Answer: 0.9032
Step-by-step explanation:
Given: Mean : [tex]\mu = 20,000\text{ hours}[/tex]
Standard deviation : [tex]\sigma = 1800 \text{ hours}[/tex]
The formula to calculate z is given by :-
[tex]z=\dfrac{x-\mu}{\sigma}[/tex]
For x= 17659
[tex]z=\dfrac{17659-20000}{1800}=−1.30055555556\approx-1.3[/tex]
The P Value =[tex]P(z>-1.5)=1-P(z<1.3)=1- 0.0968005\approx0.9031995\approx 0.9032[/tex]
Hence, the probability that the life span of the monitor will be more than 17,659 hours = 0.9032
To find the probability that the life span of the monitor will be more than 17,659 hours, use the z-score formula and the standard normal distribution table. The probability is approximately 0.0968.
Explanation:To find the probability that the life span of the monitor will be more than 17,659 hours, we need to calculate the z-score and use the standard normal distribution table. The z-score is calculated as:
z = (x - μ) / σ
Where x is the value we want to find the probability for, μ is the mean, and σ is the standard deviation. In this case, x = 17,659, μ = 20,000, and σ = 1800. Plugging these values into the formula, we get:
z = (17659 - 20000) / 1800 = -1.3
Now, we can look up the probability corresponding to the z-score -1.3 in the standard normal distribution table. The probability is approximately 0.0968. Therefore, the probability that the life span of the monitor will be more than 17,659 hours is approximately 0.0968.
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Gianna is going to throw a ball from the top floor of her middle school. When she throws the ball from 32feet above the ground, the function h(t)=−16t2+48t+32 models the height, h, of the ball above the ground as a function of time, t. At what time will the ball reach a height of 64feet?
Answer:
There are two times for the ball to reach a height of 64 feet:
1 second after thrown ⇒ the ball moves upward
2 seconds after thrown ⇒ the ball moves downward
Step-by-step explanation:
* Lets explain the function to solve the problem
- h(t) models the height of the ball above the ground as a function
of the time t
- h(t) = -16t² + 48t + 32
- Where h(t) is the height of the ball from the ground after t seconds
- The ball is thrown upward with initial velocity 48 feet/second
- The ball is thrown from height 32 feet above the ground
- The acceleration of the gravity is -32 feet/sec²
- To find the time when the height of the ball is above the ground
by 64 feet substitute h by 64
∵ h(t) = -16t² + 48t + 32
∵ h = 64
∴ 64 = -16t² + 48t + 32 ⇒ subtract 64 from both sides
∴ 0 = -16t² + 48t - 32 ⇒ multiply the both sides by -1
∴ 16t² - 48t + 32 = 0 ⇒ divide both sides by 16 because all terms have
16 as a common factor
∴ t² - 3t + 2 = 0 ⇒ factorize it
∴ (t - 2)(t - 1) = 0
- Equate each bracket by zero to find t
∴ t - 2 = 0 ⇒ add 2 to both sides
∴ t = 2
- OR
∴ t - 1 = 0 ⇒ add 1 to both sides
∴ t = 1
- That means the ball will be at height 64 feet after 1 second when it
moves up and again at height 64 feet after 2 seconds when it
moves down
* There are two times for the ball to reach a height of 64 feet
1 second after thrown ⇒ the ball moves upward
2 seconds after thrown ⇒ the ball moves downward
Find an equation of the tangent line to the graph of y = g(x) at x = 6 if g(6) = −3 and g'(6) = 5. (Enter your answer as an equation in terms of y and x.)
Answer:
The equation of tangent line is [tex]y=5x-33 [/tex]
Step-by-step explanation:
We need to find out the equation of tangent line.
Given :- g(6)=−3 and g'(6)= 5
If g(6)=−3
then the point on the line for the required tangent is (6,−3)
If g'(6)= 5
then the slope of the tangent at that point is 45
The tangent line can be specified by the slope-point form of the equation:
[tex](y-y_1)=m(x-x_1)[/tex]
which in this case is
[tex](y-(-3))=5(x-6)[/tex]
[tex](y+3)=(5x-30)[/tex]
subtract both the sides by 3,
[tex]y+3-3=5x-30-3[/tex]
[tex]y=5x-33[/tex]
Therefore, the equation of tangent line is [tex]y=5x-33[/tex]
Final answer:
The equation of the tangent line to the graph of y = g(x) at the point where x = 6 is y = 5x - 33, using the point-slope form and the given point (6, -3) with the slope of 5.
Explanation:
To find the equation of the tangent line to the graph at a particular point, we use the point-slope form of a line, given by y - y1 = m(x - x1), where (x1, y1) is the point on the graph and m is the slope at that point. Given that g(6) = -3 and g'(6) = 5, we can substitute these values into the point-slope form to get the equation of the tangent line. The equation is then y + 3 = 5(x - 6), which simplifies to y = 5x - 33.
Nico is saving money for his college education. He invests some money at 7%, and $1200 less than that amount at 3%. The investments produced a total of $174 interest in 1 yr. How much did he invest at each rate?
He invested $____at 7% and _____ at 3%.
Answer:
Nico invest [tex]\$2,100[/tex] at 7% and [tex]x=\$900[/tex] at 3%
Step-by-step explanation:
we know that
The simple interest formula is equal to
[tex]I=P(rt)[/tex]
where
I is the Final Interest Value
P is the Principal amount of money to be invested
r is the rate of interest
t is Number of Time Periods
in this problem we have
At 7%
[tex]t=1\ years\\ P=\$x\\r=0.07[/tex]
substitute in the formula above
[tex]I1=x(0.07*1)[/tex]
[tex]I1=0.07x[/tex]
At 3%
[tex]t=1\ years\\ P=\$(x-1,200)\\r=0.03[/tex]
substitute in the formula above
[tex]I2=(x-1,200)(0.03*1)[/tex]
[tex]I2=0.03x-36[/tex]
The total interest is equal to
I=I1+I2
I=$174
substitute
[tex]174=0.07x+0.03x-36[/tex]
[tex]0.10x=174+36[/tex]
[tex]0.10x=210[/tex]
[tex]x=\$2,100[/tex]
[tex]x-1,200=2,100-1,200=\$900[/tex]
therefore
Nico invest [tex]\$2,100[/tex] at 7% and [tex]x=\$900[/tex] at 3%
A boat leaves a dock at 7:00 PM and travels due south at a speed of 20 km/h. Another boat has been heading due east at 15 km/h and reaches the same dock at 8:00 PM. How many minutes after 7:00 PM were the two boats closest together? (Round your answer to the nearest minute.) min
Answer:
22 minutes after 7:00 P.M. they will be closest.
Step-by-step explanation:
A boat heading south travelling for t hours at the rate of 20 km/h, so the distance x = 20t
The another boat will reach the dock after travelling another 1-t hours at the rate of 15 km/h, so the distance =
y = 15 - 15t
D = d² = x² + y²
D = (20t)² + (15 - 15t)²
dD/dt = -2(15² )( 1-t ) +2 × 20² × t
dD/dt = 2 (15² + 20²) × t -2 ( 15 )² = 0
t = [tex]\frac{2(15)^{2}}{(2\times15^{2}+2\times20^{2})}[/tex]
t = 0.36 hours = 0.36 × 60 = 21.6 minutes ≈ 22 minutes
Therefore, the distance is minimized 22 minutes after 7 pm.
The two boats were closest together 12 minutes after 7:00 PM.
Explanation:To find the time when the two boats were closest together, we can first determine the position of each boat at 8:00 PM. The boat traveling south will have traveled for 1 hour at a speed of 20 km/h, so it would be 20 km south of the dock. The boat traveling east will have traveled for 1 hour at a speed of 15 km/h, so it would be 15 km east of the dock. We can then calculate the distance between the two boats by using the Pythagorean theorem. The distance is the square root of the sum of the squares of the distances traveled south and east, which is approximately 25 km. Since both boats started at the dock at 7:00 PM, to find the time when they were closest together, we can subtract the time traveled by the boat heading south until it reaches the closest point to the other boat from 60 minutes. The boat heading south will have traveled (20/25) * 60 minutes, which is 48 minutes. So, the two boats were closest together 12 minutes after 7:00 PM.
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Find the degree 3 Taylor polynomial P3(x) of function f(x)=(-3x+15)^(3/2) at a=2
Take up to the third-order derivative:
[tex]f(x)=(-3x+15)^{3/2}[/tex]
[tex]f'(x)=\dfrac32(-3x+15)^{1/2}(-3)=-\dfrac92(-3x+15)^{1/2}[/tex]
[tex]f''(x)=-\dfrac94(-3x+15)^{-1/2}(-3)=\dfrac{27}4(-3x+15)^{-1/2}[/tex]
[tex]f'''(x)=-\dfrac{27}8(-3x+15)^{-3/2}(-3)=\dfrac{81}8(-3x+15)^{-3/2}[/tex]
Evaluate each derivative at [tex]x=a=2[/tex]:
[tex]f(2)=9^{3/2}=27[/tex]
[tex]f'(2)=-\dfrac929^{1/2}=-\dfrac{27}2[/tex]
[tex]f''(2)=\dfrac{27}4\dfrac1{9^{1/2}}=\dfrac94[/tex]
[tex]f'''(2)=\dfrac{81}8\dfrac1{9^{3/2}}=\dfrac38[/tex]
Then the Taylor polynomial is
[tex]P_3(x)=f(2)+f'(2)(x-2)+\dfrac{f''(2)}2(x-2)^2+\dfrac{f'''(2)}6(x-2)^3[/tex]
[tex]P_3(x)=27-\dfrac{27}2(x-2)+\dfrac98(x-2)^2+\dfrac1{16}(x-2)^3[/tex]
Translate the Phrase : "" Nine times the difference of a number and 3"" into an Algebraic Expression and then Simplify . Let X represent the real number .
Step-by-step explanation:
[tex]x-\text{the number}\\\\\text{Nine times the difference of a number and 3}:\\\\\boxed{9\times(x-3)=9(x-3)}[/tex]
[tex]9(x-3)\qquad\text{use the distributive property}\ a(b+c)=ab+ac\\\\=9x+(9)(-3)\\\\=9x-27[/tex]
Consider two sizes of disk, both of mass M. One size of disk has radius R; the other has radius 4R. System A consists of two of the larger disks rigidly connected to each other with a common axis of rotation. System B consists of one of the larger disks and a number of the smaller disks rigidly connected with a common axis of rotation. If the moment of inertia for system A = the moment of inertia for system B, how many of the smaller disks are in system B? 1 4 10 16
Answer:
16
Step-by-step explanation:
Moment of inertia of a disk is proportional to its mass and to the square of its radius. For two disks with the same mass, the larger one will have a moment of inertia that is (4R/R)^2 = 16 times that of the smaller one.
It will take 16 smaller disks to make the systems have the same moment of inertia.
e. Which of the following is NOT a possible probability? a. 25/100 b. 1.25 c. 1 d. 0
Answer:
B. 1.25
Step-by-step explanation:
Probability is as below
[tex]0 \leqslant p(a) \leqslant 1[/tex]
When P(A) = 0, it is an unlikely event
When P(A) = 1, it is a certain event
David, a platform diver, dives into the pool during practice. The height of David above the water at any given time, s, can be modeled by the quadratic function h(s).
Each of the following functions is a different form of the quadratic model for the situation above. Which form would be the most helpful if attempting to determine the time required for David to enter the water?
A. h(s) = -4.9(s - 2)(s + 1)
B. h(s) = -4.9s(s - 1) + 9.8
C. h(s) = -4.9(s - 0.5)2 + 11.025
D. h(s) = -4.9s2 + 4.9s + 9.8
Check the pictures below.
if we knew the roots/solutions of the equation, we can set h(s) = 0 and solve for "s" to find out how many seconds is it when the height is 0.
if you notice in the first picture, when f(x) = 0, is when the parabola hits a root/solution or the ground, for David he'll be hitting the water surface, and the equation that has both of those roots/solutions conspicuous is
h(s) = -4.9(s - 2)(s + 1).
Consider a bag that contains 220 coins of which 6 are rare Indian pennies. For the given pair of events A and B, complete parts (a) and (b) below. A: When one of the 220 coins is randomly selected, it is one of the 6 Indian pennies. B: When another one of the 220 coins is randomly selected (with replacement), it is also one of the 6 Indian pennies. a. Determine whether events A and B are independent or dependent. b. Find P(A and B), the probability that events A and B both occur.
Answer:
a. The two events are dependent.
b. [tex]P(A\cap B)[/tex]= [tex]\frac{1}{220}[/tex].
Step-by-step explanation:
Given
Total coins =220
Number of Indian pennies= 6
A: When one of the 220 coins is randomly selected, it is one of the Indian pennies.
Therefore , the probability of getting an Indian pennies=[tex]\frac{6}{220 }[/tex]
By using formula of probability=[tex]\frac{Number \; of\; favourable\; cases}{total\; number \; of \;cases}[/tex]
Probability of getting an Indian pennies=[tex]\frac{3}{110}[/tex]
B: When another one of the 220 coins is randomly selected( with replacement) , It is also one of the Indian pennies.
Therefore, probability of getting an Indian pennies=[tex]\frac{6}{220}[/tex]
Probability of getting an Indian pennies =[tex]\frac{3}{110}[/tex]
[tex]A\cap B[/tex]: 1
[tex]P(A\cap B)=\frac{1}{220}[/tex]
If two events are independent. Then
[tex]P(A\cap B)= P(A)\times p(B)[/tex]
P(A).P(B)= [tex]\frac{3}{110} \times \frac{3}{110}[/tex]=[tex]\frac{9}{12100}[/tex]
Hence, [tex]P(A\cap B)\neq P(A).P(B)[/tex]
Therefore, the two events are dependent.
b. Probability that events A and B both occur
Number of favourable cases when both events A and B occur=1
Total coins=220
Probability=[tex]\frac{Number \; of\; favourable \; cases}{Total\; number\; of\; cases}[/tex]
[tex]P(A\cap B)=\frac{1}{220}[/tex]
A box has 14 camera of which 6 are refurbished and 8 are new. If four of these 14 cameras are selected at random without replacement, what is the probability that (i) one new camera will be selected? (ii) at most one new camera will be selected?
Answer:
160/1001, 175/1001
Step-by-step explanation:
i) There are:
₈C₁ ways to choose 1 new camera from 8 new cameras
₆C₃ ways to choose 3 refurbished cameras from 8 refurbished cameras
₁₄C₄ ways to choose 4 cameras from 14 cameras
The probability is:
P = ₈C₁ ₆C₃ / ₁₄C₄
P = 8×20 / 1001
P = 160 / 1001
P ≈ 0.160
ii) At most one new camera means either one new camera or no new cameras. We already found the probability of one new camera. The probability of no new cameras is the same as the probability of choosing 4 refurbished cameras:
P = ₆C₄ / ₁₄C₄
P = 15 / 1001
So the total probability is:
P = 160/1001 + 15/1001
P = 175/1001
P ≈ 0.175
To find the probability that one new camera will be selected, use the binomial coefficient and calculate the probability of selecting one new camera and three cameras that are not new. To find the probability of at most one new camera, calculate the probabilities of selecting zero new cameras and one new camera and add them together.
Explanation:To find the probability that one new camera will be selected, we need to calculate the probability of selecting one new camera and three cameras that are not new. The total number of ways to select four cameras from 14 without replacement is given by the binomial coefficient 14 choose 4, which is equal to 14!/(4!(14-4)!). The probability of selecting one new camera is given by the product of the probability of selecting one new camera (8/14) and the probability of selecting three cameras that are not new (6/13 * 5/12 * 4/11). To find the probability that at most one new camera will be selected, we need to calculate the probabilities of selecting zero new cameras and one new camera and add them together.
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Two automobiles left simultaneously from cities A and B heading towards each other and met in 5 hours. The speed of the automobile that left city A was 10 km/hour less than the speed of the other automobile. If the first automobile had left city A 4 1 2 hours earlier than the other automobile left city B, then the two would have met 150 km away from B. Find the distance between A and B.
Answer:
450 km
Step-by-step explanation:
Let's say Va is the speed of the car from city A, Ta is the time it spent traveling, and Da is the distance it traveled.
Similarly, Vb is the speed of the car from city B, Tb is the time it spent traveling, and Db is the distance it traveled.
Given:
Va = Vb - 10
Ta₁ = Tb₁ = 5
Ta₂ = Tb₂ + 4.5
Db₂ = 150
Find:
D = Da₁ + Db₁ = Da₂ + Db₂
Distance = rate × time
In the first scenario:
Da₁ = Va Ta₁
Da₁ = (Vb - 10) (5)
Da₁ = 5Vb - 50
Db₁ = Vb Tb₁
Db₁ = Vb (5)
Db₁ = 5Vb
So:
D = Da₁ + Db₁
D = 10Vb - 50
In the second scenario:
Da₂ = Va Ta₂
Da₂ = (Vb - 10) (Tb₂ + 4.5)
Da₂ = Vb Tb₂ + 4.5Vb - 10Tb₂ - 45
Db₂ = Vb Tb₂
150 = Vb Tb₂
Substituting:
Da₂ = 150 + 4.5Vb - 10Tb₂ - 45
Da₂ = 105 + 4.5Vb - 10Tb₂
Da₂ = 105 + 4.5Vb - 10 (150 / Vb)
Da₂ = 105 + 4.5Vb - (1500 / Vb)
So:
D = Da₂ + Db₂
D = 105 + 4.5Vb - (1500 / Vb) + 150
D = 255 + 4.5Vb - (1500 / Vb)
Setting this equal to the equation we found for D from the first scenario:
10Vb - 50 = 255 + 4.5Vb - (1500 / Vb)
5.5Vb - 305 = -1500 / Vb
5.5Vb² - 305Vb = -1500
5.5Vb² - 305Vb + 1500 = 0
11Vb² - 610Vb + 3000 = 0
(Vb - 50) (11Vb - 60) = 0
Vb = 50, 5.45
Since Vb > 10, Vb = 50 km/hr.
So the distance between the cities is:
D = 10Vb - 50
D = 10(50) - 50
D = 450 km
According to a study, 80% of K-12 schools or districts in a country use digital content such as ebooks, audio books, and digital textbooks. Of these 80%, 5 out of 10 use digital content as part of their curriculum. Find the probability that a randomly selected school or district uses digital content and uses it as part of their curriculum.
Answer: There is a probability of 40% of getting a school or district uses digital content and uses it as part of their curriculum.
Step-by-step explanation:
Since we have given that
Probability that schools or districts in a country use digital content = 80%
Probability that schools uses digital content as a part of their curriculum out of 80% = [tex]\dfrac{5}{10}[/tex]
So, the probability that a selected school or district uses digital content and uses it as a part of their curriculum is given by
[tex]\dfrac{80}{100}\times \dfrac{5}{10}\\\\=0.8\times 0.5\\\\=0.4\\\\=40\%[/tex]
Hence, there is a probability of 40% of getting a school or district uses digital content and uses it as part of their curriculum.
The probability that a randomly selected school or district uses digital content and uses it as part of their curriculum is 40%.
Explanation:To find the probability that a randomly selected school or district uses digital content and uses it as part of their curriculum, we need to multiply the probabilities of these events occurring.
Given that 80% of K-12 schools or districts use digital content and 5 out of 10 of these schools use it as part of their curriculum, we can calculate the probability as:
P(Uses digital content and uses it as part of curriculum) = P(Uses digital content) x P(Uses it as part of curriculum | Uses digital content)
Substituting the values, we have:
P(Uses digital content and uses it as part of curriculum) = 0.80 x 0.50 = 0.40 or 40%
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Jessica is deciding on her schedule for next semester. She must take each of the following classes: English 101, Spanish 102, Biology 102, and College Algebra. If there are 15 sections of English 101, 9 sections of Spanish 102, 11 sections of Biology 102, and 15 sections of College Algebra, how many different possible schedules are there for Jessica to choose from? Assume there are no time conflicts between the different classes.
Jessica has a total of 22,275 different possible schedules to choose from for her next semester given the number of sections for each class and assuming there are no time conflicts.
Explanation:Jessica is creating her semester schedule and there are 15 sections of English 101, 9 sections of Spanish 102, 11 sections of Biology 102, and 15 sections of College Algebra. To figure out how many different possible schedules are available, we need to multiply the number of sections for each class.
Therefore, the total number of different possible schedules Jessica can choose is calculated as follows:
15 (English 101) * 9 (Spanish 102) * 11 (Biology 102) * 15 (College Algebra) = 22,275 possible schedules.
This is under the assumption that there are no time conflicts between the different classes.
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which of the following is the quotient of .4375 divided by .35
Answer:
Quotient will be 1.25
Step-by-step explanation:
First we convert decimal numbers to fractions. So write down the decimal divided by 1 and then multiply both top and bottom with 10 for every number after decimal point.
Here we found for .4375 = [tex]\frac{4375}{10000}[/tex]
and .35 = [tex]\frac{35}{100}[/tex]
Now we divide both the numbers as
= [tex]\frac{\frac{4375}{1000} }{\frac{35}{100} }[/tex]
= [tex]\frac{4375}{1000}[/tex] × [tex]\frac{100}{35}[/tex]
= [tex]\frac{125}{100}[/tex]
= 1.25
Quotient will be 1.25
Final answer:
The quotient of 0.4375 divided by 0.35 is 1.25, which rounded to the tenths place is 1.3.
Explanation:
The student is asking to find the quotient of two decimal numbers, which is a basic arithmetic operation involving division. The numbers are 0.4375 and 0.35. To find the quotient, simply divide 0.4375 by 0.35.
Using a calculator or performing the division manually, you would proceed as follows:
Adjust the decimals by multiplying both numbers by 100 to make them whole numbers, resulting in 43.75 divided by 35.
Perform the division to get the preliminary result: 43.75 / 35 = 1.25.
Since we need to round the final answer to the tenths place based on the least precise number given (35.5 g), round 1.25 to one decimal place, which is 1.3 (1.25 rounds up because the next digit, 5, is equal to or greater than 5).
Therefore, the quotient of 0.4375 divided by 0.35, rounded to the tenths place, is 1.3.
The SAT scores have an average of 1200 with a standard deviation of 60. A sample of 36 scores is selected.What is the probability that the sample mean will be larger than 1224? Round your answer to three decimal places.
Answer: 0.008
Step-by-step explanation:
Given: Mean : [tex]\mu=1200[/tex]
Standard deviation : [tex]\sigma = 60[/tex]
Sample size : [tex]n=36[/tex]
The formula to calculate z-score is given by :_
[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
For x= 1224, we have
[tex]z=\dfrac{1224-1200}{\dfrac{60}{\sqrt{36}}}=2.4[/tex]
The P-value = [tex]P(z>2.4)=1-P(z<2.4)=1-0.9918024=0.0081976\approx0.008[/tex]
Hence, the probability that the sample mean will be larger than 1224 =0.008
A man started walking a kilometer a day. If, instead of walking a kilometer everyday, he walked 90% of the distance he walked the previous day, after how many days would he be walking less than 0.5km each day?
PLEASE give an explanation with your answer!!
Answer:
After the 6th day, beginning with the 7th day, he is walking less than 0.5 km.
Step-by-step explanation:
To find a percent of a number, change the percent to a decimal and multiply by the number. To find 90% of a number, change 90% to a decimal and multiply by the number.
90% = 0.9
He first walks 1 km. The next day, he walks 90% of 1 km. To find 90% of 1 km, multiply 0.9 by 1 km. It is 0.9 km. For the next day, he walks 90% of 0.9 km, which is 0.9 * 0.9 km = 0.81 km. To find how much he walks each day, multiply what he walked on the previous day by 0.9.
Now you can find out how much he walks each day until you see he walks less than 0.5 km.
Day 1: 1 km
Day 2: 1 km * 0.9 = 0.9 km
Day 3: 0.9 km * 0.9 = 0.81 km
Day 4: 0.81 km * 0.9 = 0.729 km
Day 4: 0.729 km * 0.9 = 0.6561
Day 5: 0.6561 km * 0.9 = 0.59049 km
Day 6: 0.59049 km * 0.9 = 0.531441 km
Day 7: 0.531441 km * 0.9 = 0.4782969 km
On the 6th day, he is still walking more than 0.5 km, but by the 7th day, he is walking less than 0.5 km.
Answer: After the 6th day, beginning with the 7th day, he is walking less than 0.5 km.
After the 6th day, beginning with the 7th day, he exists walking less than 0.5 km.
How to find the percent of a number?To find a percent of a number, change the percent to a decimal and multiply by the number. To discover 90% of a number, change 90% to a decimal and multiply by the number.
90% = 0.9
He first walks 1 km. The next day, he walks 90% of 1 km.
To discover 90% of 1 km, multiply 0.9 by 1 km. It is 0.9 km.
For the subsequent day, he walks 90% of 0.9 km, which exists
0.9 [tex]*[/tex] 0.9 km = 0.81 km.
To find how much he walks each day, multiply what he walked on the last day by 0.9.
Now find out how much he walks each day until sees he walks less than 0.5 km.
Day 1: 1 km
Day 2: 1 km [tex]*[/tex] 0.9 = 0.9 km
Day 3: 0.9 km [tex]*[/tex] 0.9 = 0.81 km
Day 4: 0.81 km [tex]*[/tex] 0.9 = 0.729 km
Day 4: 0.729 km [tex]*[/tex] 0.9 = 0.6561
Day 5: 0.6561 km [tex]*[/tex] 0.9 = 0.59049 km
Day 6: 0.59049 km [tex]*[/tex] 0.9 = 0.531441 km
Day 7: 0.531441 km [tex]*[/tex] 0.9 = 0.4782969 km
On the 6th day, he stands still walking more than 0.5 km, but by the 7th day, he exists walking less than 0.5 km.
Answer: After the 6th day, beginning with the 7th day, he exists walking less than 0.5 km.
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Devise the exponential growth function that fits the given data, then answer to accompanying question. Be sure to identify the references point (t = 0) and The current population of a town is 70,000 and is growing exponentially. If the population to be 75,000 in 10 years, then what will be the populations 20 years What is the reference point (t = 0)? the initial population 70,000 the current year the population in 10 years, 75,000 What are the units of time? percent people decades years Write the exponential growth function. Round any numerical values to three decimal places as needed. y(t) = What is the population 20 years from now?
Answer:
In 20 years, the population will be about 80.3 thousand people
Step-by-step explanation:
If our first time is 0 and the population that goes along with that time is 70,000, we have a coordinate point where x is the time (0), and y is the population at that time (70). Our next time is 10 years later, when the population is 75,000. The coordinate point for that set of data is (10, 75). Now we will use those 2 points in the standard form of an exponential equation to write the model for this particular situation.
Exponential equations are of the form
[tex]y=a(b)^x[/tex]
where x and y are the coordinates from our points, one at a time; a is the initial value, and b is the growth rate. Filling in an equation with the first set of data:
[tex]70=a(b)^0[/tex]
Anything raised to the power of 0 = 1, so b to the power of 0 = 1 and we simply have that a = 70.
Now we use that value of a along with the x and y from the next coordinate pair to solve for b:
[tex]75=70(b)^{10}[/tex]
Begin by dividing both sides by 70 to get
[tex]1.071428571=b^{10}[/tex]
Undo the power of 10 on the right by taking the 10th root of both sides:
[tex](1.071428571)^{\frac{1}{10}}=(b^{10})^{\frac{1}{10}}[/tex]
On the right side we simply have b now, and on the left we have
1.006923142=b
Now we have a and b to write the model for this situation:
[tex]y=70(1.006923142)^x[/tex]
We need to find y, the population, in x = 20 years:
[tex]y=70(1.006923142)^{20}[/tex]
Raise the parenthesis to the 20th power giving you
y = 70(1.147959784) and
y = 80.3 thousand people
Final answer:
The exponential growth function that fits the given data is y(t) = a * (1 + r)^t. Using this function, we can find the population 20 years from now.
Explanation:
The exponential growth function that fits the given data is:
y(t) = a * (1 + r)^t
where:
a represents the initial population (70,000)r represents the growth rate per yeart represents the time in yearsTo find the growth rate per year, we can use the formula: r = (P/P0)^(1/t) - 1
Given that the population is projected to be 75,000 in 10 years, we can substitute these values into the formula to find the growth rate:
r = (75,000/70,000)¹/¹⁰ - 1 ≈ 0.035
The exponential growth function becomes:
y(t) = 70,000 * (1 + 0.035)^t
To find the population 20 years from now, we can substitute t = 20 into the exponential growth function:
y(20) = 70,000 * (1 + 0.035)²⁰ ≈ 95,212
Analyze the diagram below and answer the question that follows.
Answer:
The right statement is sin(J) = cos(L) ⇒ answer D
Step-by-step explanation:
* Lets describe the figure
- LKJ is a right triangle, where K is a right angle
∵ m∠K = 90°
∵ LJ is opposite to angle K
∴ LJ is the hypotenuse
∵ LJ = 219
∵ KJ = 178
- By using Pythagoras Theorem
∵ (LJ)² = (LK)² + (KJ)²
∴ (219)² = (LK)² + (178)² ⇒ subtract (178)² from both sides
∴ (LK)² = (219)² - (178)²
∴ (LK)² = 16277
∴ LK = √16277 = 127.58
* Lets revise how to find the trigonometry function
# sin Ф = opposite/hypotenuse
# cos Ф = adjacent/hypotenuse
# tan Ф = opposite/adjacent
∵ LK is the opposite side to angle J
∵ LJ is the hypotenuse
∵ sin(J) = LK/LJ
∵ LK = 127.58 , LJ = 219
∴ sin(J) = 127.58/219 = 0.583
∵ LK is the adjacent side to angle L
∵ LJ is the hypotenuse
∵ cos(L) = LK/LJ
∵ LK = 127.58 , LJ = 219
∴ cos(L) = 127.58/219 = 0.583
∴ sin(J) = cos(L)
* The right statement is sin(J) = cos(L)
A packet of sour worms contains four strawberry, four lime, two black currant, two orange sour, and three green apple worms. What is the probability that Dustin will choose a green apple sour worm, P(green apple)?
Answer:
3/15 or 0.2
Step-by-step explanation:
Consider the daily market for hot dogs in a small city. Suppose that this market is in long-run competitive equilibrium with many hot dog stands in the city, each one selling the same kind of hot dogs. Therefore, each vendor is a price taker and possesses no market power.
Answer:IF each vendor has his own price or (ppower) so far every single vendor will have his own price.
Step-by-step explanation:
The graph show\ing the demand (D) and supply (S = MC) curves in the market for hot dogs indicate: Competitive market.
Competitive marketIn a market were their is competition, when demand and supply curves intersect this indicate market equilibrium.
Based on the graph the market equilibrium price will be $1.50 per hot dog while on the other hand the market equilibrium quantity will be 250 hot dogs which is the point were demand and supply intersect.
Inconclusion the market for hot dogs indicate: Competitive market.
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Consider the function f(x)=3−2x2,−3≤x≤1 The absolute maximum value is? and this occurs at x equals? The absolute minimum value is? and this occurs at x equals?
Answer:
Step-by-step explanation:
so u do 2+2 4=243==32===3=424=4=234=234=32=43=4=34
A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 10.0 m/s and accelerates at the rate of 0.500 m/s2. If the racer was 300 m from the finish when starting to accelerate, what is the racer’s final velocity in m/s
Answer:
20m/s
Step-by-step explanation:
This can be solved using the acceleration / velocity equations.
Specifically,
v² = u² + 2as
Where
v = final velocity = what we need to find
u = initial velocity = given as 10.0m/s
a = acceleration = given as 0.5m/s²
s = distance = 300m
Hence,
v² = 10² + (2)(0.5) (300)
= 100 + 300
=400
v = √400 = 20 m/s
PLEASE GIVE AN EXPLANATION WITH YOUR ANSWER! The table below shows the change in the value of shares over the last three years. Calculate the percentage change in shares from the start of 2013 to the end of 2015.
First, lets convert them into multipliers:
The multiplier for a:
25% increase = 1.25
40% decrease = 0.6
40% increase = 1.4
Now to work out the overall percentage change, we just times all of the multipliers together, and convert it back to a percentage:
1.25 x 0.6 x 1.4 = 1.05
So the overall multiplier is 1.05
And a multiplier of 1.05 = a 5% increase.
That means that the percentage change is + 5%
_________________________________________
Answer:
The percentage change in shares from the start of 2013 to the end of 2015 is:
+ 5%
_______________________________________
Note: if you haven't been taught multipliers - then ask and I'll try my best to explain!
Gravel is being dumped from a conveyor belt at a rate of 40 ft3/min. It forms a pile in the shape of a right circular cone whose base diameter and height are always the same. How fast is the height of the pile increasing when the pile is 13 ft high?
Answer:
[tex]\frac{dh}{dt}=\frac{160}{169\pi } ft/min[/tex]
Step-by-step explanation:
This is a classic related rates problem. Gotta love calculus!
Start out with the formula for the volume of a cone, which is
[tex]V=\frac{1}{3}\pi r^2h[/tex]
and with what we know, which is [tex]\frac{dV}{dt}=40[/tex]
and the fact that the diameter = height (we will come back to that in a bit).
We need to find [tex]\frac{dh}{dt}[/tex] when h = 13
The thing we need to notice now is that there is no information given to us that involves the radius. It does, however, give us a height. We need to replace the r with something in terms of h. Let's work on that first.
We know that d = h. Because d = 2r, we can say that 2r = h, and solving for r gives us that [tex]r=\frac{h}{2}[/tex].
Now we can rewrite the formula with that replacement:
[tex]V=\frac{1}{3}\pi (\frac{h}{2})^2h[/tex]
Simplify that all the way down to
[tex]V=\frac{1}{12}\pi h^3[/tex]
The derivative of that function with respect to time is
[tex]\frac{dV}{dt}=\frac{1}{12}\pi(3h^2)\frac{dh}{dt}[/tex]
Filling in what we have gives us this:
[tex]40=\frac{1}{12}\pi (3)(13)^2\frac{dh}{dt}[/tex]
Solve that for the rate of change of the height:
[tex]\frac{dh}{dt}=\frac{160}{169\pi } \frac{ft}{min}[/tex]
or in decimal form:
[tex]\frac{dh}{dt}=.95\pi \frac{ft}{min}[/tex]
This involves relationship between rates using Calculus.
dh/dt = 0.3 ft/min
We are given;Volumetric rate; dv/dt = 40 ft³/min
height of pile; h = 13 ft
We are not given the diameter here but as we are dealing with a right circular cone, we will assume that the diameter is equal to the height.
Thus; diameter; d = 13 ft
radius; r = h/2 = d/2 = 13/2
radius; r= 6.5 ft
Formula for volume of a cone is;V = ¹/₃πr²h
We want to find how fast the height is increasing and this is dh/dt.
Thus, we will need to express r in the volume formula in terms of h;
V = ¹/₃π(h/2)²h
V = ¹/₃π(h²/4)h
V = ¹/₁₂πh³
differentiating both sides with respect to time t gives;dV/dt = 3(¹/₁₂πh²)dh/dt
dV/dt = ¹/₄πh²(dh/dt)
Plugging in the relevant values, we have;
40 = ¹/₄π × 13² × (dh/dt)
dh/dt = (40 × 4)/(π × 13²)
dh/dt = 0.3 ft/min
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At a Shop on Times Square three "" I LOVE NY"" T- Shirts Sell every 10 minutes for $ 19.95 each . Every 45 minutes one Yankee's Hat sells for $ 24.95 . The Shop is Open from 9 Am to 9 Pm Every day .So,the Question is How Many T-Shirts Are Sold in a Week ???
The shop sells 216 "I LOVE NY" T-shirts each day. Therefore, over the course of one week, the shop sells 1512 T-shirts.
Explanation:The shop in Times Square is open from 9 am to 9 pm, which means the shop operates for 12 hours. Since there are 60 minutes in an hour, this shop is open for a total of 720 minutes each day.
Three "I LOVE NY" T-shirts are sold every 10 minutes. So, in 720 minutes, the number of T-shirts sold would be 720 ÷ 10 = 72 sets of three T-shirts. Therefore, 72 sets x 3 shirts = 216 T-shirts are sold per day.
Finally, to calculate the weekly total, it is necessary to multiply the daily total by 7 (the number of days in a week). So, 216 T-shirts x 7 days = 1512 T-shirts sold in one week.
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Use the formula and the values given to find the value of the remaining variable.
C = 2€r; C = 22€
R=_______
Answer:
r = 11
Step-by-step explanation:
Substituting the given value for C, we have ...
22€ = 2€·r
Dividing by the coefficient of r, we get
22€/(2€) = r = 11
To find the value of 'r' using the equation C = 2€r; C = 22€, you simply need to substitute the given value of C into the equation and solve for r. In this case, r equals 11.
Given : C = 2€r; C = 22€
To find the value of r, substitute C = 22€ into the equation:
22€ = 2€r
Divide both sides by 2€ to isolate r:
r = 11
What is the difference?
X/x2+3x+2 - 1/(X+ 2)(x+1)
Answer:
D
Step-by-step explanation:
[tex]\[\frac{x}{x^{2} +3x+2} -\frac{1}{(x+2)(x+1)} =\frac{x}{x^2+3x+2} -\frac{1}{x(x+1)+2(x+1)} =\frac{x}{x^{2}+3x+2 } -\frac{1}{x^{2} +2x+x+2} =\frac{x}{x^{2} +3x+2} -\frac{1}{x^{2} +3x+2} =\frac{x-1}{x^{2} +3x+2} \][/tex]
Answer:
he is correct
Step-by-step explanation:
dddddddddddddddddddddddddddd
This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the extreme values of the function subject to the given constraint. f(x1, x2, ..., xn) = x1 + x2 + ... + xn; x12 + x22 + ... + xn2 = 4
[tex]f(x_1,\ldots,x_n)=x_1+\cdots+x_n=\displaystyle\sum_{i=1}^nx_i[/tex]
[tex]{x_1}^2+\cdots+{x_n}^2=\displaystyle\sum_{i=1}^n{x_i}^2=4[/tex]
The Lagrangian is
[tex]L(x_1,\ldots,x_n,\lambda)=\displaystyle\sum_{i=1}^nx_i+\lambda\left(\sum_{i=1}^n{x_i}^2-4\right)[/tex]
with partial derivatives (all set equal to 0)
[tex]L_{x_i}=1+2\lambda x_i=0\implies x_i=-\dfrac1{2\lambda}[/tex]
for [tex]1\le i\le n[/tex], and
[tex]L_\lambda=\displaystyle\sum_{i=1}^n{x_i}^2-4=0[/tex]
Substituting each [tex]x_i[/tex] into the second sum gives
[tex]\displaystyle\sum_{i=1}^n\left(-\frac1{2\lambda}\right)^2=4\implies\dfrac n{4\lambda^2}=4\implies\lambda=\pm\frac{\sqrt n}4[/tex]
Then we get two critical points,
[tex]x_i=-\dfrac1{2\frac{\sqrt n}4}=-\dfrac2{\sqrt n}[/tex]
or
[tex]x_i=-\dfrac1{2\left(-\frac{\sqrt n}4\right)}=\dfrac2{\sqrt n}[/tex]
At these points we get a value of [tex]f(x_1,\cdots,x_n)=\pm2\sqrt n[/tex], i.e. a maximum value of [tex]2\sqrt n[/tex] and a minimum value of [tex]-2\sqrt n[/tex].
An inner city revitalization zone is a rectangle that is twice as long as it is wide. The width of the region is growing at a rate of 40 m per year at a time when the region is 290 m wide. How fast is the area changing at that point in time?
Answer:
20
Step-by-step explanation: