Answer:
x ≥ -2
Step-by-step explanation:
We need to solve 3 x plus 10 greater than or equal to 4
3x + 10 ≥ 4
Solving and finding the value of x
Adding -10 on both sides
3x + 10 -10 ≥ 4 -10
3x ≥ -6
Divide by 3
3x/3 ≥ -6/3
x ≥ -2
So, the solution is x ≥ -2
Answer:
x [tex]\geq[/tex] - 2
Step-by-step explanation:
We need to solve 3 x plus 10 greater than or equal to 4
3x + 10 ≥ 4
Solving and finding the value of x
Adding -10 on both sides
3x + 10 -10 ≥ 4 -10
3x ≥ -6
Divide by 3
3x/3 ≥ -6/3
x ≥ -2
So, the solution is x ≥ -2
Mandy has an IQ of 115. We know that the mean () IQ is 100 with a standard deviation of 15. There are 100 people in Mandy’s Alcoholics Anonymous meeting. Taken at random, how many members are smarter than Mandy
Answer: 16
Step-by-step explanation:
Given : Mean : [tex]\mu=100[/tex]
Standard deviation : [tex]\sigma =15[/tex]
The value of z-score is given by :-
[tex]z=\dfrac{x-\mu}{\sigma}[/tex]
For x= 115
[tex]z=\dfrac{115-100}{15}=1[/tex]
The p-value : [tex]P(z>1)=1-P(z<1)=1- 0.8413447=0.1586553[/tex]
Now, the number of people smarter than Many is given by :_
[tex]100\times0.1586553=15.86553\approx16[/tex]
Hence, there are 16 members smarter than Mandy.
Find the unpaid balance on the debt. (Round your answer to the nearest cent.)
After 6 years of monthly payments on $190,000 at 3% for 25 years.
Answer:
unpaid balance is $156439.86
Step-by-step explanation:
present value= $190000
rate = 3% monthly = 3/12 = 0.25 % = 0.0025
time = 25 year = 25 × 12 300
to find out
the unpaid balance
solution
first we calculate monthly payment
payment = principal ( 1- (1+r)^(-t) /r )
and we put all value here
190000 = principal ( 1- (1+0.0025)^(-300) / 0.0025 )
principal = 190000 / ( 1- (1+0.0025)^(-300) / 0.0025 )
principal = 901
so after 6 year unpaid is
present value = principal ( 1- (1+r)^(-t) /r )
put here principal 901 and rate 0.0025 and time 300 - (year )
time = 300 - (6*12) = 300 - 96 = 228
payment = 901 ( 1- (1+0.0025)^(-228) / 0.0025 )
payment = 901 ( 713.63)
so unpaid balance is $156439.86
Solve this differential Equation by using power series
y''-x^2y=o
We're looking for a solution
[tex]y=\displaystyle\sum_{n=0}^\infty a_nx^n[/tex]
which has second derivative
[tex]y''=\displaystyle\sum_{n=2}^\infty n(n-1)a_nx^{n-2}=\sum_{n=0}^\infty(n+2)(n+1)a_{n+2}x^n[/tex]
Substituting these into the ODE gives
[tex]\displaystyle\sum_{n=0}^\infty(n+2)(n+1)a_{n+2}x^n-\sum_{n=0}^\infty a_nx^{n+2}=0[/tex]
[tex]\displaystyle\sum_{n=0}^\infty(n+2)(n+1)a_{n+2}x^n-\sum_{n=2}^\infty a_{n-2}x^n=0[/tex]
[tex]\displaystyle2a_2+6a_3x+\sum_{n=2}^\infty(n+2)(n+1)a_{n+2}x^n-\sum_{n=2}^\infty a_{n-2}x^n=0[/tex]
[tex]\displaystyle2a_2+6a_3x+\sum_{n=2}^\infty\bigg((n+2)(n+1)a_{n+2}-a_{n-2}\bigg)x^n=0[/tex]
Right away we see [tex]a_2=a_3=0[/tex], and the coefficients are given according to the recurrence
[tex]\begin{cases}a_0=y(0)\\a_1=y'(0)\\a_2=0\\a_3=0\\n(n-1)a_n=a_{n-4}&\text{for }n\ge4\end{cases}[/tex]
There's a dependency between terms in the sequence that are 4 indices apart, so we consider 4 different cases.
If [tex]n=4k[/tex], where [tex]k\ge0[/tex] is an integer, then[tex]k=0\implies n=0\implies a_0=a_0[/tex]
[tex]k=1\implies n=4\implies a_4=\dfrac{a_0}{4\cdot3}=\dfrac2{4!}a_0[/tex]
[tex]k=2\implies n=8\implies a_8=\dfrac{a_4}{8\cdot7}=\dfrac{6\cdot5\cdot2}{8!}a_0[/tex]
[tex]k=3\implies n=12\implies a_{12}=\dfrac{a_8}{12\cdot11}=\dfrac{10\cdot9\cdot6\cdot5\cdot2}{12!}a_0[/tex]
and so on, with the general pattern
[tex]a_{4k}=\dfrac{a_0}{(4k)!}\displaystyle\prod_{i=1}^k(4i-2)(4i-3)[/tex]
If [tex]n=4k+1[/tex], then[tex]k=0\implies n=1\implies a_1=a_1[/tex]
[tex]k=1\implies n=5\implies a_5=\dfrac{a_1}{5\cdot4}=\dfrac{3\cdot2}{5!}a_1[/tex]
[tex]k=2\implies n=9\implies a_9=\dfrac{a_5}{9\cdot8}=\dfrac{7\cdot6\cdot3\cdot2}{9!}a_1[/tex]
[tex]k=3\implies n=13\implies a_{13}=\dfrac{a_9}{13\cdot12}=\dfrac{11\cdot10\cdot7\cdot6\cdot3\cdot2}{13!}a_1[/tex]
and so on, with
[tex]a_{4k+1}=\dfrac{a_1}{(4k+1)!}\displaystyle\prod_{i=1}^k(4i-1)(4i-2)[/tex]
If [tex]n=4k+2[/tex] or [tex]n=4k+3[/tex], then[tex]a_2=0\implies a_6=a_{10}=\cdots=a_{4k+2}=0[/tex]
[tex]a_3=0\implies a_7=a_{11}=\cdots=a_{4k+3}=0[/tex]
Then the solution to this ODE is
[tex]\boxed{y(x)=\displaystyle\sum_{k=0}^\infty a_{4k}x^{4k}+\sum_{k=0}^\infty a_{4k+1}x^{4k+1}}[/tex]
Which established actor stared in The Godfather as Don Vito? a. Marlon Brando b. John Wayne c. Henry Fonda d. None of the above
Answer:
a) Marlon Brando
Step-by-step explanation:
Don Vito Corleone was played by Marlon Brando in the film Godfather released in the year 1972 directed by Francis Ford Coppola based on the 1969 book by Mario Puzo. Paramount studios was against the casting of Marlon Brando but the director insisted on giving him a screen test. The screen test convinced the studio to cast him in the role.
Marlon Brando starred as Don Vito Corleone in The Godfather, earning an Academy Award for his performance.
Explanation:The established actor who starred in The Godfather as Don Vito Corleone is Marlon Brando. Brando's performance in this iconic film is widely recognized as one of his greatest roles. Released in 1972, The Godfather is a crime film directed by Francis Ford Coppola and based on the novel of the same name by Mario Puzo. Marlon Brando's portrayal of the mafia patriarch earned him an Academy Award for Best Actor, which further solidified his legacy as one of Hollywood's legendary actors. The other actors listed, John Wayne and Henry Fonda, were not part of the cast of The Godfather.
Assume that females have pulse rates that are normally distributed with a mean of mu equals 73.0 beats per minute and a standard deviation of sigma equals 12.5 beats per minute. Complete parts (a) through (c) below. a. If 1 adult female is randomly selected, find the probability that her pulse rate is between 69 beats per minute and 77 beats per minute. The probability is nothing. (Round to four decimal places as needed.)
The probability that her pulse rate is between 69 beats per minute and 77 beats per minute is 25.1%
What is z score?Z score is used to determine by how many standard deviations the raw score is above or below the mean.
It is given by:
z = (raw score - mean) / standard deviation
Mean = 73, standard deviation = 12.5
For x = 69:
z = (69 - 73) / 12.5 = -0.32
For x = 77:
z = (77 - 73) / 12.5 = 0.32
P(-0.32 <z < 0.32) = P(z < 0.32) - P(z < -0.32) = 0.6255 - 0.3745 = 0.251
The probability that her pulse rate is between 69 beats per minute and 77 beats per minute is 25.1%
Find out more on z score at: https://brainly.com/question/25638875
To find the probability that a randomly selected female has a pulse rate between 69 and 77 beats per minute, we calculate the z-scores for these values and use the standard normal distribution table. The probability is approximately 0.2481.
Explanation:To find the probability that a randomly selected female has a pulse rate between 69 and 77 beats per minute, we need to calculate the z-scores for these values and use the standard normal distribution table.
First, we calculate the z-score for 69 using the formula: z = (x - mu) / sigma, where x is the value, mu is the mean, and sigma is the standard deviation. Plugging in the values, we get z = (69 - 73) / 12.5 = -0.32.
Next, we calculate the z-score for 77: z = (77 - 73) / 12.5 = 0.32.
From the standard normal distribution table, we find that the probability of a z-score between -0.32 and 0.32 is approximately 0.2481.
Learn more about Probability here:https://brainly.com/question/32117953
#SPJ3
1) For the following problem: Let f={(-2,3),(-1,1),(0,0),(1,-1), (2,-3)} and let g-{-3,1),(-1,-2), (0, 2),(2, 2),(3,1)}.Find the following a) f(1) and g-1) b) (gof (1) c) (gofof)(-1) d) (fog)(3)
Answer:
1. f(1)=-1 and g(-1)=-2.
2. (gof)(1)=-2.
3. (gofof)(-1)=-2
4. (fog)(3)=-1
Step-by-step explanation:
The given functions are defined as
f={(-2,3),(-1,1),(0,0),(1,-1), (2,-3)}
g={(-3,1),(-1,-2), (0, 2),(2, 2),(3,1)}
1.
The value of function f at x=1 is -1, So, f(1)=-1.
The value of function g at x=-1 is -2, So, g(-1)=-2.
Therefore the value of f(1) is -1 and the value of g(-1) is -2.
2.
[tex](g\circ f)(1)=g(f(1))[/tex] [tex][\because (g\circ f)(x)=g(f(x))][/tex]
[tex](g\circ f)(1)=g(-1)[/tex] [tex][\because f(1)=-1][/tex]
[tex](g\circ f)(1)=-2[/tex] [tex][\because g(-1)=-2][/tex]
Therefore the value of (gof)(1) is -2.
3.
[tex](g\circ f\circ f)(-1)=(g\circ f)(f(-1))[/tex] [tex][\because (g\circ f)(x)=g(f(x))][/tex]
[tex](g\circ f\circ f)(-1)=(g\circ f)(1)[/tex] [tex][\because f(-1)=1][/tex]
[tex](g\circ f\circ f)(-1)=-2[/tex] [tex][\because \text{From part 2}, (g\circ f)(1)=-2][/tex]
Therefore the value of (gofof)(1) is -2.
4.
[tex](f\circ g)(3)=f(g(3))[/tex] [tex][\because (f\circ g)(x)=f(g(x))][/tex]
[tex](f\circ g)(3)=f(1)[/tex] [tex][\because g(3)=1][/tex]
[tex](f\circ g)(3)=-1[/tex] [tex][\because f(1)=-1][/tex]
Therefore the value of (fog)(3) is -1.
Real estate ads suggest that 75 % of homes for sale have garages, 29 % have swimming pools, and 13 % have both features. What is the probability that a home for sale has a) a pool or a garage? b) neither a pool nor a garage? c) a pool but no garage?
Answer: a) 91% b) 9% c) 16%
Step-by-step explanation:
Let A be the event that homes for sale have garages and B be the event that homes for sale have swimming pools.
Now, given :[tex]P(A)=75\%=0.75[/tex]
[tex]P(B)=29\%=0.29[/tex]
[tex]P(A\cap B)=13\%=0.13[/tex]
a) [tex]P(A\cup B)=P(A)+P(B)-P(A\cap B)\\\\\Rightarrow\ P(A\cup B)=0.75+0.29-0.13=0.91[/tex]
Hence, the probability that a home for sale has a pool or a garage is 91%.
b) The probability that a home for sale has neither a pool nor a garage is given by :-
[tex]1-P(A\cup B)=1-0.91=0.09[/tex]
Hence, the probability that a home for sale has neither a pool nor a garage is 9%.
c) The probability that a home for sale has a pool but no garage is given by :-
[tex]P(B)-P(A\cap B)=0.29-0.13=0.16[/tex]
Hence, the probability that a home for sale has a pool but no garage is 16%.
Final answer:
The probability a home for sale has a pool or garage is 91%, the probability of having neither feature is 9%, and the probability of having a pool but no garage is 16%.
Explanation:
The question involves solving problems related to probability and set theory. To find out the probability of events involving homes for sale with certain features, we can use the principle of inclusion and exclusion for probabilities.
Answering Part a)
The probability that a home has a pool or a garage is given by the formula P( A or B ) = P( A ) + P( B ) - P( A and B ). Thus, the probability is:
P( Pool or Garage ) = P( Pool ) + P( Garage ) - P( Pool and Garage )
= 0.29 + 0.75 - 0.13
= 0.91 or 91%
Answering Part b)
To find a home with neither a pool nor a garage, we subtract the probability of having either from 1. Therefore:
P( Neither ) = 1 - P( Pool or Garage )
= 1 - 0.91
= 0.09 or 9%
Answering Part c)
The probability of a home having a pool but no garage is calculated by taking the probability of having a pool and subtracting the probability of having both a pool and a garage:
P( Pool but no Garage ) = P( Pool ) - P( Pool and Garage )
= 0.29 - 0.13
= 0.16 or 16%
6. Raw Data: 3,5,7,4,3, 8, 6, 6,9, 6,7,8,9,3, 3, 9 There are 16 data items Find: a) Mean b) Median c) Midrange d) Mode 7. Find the standard deviation of the data in question 6
Answer:
6.a. Mean= 6
b. Median=6
c Midrange=6
d.Mode=4
7.Standard deviation=2.2079
Step-by-step explanation:
Given data
3,3,3,3,4,5,6,6,6,7,7,8,8,9,9,9
Total data items,n=16
Sum o data items=96
a. Mean=[tex]\frac{sum\;of\;data\;items}{total\;data\;items}[/tex]
Mean=[tex]\frac{96}{16}[/tex]
Mean=6
b.If total number of items are even then
Median=[tex]\frac{\frac{n}{2}^{th}\;observation+\left(\frac{n}{2}+1\right)^{th}}{2}[/tex]
Median=[tex]\frac{\frac{16}{2}^{th} observation+\left(\frac{16}{2}+1\right)^{th} observation}{2}[/tex]
Median=[tex]\frac{8^{th} observation+9^{th} observation}{2}[/tex]
Median= [tex]\frac{6+6}{2}[/tex]
Median= [tex]\frac{12}{2}[/tex]
Median=6
c. Midrange=[tex]\frac{lower\;value+highest\;value}{2}[/tex]
Lower data item=3
Highest data item=9
Midrange= [tex]\frac{3+9}{2}[/tex]
Midrange= 6
d.Mode : It is defines as a number that appear most often in a set of numbers.
Mode=3
7. Mean[tex]\bar x=6[/tex]
[tex]\mid x-\bar x\mid[/tex] [tex]{\mid x-\bar x\mid}^2[/tex]
3 9
3 9
3 9
3 9
2 4
1 1
0 0
0 0
0 0
1 1
1 1
2 4
2 4
3 9
3 9
3 9
[tex]\sum{\mid x-\bar x\mid}^2=78[/tex]
n=16
Standard deviation=[tex]\sqrt{\frac{\sum{\mid x-\bar x}^2}{n}}[/tex]
Standard devaition=[tex]\sqrt{\frac{78}{16}}[/tex]
Standard deviation=[tex]\sqrt{4.875}[/tex]
Standard deviation of data =2.2079
When Bill makes a sandwich, he may choose from among 3 kinds of rolls, 4 varieties of meat, and 2 types of sliced cheese. If he chooses one roll, one meat, and one type of cheese, how many different kinds of sandwiches can he make?
Answer: Bill can make 24 different kinds of sandwiches
Step-by-step explanation:
Given : The number of kinds of rolls = 3
The number of varieties of meat = 4
The number of types of sliced cheese = 2
If he chooses one roll, one meat, and one type of cheese, then the number of different kinds of sandwiches he can make is given by :-
[tex]3\times2\times4=24[/tex]
Hence, Bill can make 24 different kinds of sandwiches.
Water is leaking out the bottom of a hemispherical tank of radius 9 feet at a rate of 2 cubic feet per hour. The tank was full at a certain time. How fast is the water level changing when its height h is 6 feet? Note: the volume of a segment of height h in a hemisphere of radius r is pi h squared left bracket r minus left parenthesis h divided by 3 right parenthesis right bracket.
Answer:
The water level changing by the rate of -0.0088 feet per hour ( approx )
Step-by-step explanation:
Given,
The volume of a segment of height h in a hemisphere of radius r is,
[tex]V=\pi h^2(r-\frac{h}{3})[/tex]
Where, r is the radius of the hemispherical tank,
h is the water level, ( in feet )
Here, r = 9 feet,
[tex]\implies V=\pi h^2(9-\frac{h}{3})[/tex]
[tex]V=9\pi h^2-\frac{\pi h^3}{3}[/tex]
Differentiating with respect to t ( time ),
[tex]\frac{dV}{dt}=18\pi h\frac{dh}{dt}-\frac{3\pi h^2}{3}\frac{dh}{dt}[/tex]
[tex]\frac{dV}{dt}=\pi h(18-h)\frac{dh}{dT}[/tex]
Here, [tex]\frac{dV}{dt}=-2\text{ cubic feet per hour}[/tex]
And, h = 6 feet,
Thus,
[tex]-2=\pi 6(18-6)\frac{dh}{dt}[/tex]
[tex]\implies \frac{dh}{dt}=\frac{-2}{72\pi}=-0.00884194128288\approx -0.0088[/tex]
solve y' -x^2y = 0 using power series and write the first four terms of the power series
We're looking for a solution of the form
[tex]y=\displaystyle\sum_{n\ge0}a_nx^n[/tex]
with derivative
[tex]y'=\displaystyle\sum_{n\ge0}(n+1)a_{n+1}x^n[/tex]
Note that [tex]x=0\implies y(0)=a_0[/tex].
Substituting into the ODE gives
[tex]\displaystyle\sum_{n\ge0}(n+1)a_{n+1}x^n-\sum_{n\ge0}a_nx^{n+2}=0[/tex]
The first series starts with a constant term, while the second starts with a quadratic term, so we should pull out the first two terms of the first series and have it start at [tex]n=2[/tex], then shift the index on the second series to achieve the same effect, which allows us to condense the left side as
[tex]a_1+2a_2x+\displaystyle\sum_{n\ge2}\bigg((n+1)a_{n+1}-a_{n-2}\bigg)x^n=0[/tex]
so that the series solution's coefficients are given according to the recurrence
[tex]\begin{cases}a_0=a_0\\a_1=a_2=0\\(n+1)a_{n+1}-a_{n-2}=0&\text{for }n\ge2\end{cases}[/tex]
We can simplify the latter equation somewhat to get it in terms of [tex]a_n[/tex]:
[tex]a_n=\dfrac{a_{n-3}}n\text{ for }n\ge3[/tex]
This shows dependency between coefficients that are 3 indices apart, so we check 3 cases:
If [tex]n=3k+1[/tex], where [tex]k\ge0[/tex] is an integer, then[tex]k=0\implies n=1\implies a_1=0[/tex]
[tex]k=1\implies n=4\implies a_4=\dfrac{a_1}4=0[/tex]
and so on for all such [tex]n[/tex], giving
[tex]a_{3k+1}=0[/tex]
If [tex]n=3k+2[/tex], then[tex]k=0\implies n=2\implies a_2=0[/tex]
and we get the same conclusion as before,
[tex]a_{3k+2}=0[/tex]
If [tex]n=3k[/tex], then[tex]k=0\implies n=0\impiles a_0=a_0[/tex]
[tex]k=1\implies n=3\implies a_3=\dfrac{a_0}3[/tex]
[tex]k=2\implies n=6\implies a_6=\dfrac{a_3}6=\dfrac{a_0}{3\cdot6}=\dfrac{a_0}{3^2(2\cdot1)}[/tex]
[tex]k=3\implies n=9\implies a_9=\dfrac{a_6}9=\dfrac{a_0}{3^3(3\cdot2\cdot1)}a_0[/tex]
and so on, with the general pattern
[tex]a_{3k}=\dfrac{a_0}{3^kk!}[/tex]
Then the series solution is
[tex]y=\displaystyle\sum_{k\ge0}\bigg(a_{3k}x^{3k}+a_{3k+1}x^{3k+1}+a_{3k+2}x^{3k+2}\bigg)[/tex]
[tex]y=\displaystyle a_0\sum_{k\ge0}\frac{x^{3k}}{3^kk!}[/tex]
[tex]y=\displaystyle a_0\sum_{k\ge0}\frac{\left(\frac{x^3}3\right)^k}{k!}[/tex]
whose first four terms are
[tex]\boxed{a_0\left(1+\dfrac{x^3}3+\dfrac{x^6}{18}+\dfrac{x^9}{162}\right)}[/tex]
Pulam, Inc. prepared the following master budget items for July:
Production and sales 36,000 units
Variable manufacturing costs:
Direct materials $ 36,000
Direct labor $ 72,000
Variable manufacturing overhead $ 72,000
Fixed manufacturing costs $ 180,000
Total manufacturing costs $ 360,000
During July, Pulam actually sold 42,000 units. Prepare a flexible budget for Pulam based on actual sales.(Do not round your intermediate calculations).
Production and sales units
Variable manufacturing costs:
Direct materials $
Direct labor $
Variable manufacturing overhead $
Fixed manufacturing costs $
Total manufacturing costs $
Answer:
During July, Pulam actually sold 42,000 units.
Variable Manufacturing Costs for 42,000 units are-
1. Direct Material : [tex](36000/36000)\times42000 =42000[/tex] dollars
2. Direct Labor: [tex](72000/36000)\times42000 =84000[/tex] dollars
3. Variable Manufacturing Overhead : [tex](72000/36000)\times42000 =84000[/tex] dollars
4. Fixed manufacturing costs : $ 180000
5. Total manufacturing costs : [tex]42000+84000+84000+180000=390000[/tex] dollars
Express the confidence interval left parenthesis 0.023 comma 0.085 right parenthesis(0.023,0.085) in the form of ModifyingAbove p with caret minus Upper E less than p less than ModifyingAbove p with caret plus Upper Ep−E
Final answer:
To express the confidence interval (0.023, 0.085) in the requested form, calculate the midpoint (p') as 0.054 and the margin of error (EBP) as 0.031. The confidence interval will thus be expressed as 0.054 - 0.031 < p < 0.054 + 0.031.
Explanation:
The question asks to express the confidence interval (0.023, 0.085) in the form of ModifyingAbove p with caret minus Upper E less than p less than ModifyingAbove p with caret plus Upper E. To do this, we first need to calculate the midpoint (p') and the margin of error (EBP) of the given interval.
The midpoint (p') is the average of the lower and upper bounds of the interval, calculated as (0.023 + 0.085) / 2 = 0.054. The margin of error (EBP) is the difference between p' and either boundary of the interval, calculated as 0.085 - 0.054 = 0.031 or 0.054 - 0.023 = 0.031. Therefore, the interval in the requested format is 0.054 - 0.031 < p < 0.054 + 0.031.
The confidence interval (0.023, 0.085) can be expressed in the form [tex]\( 0.054 - 0.031 < p < 0.054 + 0.031 \)[/tex].
To express the confidence interval (0.023, 0.085) in the form [tex]\( \hat{p} - E < p < \hat{p} + E \),[/tex] where [tex]\( \hat{p} \)[/tex] is the point estimate and ( E ) is the margin of error, we need to find the point estimate and the margin of error.
Step 1: Find the point estimate [tex]\( \hat{p} \)[/tex].
The point estimate is the midpoint of the confidence interval. We can calculate it using the formula:
[tex]\[ \hat{p} = \frac{\text{Lower bound} + \text{Upper bound}}{2} \][/tex]
Substitute the values:
[tex]\[ \hat{p} = \frac{0.023 + 0.085}{2} \][/tex]
[tex]\[ \hat{p} = \frac{0.108}{2} \][/tex]
[tex]\[ \hat{p} = 0.054 \][/tex]
So, the point estimate [tex]\( \hat{p} \)[/tex] is 0.054.
Step 2: Find the margin of error \( E \).
The margin of error is half the width of the confidence interval. We can calculate it using the formula:
[tex]\[ E = \frac{\text{Upper bound} - \text{Lower bound}}{2} \][/tex]
Substitute the values:
[tex]\[ E = \frac{0.085 - 0.023}{2} \][/tex]
[tex]\[ E = \frac{0.062}{2} \][/tex]
[tex]\[ E = 0.031 \][/tex]
So, the margin of error ( E ) is 0.031.
Step 3: Express the confidence interval in the required form.
Now, we can write the confidence interval in the form [tex]\( \hat{p} - E < p < \hat{p} + E \):[/tex]
[tex]\[ 0.054 - 0.031 < p < 0.054 + 0.031 \][/tex]
[tex]\[ 0.023 < p < 0.085 \][/tex]
Therefore, the confidence interval (0.023, 0.085) can be expressed in the form [tex]\( 0.054 - 0.031 < p < 0.054 + 0.031 \)[/tex].
Find the y -intercept and the slope of the line.
Write your answers in simplest form
5x - 2y = 2
Answer:
The slope is 5/2 and the y intercept is -1
Step-by-step explanation:
To find the slope and the y intercept, we will write the equation in slope intercept form, y =mx+b where m is the slope and b is the y intercept
5x -2y =2
Add 2y to each side
5x-2y+2y =2 +2y
5x = 2+2y
Subtract 2 from each side
5x-2 = 2y+2-2
5x-2 =2y
Divide each side by 2
5x/2 -2/2 = 2y/2
5/2x -1 = y
y = 5/2x -1
The slope is 5/2 and the y intercept is -1
The Vertex of a parabola is at (8-1), and it's why intercept is negative 17, which function represents the parabola
Answer:
The function is equal to [tex]y=-(1/4)(x-8)^{2}-1[/tex]
Step-by-step explanation:
we know that
The equation of a vertical parabola in vertex form is equal to
[tex]y=a(x-h)^{2}+k[/tex]
where
a is a coefficient
(h,k) is the vertex
In this problem we have
(h,k)=(8,-1)
substitute
[tex]y=a(x-8)^{2}-1[/tex]
Find the value of a
Remember that we have the y-intercept
The y-intercept is the point (0,-17)
substitute
x=0,y=-17
[tex]-17=a(0-8)^{2}-1[/tex]
[tex]-17=64a-1[/tex]
[tex]64a=-17+1[/tex]
[tex]64a=-16[/tex]
[tex]a=-16/64[/tex]
[tex]a=-1/4[/tex]
therefore
The function is equal to
[tex]y=-(1/4)(x-8)^{2}-1[/tex]
see the attached figure to better understand the problem
(1.)Find the slope of the line that passes through the given pair of points. (If an answer is undefined, enter UNDEFINED.) (?a + 3, b ? 3) and (a + 3, ?b) *******(2.)If the line passing through the points (a, 1) and (6, 5) is parallel to the line passing through the points (2, 7) and (a + 2, 1), what is the value of a?
Answer:
1. The slope of the line is [tex]m=\frac{-2b+3}{2a}[/tex].
2. The value of a is 18.
Step-by-step explanation:
If a line passes through two points, then the slope of the line is
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
(1)
It is given that the line passes through the points (-a + 3, b - 3) and (a + 3, -b). So, the slope of the line is
[tex]m=\frac{-b-(b-3)}{a+3-(-a+3)}[/tex]
[tex]m=\frac{-b-b+3}{a+3+a-3)}[/tex]
[tex]m=\frac{-2b+3}{2a}[/tex]
The slope of the line is [tex]m=\frac{-2b+3}{2a}[/tex].
(2)
If the line passing through the points (a, 1) and (6, 5), then the slope of the line is
[tex]m_1=\frac{5-1}{6-a}=\frac{4}{6-a}[/tex]
If the line passing through the points (2, 7) and (a + 2, 1), then the slope of the line is
[tex]m_2=\frac{1-7}{a+2-2}=\frac{-6}{a}[/tex]
The slopes of two parallel lines are same.
[tex]m_1=m_2[/tex]
[tex]\frac{4}{6-a}=\frac{-6}{a}[/tex]
On cross multiplication we get
[tex]4a=-6(6-a)[/tex]
[tex]4a=-36+6a[/tex]
[tex]4a-6a=-36[/tex]
[tex]-2a=-36[/tex]
Divide both sides by -2.
[tex]a=18[/tex]
Therefore the value of a is 18.
Let A, B, and C be arbitrary sets within a universal set, U. For each of the following statements, either prove that the statement is always true or show a counterexample to prove it is not always true. When giving a counterexample, you should define the three sets explicitly and say what the left-hand and right-hand sides of the equation are for those sets, to make it clear that they are not equal.
(A \ B) × C = (A × C) \ (B × C)
Answer with Step-by-step explanation:
Let A, B and C are arbitrary sets within a universal set U.
We have to prove that [tex]( A/B)\times C=(A\times C)/(B\times C)[/tex] is always true.
Let [tex](x,y)\in (A/B)\times C[/tex]
Then [tex] x\in(A/B) [/tex] and [tex] y\in C[/tex]
Therefore, [tex] x\in A[/tex] and [tex] x\notin B[/tex]
Then, (x,y) belongs to [tex] A\times C[/tex]
and (x,y) does not belongs to [tex] B\times C[/tex]
Hence,[tex] (x,y)\in(A\times C)/(B\times C)[/tex]
Conversely ,Let (x ,y)belongs to [tex] (A\times C)/(B\times C)[/tex]
Then [tex] (x,y)\in (A\times C)[/tex] and [tex] (x,y)\notin (B\times C)[/tex]
Therefore,[tex] x\in A,y\in C[/tex] and [tex] x\notin B,y\in C[/tex]
[tex] x\in(A/B)[/tex] and [tex]y\in C[/tex]
Hence, [tex] (x,y)\in(A/B)\times C[/tex]
Therefore,[tex] (A/B)\times C=(A\times C)/(B\times C)[/tex] is always true.
Hence, proved.
Factor completely 3x4 − 48. 3(x2 − 4)(x2 + 4) 3(x − 2)(x + 2)(x + 2)(x + 2) 3(x − 2)(x + 2)(x2 + 4) 3(x − 2)(x + 2)(x2 − 4)
Answer:
3 (x-2) (x+2) (x^2+4)
Step-by-step explanation:
3x^4 − 48
Factor out a 3
3(x^4 -16)
Inside the parentheses is the difference of squares (a^2 - b^2) = (a-b) (a+b)
where a = x^2 and b = 4
3 (x^2-4) (x^2+4)
Inside the first parentheses is the difference of squares where a = x and b=2
3 (x-2) (x+2) (x^2+4)
Answer:
3 (x-2) (x+2) (x^2+4)
Step-by-step explanation:
The function f(x) = 2x + 510 represents the number of calories burned when exercising, where x is the number of hours spent exercising.
The function g(x) = 200x − 125 represents the calorie deficit that occurs when combining diet with exercise, where x is the number of hours spent exercising.
What is (f + g)(2)? Explain.
514 calories burned while combining diet with 2 hours of exercise
789 calories burned while combining diet with 2 hours of exercise
514 calories burned while exercising for 2 hours
789 calories burned while exercising for 2 hours
Answer: 789 calories burned while combining diet with 2 hours of exercise
Step-by-step explanation:
we have that
[tex]f(x)=2x+510[/tex]
[tex]g(x)=200x-125[/tex]
we know that
[tex](f+g)(x)=f(x)+g(x)[/tex]
substitute
[tex](f+g)(x)=2x+510+200x-125[/tex]
[tex](f+g)(x)=202x+385[/tex]
Find [tex](f+g)(2)[/tex]
For x=2 hours
substitute
[tex](f+g)(2)=202(2)+385[/tex]
[tex](f+g)(2)=789\ calories[/tex]
therefore
The answer is
789 calories burned while combining diet with 2 hours of exercise
Suppose a basketball player has made 282 out of 393 free throws. If the player makes the next 2 free throws, I will pay you $6. Otherwise you pay me $8.Step 1 of 2 :Find the expected value of the proposition. Round your answer to two decimal places. Losses must be expressed as negative values.
Answer:
-$ 0.79
Step-by-step explanation:
Since, the player has made 282 out of 393 free throws,
So, the probability of a free throw = [tex]\frac{282}{393}[/tex],
Thus, the probability of 2 free throws = [tex]\frac{282}{393}\times \frac{282}{393}=\frac{8836}{17161}[/tex]
And, the probability of not getting 2 free throws = [tex]1-\frac{8836}{17161}=\frac{8325}{17161}[/tex]
Given, the price of winning ( getting 2 free throws ) is $6 while the price of losing ( not getting 2 free throws ) is - $ 8 ( ∵ there is a loss of $ 8 ),
Hence, the expected value of the proposition = probability of winning × winning value + probability of losing × losing value
[tex]= \frac{8836}{17161}\times 6 + \frac{8325}{17161}\times -8[/tex]
[tex]=\frac{53016}{17161}-\frac{66600}{17161}[/tex]
[tex]=-\frac{13584}{17161}[/tex]
[tex]=-\$ 0.79156226327[/tex]
[tex]\approx -\$ 0.79[/tex]
22 Which of the following polynomials has a factor of x-1? A) p(x)=x^3 +x^2 -2x+1 B) q(x)=2x^3-x^2 +x-1 (Crx)= 3x^3-x-2 D) S(x)=-3x^3+ 3x +1
Answer:
The correct option is C.
Step-by-step explanation:
If (x-c) is a factor of a polynomial f(x), then f(c)=0.
It is given that (x-1) is a factor of the polynomial. It means the value of the function at x=1 is 0.
In option A,
The given function is
[tex]p(x)=x^3+x^2-2x+1[/tex]
Substitute x=1 in the given function.
[tex]p(1)=(1)^3+(1)^2-2(1)+1=1+1-2+1=1[/tex]
Since p(1)≠0, therefore option A is incorrect.
In option B,
The given function is
[tex]q(x)=2x^3-x^2+x-1[/tex]
Substitute x=1 in the given function.
[tex]q(1)=2(1)^3-(1)^2+(1)-1=2-1+1-1=1[/tex]
Since q(1)≠0, therefore option B is incorrect.
In option C,
The given function is
[tex]r(x)=3x^3-x-2[/tex]
Substitute x=1 in the given function.
[tex]r(1)=3(1)^3-(1)-2=3-1-2=0[/tex]
Since r(1)=0, therefore option C is correct.
In option D,
The given function is
[tex]s(x)=-3x^3+3x+1[/tex]
Substitute x=1 in the given function.
[tex]s(1)=-3(1)^3+3(1)+1=-3+3+1=1[/tex]
Since s(1)≠0, therefore option D is incorrect.
A rumor spreads among a group of 400 people. The number of people, N(t), who have heard the rumor by the time t in hours since the rumor started to spread can be approximated by a function of the form
N(t) = 400 / (1+399e^(-0.4t))
Approximately how long will it take until half the people have heard the rumor?
Answer:
about 15 hours
Step-by-step explanation:
You want to find t such that N(t)=200. Fill in the equation with that information and solve for t.
200 = 400/(1 +399e^(-0.4t))
1 +399e^(-0.4t) = 400/200 = 2 . . . . . multiply by (1+399e^(-0.4t))/200
399e^(-0.4t) = 1 . . . . . . . . . . . . . . . . . . subtract 1
e^(-0.4t) = 1/399 . . . . . . . . . . . . . . . . . .divide by 399
-0.4t = ln(1/399) . . . . . . . take the natural log
t = ln(399)/0.4 ≈ 14.972 . . . . . . . divide by -0.4, simplify
Rounded to tenths, it will take 15.0 hours for half the people to have heard the rumor.
The time it will take for half the group, or 200 people, to have heard the rumor based on the given exponential decay function is approximately 1.72 hours.
Explanation:The problem presented is a mathematical question involving a model of exponential decay. In this context, we want to know when half of the people, or 200 people, will have heard the rumor. To find out, we substitute N(t) with 200 in the equation given:
200 = 400 / (1+399e^(-0.4t))
After simplifying and solving for t, t comes out to be approximately 1.72 hours. So, it takes approximately 1.72 hours for half the people to have heard the rumor.
Learn more about Exponential Decay here:https://brainly.com/question/2193799
#SPJ3
All but 1/13 of the students enrolled at a particular elementary school participated in "Family Fun Night" activities. If a total of 396 students were involved in the evening's activities, how many students attend the school?
Answer: 429
Step-by-step explanation:
If all but 1/13 of the students were involved in the evening's activities, that means that 12/13 of the students were involved.
To calculate the total number of students, first you do a simple rule of three to know the number of students that weren't involved in the activities:
If 12/13 of the students represent 396 students, then how many students are 1/13 of the students?
[tex]\frac{\frac{1}{13} *396}{\frac{12}{13} } =\frac{396}{12} =33[/tex]
Then, you add this number to the number of students that were involved in the activities:
[tex]396+33=429[/tex]
So, 429 is the number of students that attend the school.
You can also calculate it directly by doing the following rule of three:
If 12/13 of students represent 396 students, then how many students are 13/13 of the students?
[tex]\frac{\frac{13}{13} *396}{\frac{12}{13} } =\frac{396*13}{12} =\frac{5148}{12} =429[/tex]
Answer:429
Step-by-step explanation: let the total number of students in the elementary school be X
X - X/13 = 396
12X/13=396
X=396/*13/12
X=5148/12
X=429
Please help me with this
Answer:
∠AMX=72°
Step-by-step explanation:
we know that
An isosceles triangles has two equal sides and two equal interior angles
In the isosceles triangle MAX
we have that
XA=MA
and ∠AXM= ∠AMX -----> angles base
we have that
∠AXM=72°
therefore
∠AMX=72°
The random variables X and Y have the joint PMF pX,Y(x,y)={c⋅(x+y)2,0,if x∈{1,2,4} and y∈{1,3},otherwise. All answers in this problem should be numerical. Find the value of the constant c .
In order for [tex]p_{X,Y}(x,y)[/tex] to be a valid PMF, its integral over the distribution's support must be equal to 1.
[tex]p_{X,Y}(x,y)=\begin{cases}\dfrac{c(x+y)}2&\text{for }x\in\{1,2,4\}\text{ and }y\in\{1,3\}\\\\0&\text{otherwise}\end{cases}[/tex]
There are 3*2 = 6 possible outcomes for this distribution, so that
[tex]\displaystyle\sum_{x,y}p_{X,Y}(x,y)=\sum_{x\in\{1,2,4\}}\sum_{y\in\{1,3\}}\frac{c(x+y)}2=1[/tex]
[tex]1=\displaystyle\frac c2\sum_{x\in\{1,2,4\}}((x+1)+(x+3))=\frac c2\sum_{x\in\{1,2,4\}(2x+4)[/tex]
[tex]1=\displaystyle\frac c2((2+4)+(4+4)+(8+4))[/tex]
[tex]1=13c\implies\boxed{c=\dfrac1{13}}[/tex]
Find the equation of the line that passes through the points (12,21) and (46,33). V.35r-162
Answer:
the answer is Y=(6/17)X + (285/17)
Step-by-step explanation:
1. Identify the X and Y coordinate of each point.
for example, the first point could be (12,21), therefore X1 = 12 and Y1 = 21 and the second point is then (46,33), therefore X2 = 46 and Y2 = 33
2. to find the equation of the line it is necessary to find the slope.
knowing that the equiation of the slope is M = Y2-Y1/X2-X1 we replace
[tex]m=\frac{33-21}{46-12}=\frac{12}{34}[/tex] and then we simplify the result to [tex]\frac{12}{34}[/tex]
3. Using the Point-slope equation which is Y-Y1=M(X-X1) we replace
[tex]Y-33=\frac{6}{17}(X-46)[/tex]
[tex]Y=\frac{6}{17}X-\frac{276}{17}+33[/tex]
[tex]Y=\frac{6}{17}X+\frac{285}{17}[/tex]
Daily low temperatures in Columbus, OH in January 2014 were approximately normally distributed with a mean of 15.45 and a standard deviation of 13.70. What percentage of days had a low temperature between 5 degrees and 10 degrees? (Enter a number without the percent sign, rounded to the nearest 2 decimal places)
Answer: 12.10
Step-by-step explanation:
Given : Mean : [tex]\mu = 15.45[/tex]
Standard deviation : [tex]\sigma = 13.70[/tex]
The formula to calculate the z-score :-
[tex]z=\dfrac{x-\mu}{\sigma}[/tex]
For x= 5 degrees
[tex]z=\dfrac{5-15.45}{13.70}=-0.7627737226\approx-0.76[/tex]
For x= 10 degrees
[tex]z=\dfrac{10-15.45}{13.70}=-0.397810218\approx-0.40[/tex]
The P-value : [tex]P(-0.76<z<-0.40)=P(z<-0.40)-P(z<-0.76)[/tex]
[tex]=0.3445783-0.2236273=0.120951\approx0.1210[/tex]
In percent , [tex]0.1210\times100=12.10\%[/tex]
Hence, the percentage of days had a low temperature between 5 degrees and 10 degrees = 12.10%
A company that makes thing-a-ma-bobs has a start up cost of $18263. It costs the company $1.14 to make each thing-a-ma-bob and the company charges $4.06 for each thing-a-ma-bob. Let x represent the number of thing-a-ma-bobs made. Write the cost function for this company. Write the revenue function for this company. Write the profit function for this company.
Answer:
c(x) = 18263 +1.14xr(x) = 4.06xp(x) = 2.92x -18263Step-by-step explanation:
cost function
Cost for x tambs is the sum of start-up cost and per-unit cost multiplied by the number of units.
c(x) = 18263 +1.14x
revenue function
Revenue from the sale of x tambs is the product of their price and the number sold.
r(x) = 4.06x
profit function
Profit from the sale of x tambs is the difference between the revenue and cost:
p(x) = r(x) -c(x)
p(x) = 4.06x -(18263 +1.14x)
p(x) = 2.92x -18263
The cost function, C(x), for the company is [tex]\( C(x) = 1.14x + 18263 \)[/tex] .The revenue function, R(x), for the company is [tex]\( R(x) = 4.06x[/tex] .The profit function, P(x), for the company is [tex]( P(x) = R(x) - C(x) = 4.06x - (1.14x + 18263) \)[/tex] .
To determine the profit function for the company, we need to calculate the total cost and total revenue for making and selling 'x' thing-a-ma-bobs, respectively, and then subtract the total cost from the total revenue.1. The cost function, C(x), is the sum of the variable cost (cost per unit times the number of units) and the fixed cost (start-up cost). Since it costs $1.14 to make each thing-a-ma-bob and the start-up cost is $18263, the cost function is:[tex]\[ C(x) = (\text{Cost per unit}) \times x + \text{Fixed cost} \]\[ C(x) = 1.14x + 182632[/tex]
. The revenue function, R(x), is the amount of money the company earns from selling 'x' thing-a-ma-bobs at $4.06 each. Therefore, the revenue function is[tex]:\[ R(x) = (\text{Selling price per unit}) \times x \][ R(x) = 4.06x3.[/tex]The profit function, P(x), is the revenue minus the cost. To find the profit for 'x' thing-a-ma-bobs, we subtract the cost function from the revenue function:[tex]\[ P(x) = R(x) - C(x) \]\[ P(x) = 4.06x - (1.14x + 18263) \]\[ P(x) = 4.06x - 1.14x - 18263 \]\[ P(x) = 2.92x - 18263 \]So, the profit function for the company is \( P(x) = 2.92x - 18263 \),[/tex]
where 'x' represents the number of thing-a-ma-bobs made and sold.
Students are collecting canned goods for a local food pantry. Last year 12 students were able to distribute 1000 flyers in nine hours. This year there are 15 students handing out the same number of flyers. How long should it take them?
Answer:
7.2 h
Step-by-step explanation:
The time required is inversely proportional to the number of students.
t = k/n or
tn = k
Let t1 and n1 represent last year
and t2 and n2 represent this year. Then
t1n1 = t2n2
Data:
t1 = 9 h; n1 = 12 students
t2 = ?; n2 = 15 students
Calculation:
9 × 12 = t2 × 15
108 = 15t2
t2 = 108/15 = 7.2 h
It will take the students 7.2 h to distribute the flyers.
There are 40 students in a course: 15 freshmen, 8 sophomores, 12 juniors, and 5 seniors. One student is selected at random to attend a workshop and report back to the class. What is the probability that the chosen student is a junior
Answer:
The probability that the chosen student is a junior is [tex]\frac{3}{10}[/tex]
Step-by-step explanation:
Given,
The total number of students = 40,
In which, freshmen = 15,
Sophomores = 8,
Juniors = 12,
And, seniors = 5,
Thus, the probability that junior student when a student is chosen
[tex]=\frac{\text{Total possible arrangement of 1 junior student }}{\text{Total possible arrangement of 1 student}}[/tex]
[tex]=\frac{^{12}C_1}{^{40}C_1}[/tex]
[tex]=\frac{12}{40}[/tex]
[tex]=\frac{3}{10}[/tex]
The probability of selecting a junior student is 0.3 or 30%.
Explanation:To find the probability of selecting a junior student, we need to divide the number of junior students by the total number of students. There are 12 junior students out of a total of 40 students, so the probability can be calculated as:
Probability = Number of junior students / Total number of students
Probability = 12 / 40
Probability = 0.3
Therefore, the probability of selecting a junior student is 0.3 or 30%.
Learn more about probability here:https://brainly.com/question/32117953
#SPJ3