Answer:
[tex]\boxed{\sf \ \ \ [-7,0] \ \ \ }[/tex]
Step-by-step explanation:
Hello
[tex]y=x^2+7x=x(x+7) >0\\<=> x>0 \ and \ x+7 >0 \ \ or \ \ x<0 \ and \ x+7<0\\<=> x>0 \ \ or \ \ x<-7\\[/tex]
So values of x which is not in this domain is
[tex]-7\leq x\leq 0[/tex]
which is [-7,0]
hope this helps
Refer to the following wage breakdown for a garment factory:
Hourly Wages Number of employees
$4 up to $7 18
7 up to 10 36
10 up to 13 20
13 up to 16 6
What is the class interval for the preceding table of wages?
A. $4
B. $2
C. $5
D. $3
Answer:
The class interval is $3Step-by-step explanation:
The class interval is simply the difference between the lower or upper class boundary or limit of a class and the lower or upper class boundary or limit of the next class.
In this case for the class
$4 up to $7 18 and
$7 up to $10 36
The lower class boundary of the first class is $4 and the lower class boundary of the second class is $7
Hence the class interval = $7-$4= $3convert the equation y= -4x + 2/3 into general form equation and find t the values of A,B and C.
Answer:
Standard form: [tex]12x+3y-2=0[/tex]
A = 12, B = 3 and C = -2
Step-by-step explanation:
Given:
The equation:
[tex]y= -4x + \dfrac{2}3[/tex]
To find:
The standard form of given equation and find A, B and C.
Solution:
First of all, let us write the standard form of an equation.
Standard form of an equation is represented as:
[tex]Ax+By+C=0[/tex]
A is the coefficient of x and can be positive or negative.
B is the coefficient of y and can be positive or negative.
C can also be positive or negative.
Now, let us consider the given equation:
[tex]y= -4x + \dfrac{2}3[/tex]
Multiplying the whole equation with 3 first:
[tex]3 \times y= 3 \times -4x + 3 \times \dfrac{2}3\\\Rightarrow 3y=-12x+2[/tex]
Now, let us take all the terms on one side:
[tex]\Rightarrow 3y+12x-2=0\\\Rightarrow 12x+3y-2=0[/tex]
Now, let us compare with [tex]Ax+By+C=0[/tex].
So, A = 12, B = 3 and C = -2
An experiment involves 17 participants. From these, a group of 3 participants is to be tested under a special condition. How many groups of 3 participants can
be chosen, assuming that the order in which the participants are chosen is irrelevant?
Answer: 680
Step-by-step explanation:
When order doesn't matter,then the number of combinations of choosing r things out of n = [tex]^nC_r=\dfrac{n!}{r!(n-r)!}[/tex]
Given: Total participants = 17
From these, a group of 3 participants is to be tested under a special condition.
Number of groups of 3 participants chosen = [tex]^{17}C_3=\dfrac{17!}{3!(17-3)!}\[/tex]
[tex]^{17}C_3=\dfrac{17!}{3!(17-3)!}\\\\=\dfrac{17\times16\times15\times14!}{3\times2\times14!}\\\\=680[/tex]
Hence, there are 680 groups of 3 participants can be chosen,.
You are selling your product at a three-day event. Each day, there is a 60% chance that you will make money. What is the probability that you will make money on the first two days and lose money on the third day
Answer:
The required probability = 0.144
Step-by-step explanation:
Since the probability of making money is 60%, then the probability of losing money will be 100-60% = 40%
Now the probability we want to calculate is the probability of making money in the first two days and losing money on the third day.
That would be;
P(making money) * P(making money) * P(losing money)
Kindly recollect;
P(making money) = 60% = 60/100 = 0.6
P(losing money) = 40% = 40/100 = 0.4
The probability we want to calculate is thus;
0.6 * 0.6 * 0.4 = 0.144
A nut-raisin mix costs $5.26 a pound. Rashid buys 15.5 pounds of the mix for a party. Rashid’s estimated cost of the nut-raisin mix is A.$16 B.$22 C.$61 D.$80
Answer:
D.$80
Step-by-step explanation:
$5.26 x 15.5= $81.53
The closest amount to $81.53 is D.$80
Historically, the proportion of students entering a university who finished in 4 years or less was 63%. To test whether this proportion has decreased, 114 students were examined and 51% had finished in 4 years or less. To determine whether the proportion of students who finish in 4 year or less has statistically significantly decreased (at the 5% level of signficance), what is the critical value
Answer:
z(c) = - 1,64
We reject the null hypothesis
Step-by-step explanation:
We need to solve a proportion test ( one tail-test ) left test
Normal distribution
p₀ = 63 %
proportion size p = 51 %
sample size n = 114
At 5% level of significance α = 0,05, and with this value we find in z- table z score of z(c) = 1,64 ( critical value )
Test of proportion:
H₀ Null Hypothesis p = p₀
Hₐ Alternate Hypothesis p < p₀
We now compute z(s) as:
z(s) = ( p - p₀ ) / √ p₀q₀/n
z(s) =( 0,51 - 0,63) / √0,63*0,37/114
z(s) = - 0,12 / 0,045
z(s) = - 2,66
We compare z(s) and z(c)
z(s) < z(c) - 2,66 < -1,64
Therefore as z(s) < z(c) z(s) is in the rejection zone we reject the null hypothesis
The automatic opening device of a military cargo parachute has been designed to open when the parachute is 155 m above the ground. Suppose opening altitude actually has a normal distribution with mean value 155 and standard deviation 30 m. Equipment damage will occur if the parachute opens at an altitude of less than 100 m. What is the probability that there is equipment damage to the payload of at least one of five independently dropped parachutes
Answer:
the probability that one parachute of the five parachute is damaged is 0.156
Step-by-step explanation:
From the given information;
Let consider X to be the altitude above the ground that a parachute opens
Then; we can posit that the probability that the parachute is damaged is:
P(X ≤ 100 )
Given that the population mean μ = 155
the standard deviation σ = 30
Then;
[tex]P(X \leq 100 ) = ( \dfrac{X- \mu}{\sigma} \leq \dfrac{100- \mu}{\sigma})[/tex]
[tex]P(X \leq 100 ) = ( \dfrac{X- 155}{30} \leq \dfrac{100- 155}{30})[/tex]
[tex]P(X \leq 100 ) = (Z \leq \dfrac{- 55}{30})[/tex]
[tex]P(X \leq 100 ) = (Z \leq -1.8333)[/tex]
[tex]P(X \leq 100 ) = \Phi( -1.8333)[/tex]
From standard normal tables
[tex]P(X \leq 100 ) = 0.0334[/tex]
Hence; the probability of the given parachute damaged is 0.0334
Let consider Q to be the dropped parachute
Given that the number of parachute be n= 5
The probability that the parachute opens in each trail be p = 0.0334
Now; the random variable Q follows the binomial distribution with parameters n= 5 and p = 0.0334
The probability mass function is:
Q [tex]\sim[/tex] B(5, 0.0334)
Similarly; the event that one parachute is damaged is :
Q ≥ 1
P( Q ≥ 1 ) = 1 - P( Q < 1 )
P( Q ≥ 1 ) = 1 - P( Y = 0 )
P( Q ≥ 1 ) = 1 - b(0;5; 0.0334 )
P( Q ≥ 1 ) = [tex]1 -(^5_0)* (0.0334)^0*(1-0.0334)^5[/tex]
P( Q ≥ 1 ) = [tex]1 -( \dfrac{5!}{(5-0)!}) * (0.0334)^0*(1-0.0334)^5[/tex]
P( Q ≥ 1 ) = 1 - 0.8437891838
P( Q ≥ 1 ) = 0.1562108162
P( Q ≥ 1 ) [tex]\approx[/tex] 0.156
Therefore; the probability that one parachute of the five parachute is damaged is 0.156
Assume that there is a 6% rate of disk drive failure in a year. a. If all your computer data is stored on a hard disk drive with a copy stored on a second hard disk drive, what is the probability that during a year, you can avoid catastrophe with at least one working drive? b. If copies of all your computer data are stored on independent hard disk drives, what is the probability that during a year, you can avoid catastrophe with at least one working drive? four a. With two hard disk drives, the probability that catastrophe can be avoided is . (Round to four decimal places as needed.) b. With four hard disk drives, the probability that catastrophe can be avoided is . (Round to six decimal places as needed.)
Answer: 0.9964
Step-by-step explanation:
Consider,
P (disk failure) = 0.06
q = 0.06
p = 1- q
p = 1- 0.06,
p = 0.94
Step 2
Whereas p represents the probability that a disk does not fail. (i.e. working entire year).
a)
Step 3
a)
n = 2,
let x be a random variable for number...
Continuation in the attached document
plzzzzz helpp j + 9 - 3 < 8
Answer:
j < 2
Step-by-step explanation:
Simplify both sides of the inequality and isolating the variable would get you the answer
Please help. I’ll mark you as brainliest if correct!
Answer:
8lb of the cheaper Candy
17.5lb of the expensive candy
Step-by-step explanation:
Let the cheaper candy be x
let the costly candy be y
X+y = 25.5....equation one
2.2x +7.3y = 25.5(5.7)
2.2x +7.3y = 145.35.....equation two
X+y = 25.5
2.2x +7.3y = 145.35
Solving simultaneously
X= 25.5-y
Substituting value of X into equation two
2.2(25.5-y) + 7.3y = 145.35
56.1 -2.2y +7.3y = 145.35
5.1y = 145.35-56.1
5.1y = 89.25
Y= 89.25/5.1
Y= 17.5
X= 25.5-y
X= 25.5-17.5
X= 8
find the exact value of sin 0
Answer:
12/13
Step-by-step explanation:
First we must calculate the hypotenus using the pythagoran theorem
5²+12² = (MO)² MO = [tex]\sqrt{5^{2}+12^{2} }[/tex] MO = 13Now let's calculate sin0
sin O = 12/13So the exact value is 12/13
Answer:
C.) 12/13
Step-by-step explanation:
In a right angle triangle MN = 12, ON = 5 and; angle N = 90°
Now,
For hypotenuse we will use Pythagorean Theorem
(MO)² = (MN)² + (ON)²
(MO)² = (12)² + (5)²
(MO)² = 144 + 25
(MO)² = 169
MO = √169
MO = 13
now,
Sin O = opp÷hyp = 12÷13
The owner of a shoe store wanted to determine whether the average customer bought more than $100 worth of shoes. She randomly selected 10 receipts and identified the total spent by each customer. The totals (rounded to the nearest dollar) are given below.
Use a TI-83, TI-83 Plus, or TI-84 calculator to test whether the mean is greater than $100 and then draw a conclusion in the context of the problem. Use α=0.05.
125 99 219 65 109 89 79 119 95 135
Select the correct answer below:
A) Reject the null hypothesis. There is sufficient evidence to conclude that the mean is greater than $100.
B) Reject the null hypothesis. There is insufficient evidence to conclude that the mean is greater than $100.
C) Fail to reject the null hypothesis. There is sufficient evidence to conclude that the mean is greater than $100.
D) Fail to reject the null hypothesis. There is insufficient evidence to conclude that the mean is greater than $100.
Answer:
D) Fail to reject the null hypothesis. There is insufficient evidence to conclude that the mean is greater than $100.
Step-by-step explanation:
We are given that the owner of a shoe store randomly selected 10 receipts and identified the total spent by each customer. The totals (rounded to the nearest dollar) are given below;
X: 125, 99, 219, 65, 109, 89, 79, 119, 95, 135.
Let [tex]\mu[/tex] = average customer bought worth of shoes.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \leq[/tex] $100 {means that the mean is smaller than or equal to $100}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > $100 {means that the mean is greater than $100}
The test statistics that will be used here is One-sample t-test statistics because we don't know about population standard deviation;
T.S. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean = [tex]\frac{\sum X}{n}[/tex] = $113.4
s = sample standard deviation = [tex]\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }[/tex] = $42.78
n = sample of receipts = 10
So, the test statistics = [tex]\frac{113.4-100}{\frac{42.78}{\sqrt{10} } }[/tex] ~ [tex]t_9[/tex]
= 0.991
The value of t-test statistics is 0.991.
Now, at a 0.05 level of significance, the t table gives a critical value of 1.833 at 9 degrees of freedom for the right-tailed test.
Since the value of our test statistics is less than the critical value of t as 0.991 < 1.833, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.
Therefore, we conclude that the mean is smaller than or equal to $100.
A living room is two times as long and one and one-half times as wide as a bedroom. The amount of
carpet needed for the living room is how many times greater than the amount of carpet needed for the
bedroom?
1 1/2
2
3
3 1/2
Answer:
3
Step-by-step explanation:
let's call X the length of the bedroom, Y the wide of the bedroom, A the length of the living room and B the wide of the living room
A living room is two times as long as the bedroom, so:
A = 2X
A living room is one and one-half times as wide as a bedroom, so:
B = 1.5Y
The amount of carpet needed for the living room is A*B and the amount of carpet needed by the bedroom is X*Y
So, AB in terms of XY is:
A*B = (2X)*(1.5Y) = 3(X*Y)
It means that the amount of c arpet needed for the living room is 3 times greater than the amount of carpet needed for the bedroom.
Rewrite the equation in =+AxByC form. Use integers for A, B, and C. =−y6−6+x4
Answer:
6x + y = -18
Step-by-step explanation:
The given equation is,
y - 6 = -6(x + 4)
We have to rewrite this equation in the form of Ax + By = C
Where A, B and C are the integers.
By solving the given equation,
y - 6 = -6x - 24 [Distributive property]
y - 6 + 6 = -6x - 24 + 6 [By adding 6 on both the sides of the equation]
y = -6x - 18
y + 6x = -6x + 6x - 18
6x + y = -18
Here A = 6, B = 1 and C = -18.
Therefore, 6x + y = -18 will be the equation.
Please help with this question ASAP!
You are studying for the SAT and start the first week spending 2 hours studying. You plan to increase the amount you study by 10% each week. How many hours do you study in the 8th week?
Answer:
8w : 3.8974342 ≈ 3.9 or 4 (hope it help)
Step-by-step explanation:
1w : 2
2w : 2 + 10% = 2.2
3w : 2.2 + 10% = 2.42
4w : 2.42 + 10% = 2.662
5w : 2.662 + 10% = 2.9282
6w : 2.9282 + 10% = 3.22102
7w : 3.22102 + 10% = 3.543122
8w : 3.543122 + 10% = 3.8974342
3.8974342 ≈ 3.9 or 4
Use all the information below to find the missing x-value for the point that is on this line. m = - 1 / 3 b = 7 ( x, 4 )
Answer:
[tex]\boxed{x = 9}[/tex]
Step-by-step explanation:
m = -1/3
b = 7
And y = 4 (Given)
Putting all of the givens in [tex]y = mx+b[/tex] to solve for x
=> 4 = (-1/3) x + 7
Subtracting 7 to both sides
=> 4-7 = (-1/3) x
=> -3 = (-1/3) x
Multiplying both sides by -3
=> -3 * -3 = x
=> 9 = x
OR
=> x = 9
Answer:
x = 9
Step-by-step explanation:
m = -1/3
b = 7
Using slope-intercept form:
y = mx + b
m is slope, b is y-intercept.
y = -1/3x + 7
Solve for x:
Plug y as 4
4 = 1/3x + 7
Subtract 7 on both sides.
-3 = -1/3x
Multiply both sides by -3.
9 = x
Identify the value of the CRITICAL VALUE(S) used in a hypothesis test of the following claim and sample data:
Claim: "The average battery life (between charges) of this model of tablet is at least 12 hours."
A random sample of 80 of these tablets is selected, and it is found that their average battery life is 11.58 hours with a standard deviation of 1.93 hours. Test the claim at the 0.05 significance level.
a. -0.218
b. -1.645
c. -1.946
d. -1.667
Answer:
C
Step-by-step explanation:
The critical value we are asked to state in this question is the value of the z statistic
Mathematically;
z-score = (x- mean)/SD/√n
From the question
x = 11.58
mean = 12
SD = 1.93
n = 80
Substituting this value, we have
z= (11.58-12)/1.93/√80 = -1.946
let x = the amoun of raw sugar in tons a procesing plant is a sugar refinery process in one day . suppose x can be model as exponetial distribution with mean of 4 ton per day . The amount of raw sugar (x) has
Answer:
The answer is below
Step-by-step explanation:
A sugar refinery has three processing plants, all receiving raw sugar in bulk. The amount of raw sugar (in tons) that one plant can process in one day can be modelled using an exponential distribution with mean of 4 tons for each of three plants. If each plant operates independently,a.Find the probability that any given plant processes more than 5 tons of raw sugar on a given day.b.Find the probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day.c.How much raw sugar should be stocked for the plant each day so that the chance of running out of the raw sugar is only 0.05?
Answer: The mean (μ) of the plants is 4 tons. The probability density function of an exponential distribution is given by:
[tex]f(x)=\lambda e^{-\lambda x}\\But\ \lambda= 1/\mu=1/4 = 0.25\\Therefore:\\f(x)=0.25e^{-0.25x}\\[/tex]
a) P(x > 5) = [tex]\int\limits^\infty_5 {f(x)} \, dx =\int\limits^\infty_5 {0.25e^{-0.25x}} \, dx =-e^{-0.25x}|^\infty_5=e^{-1.25}=0.2865[/tex]
b) Probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day can be solved when considered as a binomial.
That is P(2 of the three plant use more than five tons) = C(3,2) × [P(x > 5)]² × (1-P(x > 5)) = 3(0.2865²)(1-0.2865) = 0.1757
c) Let b be the amount of raw sugar should be stocked for the plant each day.
P(x > a) = [tex]\int\limits^\infty_a {f(x)} \, dx =\int\limits^\infty_a {0.25e^{-0.25x}} \, dx =-e^{-0.25x}|^\infty_a=e^{-0.25a}[/tex]
But P(x > a) = 0.05
Therefore:
[tex]e^{-0.25a}=0.05\\ln[e^{-0.25a}]=ln(0.05)\\-0.25a=-2.9957\\a=11.98[/tex]
a ≅ 12
Find the length of the following tangent segments to the circles centered at O and O's whose radii are 5 and 3 respectively and the distance between O and O's is 12. Find segment AB
Answer:
AB = 2 sqrt(35) (or 11.83 to two decimal places)
Step-by-step explanation:
Refer to diagram.
ABO'P is a rectangle (all angles 90)
=>
PO' = AB
AB = PO' = sqrt(12^2-2^2) = sqrt(144-4) = sqrt(140) = 2sqrt(35)
using Pythagoras theorem.
A drawer contains 3 white shirts, 2 blue shirts, and 5 gray shirts. A shirt is randomly
selected from the drawer and set aside. Then another shirt is randomly selected from the
drawer.
What is the probability that the first shirt is white and the second shirt is gray?
Answer:
Probability that first shirt is white and second shirt is gray if first shirt selected is set aside = [tex]\frac{1}{4}[/tex]
Step-by-step explanation:
Given that
3 white, 2 blue and 5 gray shirts are there.
To find:
Probability that first shirt is white and second shirt is gray if first shirt selected is set aside = ?
Solution:
Here, total number of shirts = 3+2+5 = 10
First of all, let us learn about the formula of an event E:
[tex]P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}[/tex]
[tex]P(First\ White) = \dfrac{\text{Number of white shirts}}{\text {Total number of shirts left}}[/tex]
[tex]P(First\ White) = \dfrac{3}{10}[/tex]
Now, this shirt is set aside.
So, total number of shirts left are 9 now.
[tex]P(First\ White\ and\ second\ gray) = P(First White) \times P(Second\ Gray)\\\Rightarrow P(First\ White\ and\ second\ gray) = P(First White) \times \dfrac{\text{Number of gray shirts}}{\text{Total number of shirts left}}\\\\\Rightarrow P(First\ White\ and\ second\ gray) = \dfrac{3}{10} \times \dfrac{5}{9}\\\Rightarrow P(First\ White\ and\ second\ gray) = \dfrac{1}{2} \times \dfrac{1}{2}\\\Rightarrow P(First\ White\ and\ second\ gray) = \bold{\dfrac{1}{4} }[/tex]
So, the answer is:
Probability that first shirt is white and second shirt is gray if first shirt selected is set aside = [tex]\frac{1}{4}[/tex]
In which table does y vary inversely with x? A. x y 1 3 2 9 3 27 B. x y 1 -5 2 5 3 15 C. x y 1 18 2 9 3 6 D. x y 1 4 2 8 3 12
Answer:
In Table C, y vary inversely with x.
1×18 = 18
2×9 = 18
3×6 = 18
18 = 18 = 18
Step-by-step explanation:
We are given four tables and asked to find out in which table y vary inversely with x.
We know that an inverse relation has a form given by
y = k/x
xy = k
where k must be a constant
Table A:
x | y
1 | 3
2 | 9
3 | 27
1×3 = 3
2×9 = 18
3×27 = 81
3 ≠ 18 ≠ 81
Hence y does not vary inversely with x.
Table B:
x | y
1 | -5
2 | 5
3 | 15
1×-5 = -5
2×5 = 10
3×15 = 45
-5 ≠ 10 ≠ 45
Hence y does not vary inversely with x.
Table C:
x | y
1 | 18
2 | 9
3 | 6
1×18 = 18
2×9 = 18
3×6 = 18
18 = 18 = 18
Hence y vary inversely with x.
Table D:
x | y
1 | 4
2 | 8
3 | 12
1×4 = 4
2×8 = 16
3×12 = 36
4 ≠ 16 ≠ 36
Hence y does not vary inversely with x.
Evaluate the series
Answer:
the value of the series;
[tex]\sum_{k=1}^{6}(25-k^2) = 59[/tex]
C) 59
Step-by-step explanation:
Recall that;
[tex]\sum_{1}^{n}a_n = a_1+a_2+...+a_n\\[/tex]
Therefore, we can evaluate the series;
[tex]\sum_{k=1}^{6}(25-k^2)[/tex]
by summing the values of the series within that interval.
the values of the series are evaluated by substituting the corresponding values of k into the equation.
[tex]\sum_{k=1}^{6}(25-k^2) =(25-1^2)+(25-2^2)+(25-3^2)+(25-4^2)+(25-5^2)+(25-6^2)\\\sum_{k=1}^{6}(25-k^2) =(25-1)+(25-4)+(25-9)+(25-16)+(25-25)+(25-36)\\\sum_{k=1}^{6}(25-k^2) =24+21+16+9+0+(-11)\\\sum_{k=1}^{6}(25-k^2) = 59\\[/tex]
So, the value of the series;
[tex]\sum_{k=1}^{6}(25-k^2) = 59[/tex]
Use the functions m(x) = 4x + 5 and n(x) = 8x − 5 to complete the function operations listed below. Part A: Find (m + n)(x). Show your work. (3 points) Part B: Find (m ⋅ n)(x). Show your work. (3 points) Part C: Find m[n(x)]. Show your work. (4 points)
Answer:
Step-by-step explanation:
Part A
(m + n)x = 4x + 5 + 8x - 5
(m + n)x = 12x The fives cancel
Part B
(m - n)x = 4x + 5 - 8x + 5
(m - n)x = -4x + 10
Part C
The trick here is to put n(x) into m(x) wherever m(x) has an x.
m[n(x)] = 5(n(x)) + 5
m[n(x)] = 5(8x - 5) + 5
m[n(x)] = 40x - 20 + 5
m[n(x)] = 40x - 15
Find the area of the surface given by z = f(x, y) that lies above the region R. f(x, y) = 64 + x2 − y2 R = {(x, y): x2 + y2 ≤ 64}
The area of the surface above the region R is 4096π square units.
Given that:
The function: [tex]f(x, y) = 64 + x^2 - y^2[/tex]
The region R is the disk with a radius of 8 units [tex]x^2 + y^2 \le 64[/tex].
To find the area of the surface given by z = f(x, y) that lies above the region R, to calculate the double integral over the region R of the function f(x, y) with respect to dA.
The integral for the area is given by:
[tex]Area = \int\int_R f(x, y) dA[/tex]
To evaluate this integral, we need to set up the limits of integration for x and y over the region R, which is the disk cantered at the origin with a radius of 8 units.
Using polar coordinates, we can parameterize the region R as follows:
x = rcos(θ)
y = rsin(θ)
where r goes from 0 to 8, and θ goes from 0 to 2π.
Now, rewrite the integral in polar coordinates:
[tex]Area =\int\int_R f(x, y) dA\\Area = \int_0 ^{2\pi} \int_0^8(64 + r^2cos^2(\theta) - r^2sin^2(\theta)) \times r dr d \theta[/tex]
Now, we can integrate with respect to r first and then with respect to θ:
[tex]Area = \int_0^{2\pi} \int_0^8] (64r + r^3cos^2(\theta) - r^3sin^2(\theta)) dr d \theta[/tex]
Integrate with respect to r:
[tex]Area = \int_0^{2\pi}[(32r^2 + (1/4)r^4cos^2(\theta) - (1/4)r^4sin^2(\theta))]_0^8 d \theta\\Area = \int_0^{2\pi} (2048 + 256cos^2(\theta) - 256sin^2(\theta)) d \theta[/tex]
Now, we can integrate with respect to θ:
[tex]Area = [2048\theta + 128(sin(2\theta) + \theta)]_0 ^{2\pi}[/tex]
Area = 2048(2π) + 128(sin(4π) + 2π) - (2048(0) + 128(sin(0) + 0))
Area = 4096π + 128(0) - 0
Area = 4096π square units
So, the area of the surface above the region R is 4096π square units.
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Find the common ratio of the following geometric sequence:
11,55, 275, 1375, ....
Answer:
Hey there!
The common ratio is 5, because you multiply by 5 to get from one term to the next.
Hope this helps :)
Answer:
5
Step-by-step explanation:
To find the common ratio take the second term and divide by the first term
55/11 = 5
The common ratio would be 5
Mia agreed to borrow a 3 year loan with 4 percent interest to buy a motorcycle if Mia will pay a total of $444 in interest how much money did she borrow how much interest would Mia pay if the simple interest rate was 5 percent
Answer:
a) $3700
b) $555
Step-by-step explanation:
The length of the loan is 3 years.
The interest after 3 years is $444.
The rate of the Simple Interest is 4%.
Simple Interest is given as:
I = (P * R * T) / 100
where P = principal (amount borrowed)
R = rate
T = length of years
Therefore:
[tex]444 = (P * 3 * 4) / 100\\\\444 = 12P / 100\\\\12P = 444 * 100\\\\12P = 44400\\\\P = 44400 / 12\\[/tex]
P = $3700
She borrowed $3700
b) If the simple interest was 5%, then:
I = (3700 * 5 * 3) / 100 = $555
The interest would be $555.
Crime and Punishment: In a study of pleas and prison sentences, it is found that 45% of the subjects studied were sent to prison. Among those sent to prison, 40% chose to plead guilty. Among those not sent to prison, 55% chose to plead guilty.
(A) If one of the study subjects is randomly selected, find the probability of getting someone who was not sent to prison.
(B) If a study subject is randomly selected and it is then found that the subject entered a guilty plea, find the probability that this person was not sent to prison.
Answer:
(a) The probability of getting someone who was not sent to prison is 0.55.
(b) If a study subject is randomly selected and it is then found that the subject entered a guilty plea, the probability that this person was not sent to prison is 0.63.
Step-by-step explanation:
We are given that in a study of pleas and prison sentences, it is found that 45% of the subjects studied were sent to prison. Among those sent to prison, 40% chose to plead guilty. Among those not sent to prison, 55% chose to plead guilty.
Let the probability that subjects studied were sent to prison = P(A) = 0.45
Let G = event that subject chose to plead guilty
So, the probability that the subjects chose to plead guilty given that they were sent to prison = P(G/A) = 0.40
and the probability that the subjects chose to plead guilty given that they were not sent to prison = P(G/A') = 0.55
(a) The probability of getting someone who was not sent to prison = 1 - Probability of getting someone who was sent to prison
P(A') = 1 - P(A)
= 1 - 0.45 = 0.55
(b) If a study subject is randomly selected and it is then found that the subject entered a guilty plea, the probability that this person was not sent to prison is given by = P(A'/G)
We will use Bayes' Theorem here to calculate the above probability;
P(A'/G) = [tex]\frac{P(A') \times P(G/A')}{P(A') \times P(G/A') +P(A) \times P(G/A)}[/tex]
= [tex]\frac{0.55 \times 0.55}{0.55\times 0.55 +0.45 \times 0.40}[/tex]
= [tex]\frac{0.3025}{0.4825}[/tex]
= 0.63
A car travels 133 mi averaging a certain speed. If the car had gone 30 mph faster, the trip would have taken 1 hr less. Find the car's average speed.
Answer:
49.923 mph
Step-by-step explanation:
we know that the car traveled 133 miles in h hours at an average speed of x mph.
That is, xh = 133.
We can also write this in terms of hours driven: h = 133/x.
If x was 30 mph faster, then h would be one hour less.
That is, (x + 30)(h - 1) = 133, or h - 1 = 133/(x + 30).
We can rewrite the latter equation as h = 133/(x + 30) + 1
We can then make a system of equations using the formulas in terms of h to find x:
h = 133/x = 133/(x + 30) + 1
133/x = 133/(x + 30) + (x + 30)/(x + 30)
133/x = (133 + x + 30)/(x + 30)
133 = x*(133 + x + 30)/(x + 30)
133*(x + 30) = x*(133 + x + 30)
133x + 3990 = 133x + x^2 + 30x
3990 = x^2 + 30x
x^2 + 30x - 3990 = 0
Using the quadratic formula:
x = [-b ± √(b^2 - 4ac)]/2a
= [-30 ± √(30^2 - 4*1*(-3990))]/2(1)
= [-30 ± √(900 + 15,960)]/2
= [-30 ± √(16,860)]/2
= [-30 ± 129.846]/2
= 99.846/2 ----------- x is miles per hour, and a negative value of x is neglected, so we'll use the positive value only)
= 49.923
Check if the answer is correct:
h = 133/49.923 = 2.664, so the car took 2.664 hours to drive 133 miles at an average speed of 49.923 mph.
If the car went 30 mph faster on average, then h = 133/(49.923 + 30) = 133/79.923 = 1.664, and 2.664 - 1 = 1.664.
Thus, we have confirmed that a car driving 133 miles at about 49.923 mph would have arrive precisely one hour earlier by going 30 mph faster
a 12- inch ruler is duvided into 3 parts. the large part is 3 times longer than the small. the meddium part is times longer than then small, the medium part is 2 times long as the smallest .how long is the smallest part?
Answer:
2 inches
Step-by-step explanation:
x= smallest
3x=largest
2x=medium
x+3x+2x=12
6x=12
x=2
so smallest is 2
largest is 6 (3x)
medium is 4 (2x)
2+6+4=12
given sin theta=3/5 and 180°<theta<270°, find the following: a. cos(2theta) b. sin(2theta) c. tan(2theta)
I hope this will help uh.....