x = 2.63 in the z table which has an area of 0.9957.
At the U.S. Open Tennis Championship a statistician keeps track of every serve that a player hits during the tournament. The statistician reported that the mean serve speed was 100 miles per hour (mph) and the standard deviation of the serve speeds was 15 mph. Assume that the statistician also gave us the information that the distribution of serve speeds was mound- shaped and symmetric. What percentage of the player's serves were between 115 mph and 145 mph
Answer:
15.74% of the player's serves were between 115 mph and 145 mph
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
[tex]\mu = 100, \sigma = 15[/tex]
What percentage of the player's serves were between 115 mph and 145 mph
This is the pvalue of Z when X = 145 subtracted by the pvalue of Z when X = 115.
X = 145
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{145 - 100}{15}[/tex]
[tex]Z = 3[/tex]
[tex]Z = 3[/tex] has a pvalue of 0.9987
X = 115
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{115 - 100}{15}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a pvalue of 0.8413
0.9987 - 0.8413 = 0.1574
15.74% of the player's serves were between 115 mph and 145 mph
Imagine you have a rectangular wooden block with dimensions of 10 cm x 3 cm x 8 cm (L x W x H). Required:a. What is the volume of your wooden block?b. What is the density of this wooden block if it has a mass of 168 g?
Answer:
a) The volume of the wooden block is 240 cm^3.
b) The density of the wooden block is 0.7 g/cm^3.
Step-by-step explanation:
The volume of the rectangular wooden block can be calculated as the multiplication of the length in each dimension: length, wide and height.
With dimensions 10 cm x 3 cm x 8 cm, the volume is:
[tex]V=L\cdot W\cdot H = 10\cdot 3\cdot 8=240[/tex]
The volume of the wooden block is 240 cm^3.
If we know that the mass of the wooden block is 168 g, we can calculate the density as:
[tex]\rho = \dfrac{M}{V}=\dfrac{168}{240}=0.7[/tex]
The density of the wooden block is 0.7 g/cm^3.
Dr. Miriam Johnson has been teaching accounting for over 20 years. From her experience, she knows that 60% of her students do homework regularly. Moreover, 95% of the students who do their homework regularly generally pass the course. She also knows that 85% of her students pass the course.
a. What is the probability that a student will do homework regularly and also pass the course?
b. What is the probability that a student will neither do homework regularly nor will pass the course?
c. Are the events "pass the course" and "do homework regularly" mutually exclusive? Explain.
d. Are the events "pass the course" and "do homework regularly" independent? Explain.
Answer:
a) The probability that a student will do homework regularly and also pass the course = P(H n P) = 0.57
b) The probability that a student will neither do homework regularly nor will pass the course = P(H' n P') = 0.12
c) The two events, pass the course and do homework regularly, aren't mutually exclusive. Check Explanation for reasons why.
d) The two events, pass the course and do homework regularly, aren't independent. Check Explanation for reasons why.
Step-by-step explanation:
Let the event that a student does homework regularly be H.
The event that a student passes the course be P.
- 60% of her students do homework regularly
P(H) = 60% = 0.60
- 95% of the students who do their homework regularly generally pass the course
P(P|H) = 95% = 0.95
- She also knows that 85% of her students pass the course.
P(P) = 85% = 0.85
a) The probability that a student will do homework regularly and also pass the course = P(H n P)
The conditional probability of A occurring given that B has occurred, P(A|B), is given as
P(A|B) = P(A n B) ÷ P(B)
And we can write that
P(A n B) = P(A|B) × P(B)
Hence,
P(H n P) = P(P n H) = P(P|H) × P(H) = 0.95 × 0.60 = 0.57
b) The probability that a student will neither do homework regularly nor will pass the course = P(H' n P')
From Sets Theory,
P(H n P') + P(H' n P) + P(H n P) + P(H' n P') = 1
P(H n P) = 0.57 (from (a))
Note also that
P(H) = P(H n P') + P(H n P) (since the events P and P' are mutually exclusive)
0.60 = P(H n P') + 0.57
P(H n P') = 0.60 - 0.57
Also
P(P) = P(H' n P) + P(H n P) (since the events H and H' are mutually exclusive)
0.85 = P(H' n P) + 0.57
P(H' n P) = 0.85 - 0.57 = 0.28
So,
P(H n P') + P(H' n P) + P(H n P) + P(H' n P') = 1
Becomes
0.03 + 0.28 + 0.57 + P(H' n P') = 1
P(H' n P') = 1 - 0.03 - 0.57 - 0.28 = 0.12
c) Are the events "pass the course" and "do homework regularly" mutually exclusive? Explain.
Two events are said to be mutually exclusive if the two events cannot take place at the same time. The mathematical statement used to confirm the mutual exclusivity of two events A and B is that if A and B are mutually exclusive,
P(A n B) = 0.
But, P(H n P) has been calculated to be 0.57, P(H n P) = 0.57 ≠ 0.
Hence, the two events aren't mutually exclusive.
d. Are the events "pass the course" and "do homework regularly" independent? Explain
Two events are said to be independent of the probabilty of one occurring dowant depend on the probability of the other one occurring. It sis proven mathematically that two events A and B are independent when
P(A|B) = P(A)
P(B|A) = P(B)
P(A n B) = P(A) × P(B)
To check if the events pass the course and do homework regularly are mutually exclusive now.
P(P|H) = 0.95
P(P) = 0.85
P(H|P) = P(P n H) ÷ P(P) = 0.57 ÷ 0.85 = 0.671
P(H) = 0.60
P(H n P) = P(P n H)
P(P|H) = 0.95 ≠ 0.85 = P(P)
P(H|P) = 0.671 ≠ 0.60 = P(H)
P(P)×P(H) = 0.85 × 0.60 = 0.51 ≠ 0.57 = P(P n H)
None of the conditions is satisfied, hence, we can conclude that the two events are not independent.
Hope this Helps!!!
Can someone plz help me solved this problem! I’m giving you 10 points! I need help plz help me! Will mark you as brainiest!
Answer:
See the answers below.
Step-by-step explanation:
[tex]a.\:\frac{f\left(x\right)-f\left(a\right)}{x-a}=\frac{2x^2-x-5-\left(2a^2-a-5\right)}{x-a}\\\\=\frac{2x^2-x+a-2a^2}{x-a}\\\\=\frac{2\left(x+a\right)\left(x-a\right)-1\left(x-a\right)}{x-a}\\\\=\frac{\left(x-a\right)\left[2\left(x+a\right)-1\right]}{x-a}\\\\=2x+2a-1\\\\\\b.\:\frac{f\left(x+h\right)-f\left(x\right)}{h}=\frac{2\left(x+h\right)^2-\left(x+h\right)-5-\left(2x^2-x-5\right)}{h}\\\\=\frac{2\left(x^2+2xh+h^2\right)-\left(x+h\right)-5-\left(2x^2-x-5\right)}{h}\\[/tex]
Expand and simplify to get:
[tex]=\frac{2h^2+4xh-h}{h}\\\\=\frac{h\left(2h+4x-1\right)}{h}\\\\=2h+4x-1[/tex]
Best Regards!
A company makes wax candles shaped like rectangular prisms. Each candle is 7cm long, 2cm wide, and 10cm tall. If they used 5740cm^3 of wax, how many candles did they make?
Answer: 41 candles
Step-by-step explanation:
Multiply the dimensions of the candle first.
V = l*w*h
7 * 2 = 14
14 * 10 = 140
Now, divide the total amount of wax used by the amount of wax used for one candle.
5,740 / 140 = 41
Solve for x: −3x + 3 < 6
Answer:x>-1
Step-by-step explanation:
Step 1: Subtract 3 from both sides.
-3x+3-3<6-3
-3x<3
Step 2: Divide both sides by -3.
-3x/-3<3/3
X>-1
N
Write the rate as a unit rate
729 riders in 9 subway cars
А
The unit rate is
This
(Simplify your answ
riders/car
Answer:
Unit rate = 81 riders/ car.
Step-by-step explanation:
Given
729 riders in 9 cars
we have to find unit rate in terms of riders per car
let the the riders per car (i.e rate) be x.
If there are 9 cars then
total no. of riders in 9 cars = no. of cars * riders per car = 9*x = 9x
given that 729 riders in 9 cars
then
9x = 729
=> x = 729/9 = 81
Thus, riders per car = x = 81.
Unit rate is 81 riders per car.
What is the distance between (8, -3) and (4, - 7)?
Answer:
[tex]distance=\sqrt{32}[/tex] , which agrees with answer "c" in your list of possible options
Step-by-step explanation:
Use the formula for distance between two points [tex](x_1,y_1)[/tex], and [tex](x_2,y_2)[/tex] on the plane:
[tex]distance = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\distance= \sqrt{(4-8)^2+(-7-(-3))^2} \\distance= \sqrt{(-4)^2+(-4)^2} \\distance=\sqrt{16+16}\\distance=\sqrt{32}[/tex]
Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar. In a talent competition, half of the contestants are eliminated in each round. At the end of the nth round, 32 contestants remain. If there were 1,024 contestants at the start of the competition, what is the value of n? The value of n is .
Answer:
n =32
Step-by-step explanation:
If 1 contestant is eliminated each round
then of 1024contestants
32 left
1024/32=32
Answer:
n=32
Step-by-step explanation:
What is the inverse of the function f(x) =1/4 x – 12?
Step-by-step explanation:
solve f(x) by supposing it has y and and then interchange it with x .
hope this is helpful
Simplify the expression by combining like terms
15 + 12x – 5.2 + 4y - 7
9.8 +12x+y-7
2.8+12x+4y
y= -3/2x-6 x=15 plssssssssssssssssssssssss help
Answer:
-45/2 - 12/2 = -57/2
Step-by-step explanation:
Substitute 15 for x in the given equation: y = (-3/2)x - 6 becomes
y = (-3/2)(15) - 6 = -45/2 - 6 when x = 15. This is equivalent to -57/2
PROBLEM 6. 10 A histogram has mean 70 and standard deviation 5 If the histogram is not bell shaped but it is symmetric. Find the least proportion of data falls between 70 and 80 If the histogram is bell shaped. Find the proportion of data between 65 and 77
Answer:
a. 0.4772 = 47.72 %
b. 0.7605 = 76.05 %
Step-by-step explanation:
What we must do is calculate the z value for each value and thus find what percentage each represents and the subtraction would be the percentage between those two values.
We have that z is equal to:
z = (x - m) / (sd)
x is the value to evaluate, m the mean, sd the standard deviation
a. ind the least proportion of data falls between 70 and 80 If the histogram is bell shaped:
So for 70 copies we have:
z = (70 - 70) / (5)
z = 0
and this value represents 0.5
So for 80 copies we have:
z = (80 - 70) / (5)
z = 2
and this value represents 0.9772
p (70 > x > 80) = 0.9772 - 0.5
p (70 > x > 80) = 0.4772 = 47.72 %
b. Find the proportion of data between 65 and 77
So for 65 copies we have:
z = (65 - 70) / (5)
z = -1
and this value represents 0.1587
So for 77 copies we have:
z = (77 - 70) / (5)
z = 1.4
and this value represents 0.9192
p (65 > x > 77) = 0.9192 - 0.1587
p (65 > x > 77) = 0.7605 = 76.05 %
Single adults: According to a Pew Research Center analysis of census data, in 2012, 20% of American adults ages 25 and older had never been married. Suppose that we select 3 random samples of 500 adults from this population. Which of the following is most likely to occur with the three samples?
A. The number that had never been married will equal 20% in each of the three samples.B. The number that had never been married will vary in each sample due to the random selection of adults.C. The average for the three samples of the number of adults that had never been married will equal 20%.D. The number of adults that had never been married will increase for each sample because the number is generally increasing over time.
Answer:
Option B
Step-by-step explanation:
The number that had never been married will vary in each sample due to the random selection of adults.
This number will vary in each sample to the random selection process but they might or might not be as close as possible to one another after sampling.
Please answer this correctly
Answer:
30
Step-by-step explanation:
Answer:
It would decrease by 9.
Step-by-step explanation:
52 is the original mean or the initial mean.
43 is the final mean.
52-43 = 9
So 9 is the difference.
Hope this helped!
An elementary school is offering 3 language classes: one in Spanish, one in French, and one in German. The classes are open to any of the 100 students in the school. There are 28 students in the Spanish class, 26 in the French class, and 16 in the German class. There are 12 students who are in both Spanish and French, 4 who are in both Spanish and German, and 6 who are in both French and German. In addition, there are 2 students taking all 3 classes. If two students are randomly chosen, what is the probability that at exactly one of them does exactly two language classes.
Answer:
The probability that at exactly one of them does exactly two language classes is 0.32.
Step-by-step explanation:
We can model this variable as a binomial random variable with sample size n=2.
The probability of success, meaning the probability that a student is in exactly two language classes can be calculated as the division between the number of students that are taking exactly two classes and the total number of students.
The number of students that are taking exactly two classes is equal to the sum of the number of students that are taking two classes, minus the number of students that are taking the three classes:
[tex]N_2=F\&S+S\&G+F\&G-F\&S\&G=12+4+6-2=20[/tex]
Then, the probabilty of success p is:
[tex]p=20/100=0.2[/tex]
The probability that k students are in exactly two classes can be calcualted as:
[tex]P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{2}{k} 0.2^{k} 0.8^{2-k}\\\\\\[/tex]
Then, the probability that at exactly one of them does exactly two language classes is:
[tex]P(x=1) = \dbinom{2}{1} p^{1}(1-p)^{1}=2*0.2*0.8=0.32\\\\\\[/tex]
Please answer this correctly
Answer:
The number of employees classified into groups as shown below:
1 - 10: 3 6 (2companies)
11-20: 16 (1 company)
21-30: 25, 26, 27 (3 companies)
31-40: 34, 35, 35, 35, 36 (5 companies)
41-50: 41, 43, 48, 48 (4 companies)
Hope this helps!
Answer:
11-20 is 1
31-40 is 5
Step-by-step explanation:
Just count the amount
Hope that helps :D
Please help !! Correct and first answer I’ll give you brainesttttt ! What is the equation of the line?
Step-by-step explanation:
can u give image PlZzzzz ....
Answer:
Hey!
Your answer should be Y=2x+4
Step-by-step explanation:
Hope this helps!
A stone is dropped into a lake, creating a circular ripple that travels outward at a speed of 60 cmys. Find the rate at which the area within the circle is increasing after (a) 1 s, (b) 3 s, and (c) 5 s. What can you conclude?
Answer:
a) [tex]t = 1\,s[/tex], [tex]\dot A \approx 22619.467\,\frac{cm^{2}}{s}[/tex], b) [tex]t = 3\,s[/tex], [tex]\dot A \approx 67858.401\,\frac{cm^{2}}{s}[/tex], c) [tex]t = 5\,s[/tex], [tex]\dot A \approx 113097.336\,\frac{cm^{2}}{s}[/tex]. The rate at which the area within the circle is increasing linearly inasmuch as time passes by.
Step-by-step explanation:
The area of a circle is described by the following formula:
[tex]A = \pi \cdot r^{2}[/tex]
Where:
[tex]A[/tex] - Area, measured in square centimeters.
[tex]r[/tex] - Radius, measured in centimeters.
Since circular ripple is travelling outward at constant speed, radius can be described by the following equation of motion:
[tex]r (t) = \dot r \cdot t[/tex]
Where:
[tex]\dot r[/tex] - Speed of the circular ripple, measured in centimeters per second.
[tex]t[/tex] - Time, measured in seconds.
The rate of change of the circle is determined by deriving the equation of area and replacing radius with the function in terms of the speed of the circular ripple and time. That is to say:
[tex]\dot A = 2\cdot \pi \cdot r \cdot \dot r[/tex]
[tex]\dot A = 2 \cdot \pi \cdot \dot r^{2}\cdot t[/tex]
Where:
[tex]\dot A[/tex] - Rate of change of the circular area, measured in square centimeters per second.
[tex]\dot r[/tex] - Speed of the circular ripple, measured in centimeters per second.
[tex]t[/tex] - Time, measured in seconds.
If [tex]\dot r = 60\,\frac{cm}{s}[/tex], then:
a) [tex]t = 1\,s[/tex]
[tex]\dot A = 2\cdot \pi \cdot \left(60\,\frac{cm}{s} \right)^{2}\cdot (1\,s)[/tex]
[tex]\dot A \approx 22619.467\,\frac{cm^{2}}{s}[/tex]
b) [tex]t = 3\,s[/tex]
[tex]\dot A = 2\cdot \pi \cdot \left(60\,\frac{cm}{s} \right)^{2}\cdot (3\,s)[/tex]
[tex]\dot A \approx 67858.401\,\frac{cm^{2}}{s}[/tex]
c) [tex]t = 5\,s[/tex]
[tex]\dot A = 2\cdot \pi \cdot \left(60\,\frac{cm}{s} \right)^{2}\cdot (5\,s)[/tex]
[tex]\dot A \approx 113097.336\,\frac{cm^{2}}{s}[/tex]
The rate at which the area within the circle is increasing linearly inasmuch as time passes by.
The answer to – 7x + y = -10
Step-by-step explanation:
y=7x-10
Answer:
[tex]\huge \boxed{y=7x-10}[/tex]
Step-by-step explanation:
[tex]-7x+y=-10[/tex]
[tex]\sf Add \ 7x \ on \ both \ sides.[/tex]
[tex]-7x+y+7x=-10+7x[/tex]
[tex]y=7x-10[/tex]
The temperature is −18.2 Celsius in South Dakota and -9.7 Celsius Minnesota. Which one of the following inequalities correctly compares the temperatures? Choose 1 answer: Which one of the following descriptions is correct?
Answer:
The answer is A) -9.7 > -18.2
Step-by-step explanation:
This is because, when you are thinking about negative numbers, the closer they are to 0, the greater they are. So, it is warmer in Minnesota.
Answer:
A and A
Step-by-step explanation:
Use the sample data and confidence level given below to complete parts (a) through (d). A research institute poll asked respondents if they felt vulnerable to identity theft. In the poll, n equals 1036 and x equals 583 who said "yes." Use a 90 % confidence level.
Required:
a. Find the best point estimate of the population proportion p.
b. Identify the value of the margin of error E =_______
c. Construct the confidence interval.
d. Write a statement that correctly interprets the confidence interval.
1. One has 99% confidence that the sample proportion is equal to the population proportion.
2. There is a 99% chance that the true value of the population proportion will fall between the lower bound and the upper bound.
3. One has 99% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.
Answer:
a. p=0.562
b. E = 0.0253
c. The 90% confidence interval for the population proportion is (0.537, 0.587).
d. We have 90% confidence that the interval (0.537, 0.587) contains the true value of the population proportion.
Step-by-step explanation:
We have to calculate a 90% confidence interval for the proportion.
The sample proportion is p=0.562.
[tex]p=X/n=583/1038=0.562[/tex]
The standard error of the proportion is:
[tex]\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.562*0.438}{1038}}\\\\\\ \sigma_p=\sqrt{0.000237}=0.0154[/tex]
The critical z-value for a 90% confidence interval is z=1.645.
The margin of error (MOE) can be calculated as:
[tex]MOE=z\cdot \sigma_p=1.645 \cdot 0.0154=0.0253[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=p-z \cdot \sigma_p = 0.562-0.0253=0.537\\\\UL=p+z \cdot \sigma_p = 0.562+0.0253=0.587[/tex]
The 90% confidence interval for the population proportion is (0.537, 0.587).
We have 90% confidence that the interval contains the true value of the population proportion.