Answer:
[tex]Pr= 0.00725[/tex]
Step-by-step explanation:
Given
[tex]p = 0.05[/tex] ---- probability that each component fails
[tex]n = 3[/tex]
Required
[tex]P(System\ Overheats)[/tex]
We understand that the system will overheat if at least 2 component fails; Assume the components are: x, y and z
The events that the system will overheat are: xyz', xy'z, x'yz and xyz
Where ' means that the component did not fail, and the probability is 1 - p (i.e. complement rule)
So, we have:
[tex]xyz' \to 0.05 * 0.05 * (1 - 0.05) = 0.002375[/tex]
[tex]xy'z \to 0.05 * (1 - 0.05)* 0.05 = 0.002375[/tex]
[tex]x'yz \to (1 - 0.05)* 0.05 * 0.05 = 0.002375[/tex]
[tex]xyz \to 0.05 * 0.05 * 0.05 =0.000125[/tex]
So, the required probability is:
[tex]Pr= 0.002375 +0.002375 +0.002375 + 0.000125[/tex]
[tex]Pr= 0.00725[/tex]
A three-year interest rate swap has a level notional amount of 300,000. Each settlement period is one year and the variable rate is the one-year spot interest rate at the beginning of the settlement period. One year has elapsed and the one-year spot interest rate at the start of year 2 is 4.45%.
Time to Maturity 1 2 3 4 5
Price of zero coupon bond with Maturity value 1 0.97 0.93 0.88 0.82 0.75
Calculate the net swap payment by the payer at the end of the second year.
A. −400
B. −300
C. −200
D. −100
E. 0
Hint : Find the swap rate R using the table and then use R and the one-year spot rate at the start of year 2 to find the net swap payment at the end of year 2.
Answer:
A. -400
Step-by-step explanation:
We solve for the swap rate
R = (1-p3)/(p1+p2+p3)
R = 1-0.88/0.97+0.93+0.88
= 0.12/2.78
= 0.04317
Remember 4.45% is the one year spot rate for the second option
Net swap
= 300000*0.04317-300000*0.0445
= 12951-13350
= -399
This is approximately -400
So the net swap payment at the end of the second year is option a, -400
Mr. Thomas invested an amount of ₱13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be ₱3508, what was the amount invested in Scheme B?
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Answer:
₱6400
Step-by-step explanation:
Let 'b' represent the amount invested in scheme B. Then 13900-b is the amount invested in scheme A. The total interest for 2 years is then ...
14%(13900-b)(2) +11%(b)(2) = 3508
1946 -0.03b = 1754 . . . . . . divide by 2, simplify
-0.03b = -192 . . . . . . . . . subtract 1946
b = 6400 . . . . . . . . . . . divide by -0.03
The amount invested in scheme B was ₱6400.
At the end of 2 years, P dollars invested at an interest rater compounded annually increases to an amount, A dollars, given by the following formula.
A = P(1+r)?
Find the interest rate if $192 increased to $363 in 2 years. Write your answer as a percent..
-
Annual compound interest rate = % (Type an integer or a decimal.)
Answer:
37.5%
Step-by-step explanation:
A=P(1+r)^t
363=192*(1+r)^2
1.375=1+r, r=0.375=37.5%
A quality-conscious disk manufacturer wishes to know the fraction of disks his company makes which are defective.
a. Suppose a sample of 865 floppy disks is drawn. Of these disks, 112 were defective. Using the data, estimate the proportion of disks which are defective.
b. Suppose a sample of 865 floppy disks is drawn. Of these disks, 112 were defective. Using the data, construct the 95% confidence interval for the population proportion of disks which are defective.
Answer:
a) 0.1295
b) The 95% confidence interval for the population proportion of disks which are defective is (0.1071, 0.1519).
Step-by-step explanation:
Question a:
112 out of 865, so:
[tex]\pi = \frac{112}{865} = 0.1295[/tex]
Question b:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
For this problem, we have that:
[tex]n = 865, \pi = 0.1295[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1295 - 1.96\sqrt{\frac{0.1295*0.8705}{865}} = 0.1071[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1295 + 1.96\sqrt{\frac{0.1295*0.8705}{865}} = 0.1519[/tex]
The 95% confidence interval for the population proportion of disks which are defective is (0.1071, 0.1519).
Tìm GTLN và GTNN của
M=x+2y biết x^2 + 4y^2
Answer:
saya tidak tahu !)!(!)!(!)!(!)!(!)!(!)!(!)!(!)!(!)!(!)!(!)!(!)!(!qjjjjbugg7yhhuhhhuhgy fundes per grallahegeuebevhehehnjķßùèò395oìùýþŕèwqàďhĺ
rotation 90 degrees counterclockwise about the origin
I'm going to try my best to explain 90° rotation:
So, you know that if you rotate something 180°, it's completely flipped (think about spinning around half-way).
Or if you spin something 360°, you spin around the whole way and end up in the same spot that you did when you started.
Notice how 90 is actually 1/4 of 360.
So imagine spinning instead of 180, spinning half of that. so you barely rotate. That's exactly what you're doing to this shape here. and if you do it about the origin counterclockwise, the origin is (0,0) so I drew it in Quadrant III, as you can see in my attachment.
You can see that every point has been moved by 90°, I put all of the variables there so you could visualize it better!
I hope this helped, let me know if you have any questions! :)
Given a parametric curve
{x = 2 cost
{y = 4 sint 0 <= t <= π
a. Set up but do NOT evaluate an integral to find the area of the region enclosed by the curve and the x-axis.
b. Set up but do NOT evaluate an integral to find the area of the surface obtained by rotating the curve about the x-axis.
(a) The area of the region would be given by the integral
[tex]\displaystyle\int_0^\pi y(t)\left|x'(t)\right|\,\mathrm dt = 8 \int_0^\pi \sin^2(t)\,\mathrm dt[/tex]
(b) The area of the surface of revolution would be given by
[tex]\displaystyle\int_0^\pi y(t)\sqrt{x'(t)^2+y'(t)^2}\,\mathrm dt = 4\int_0^\pi\sin(t)\sqrt{4\sin^2(t)+16\cos^2(t)}\,\mathrm dt[/tex]
evaluate the expression when c= -4 and x=5
x-4c
Answer:
21
Step-by-step explanation:
Fill in x into 5 and c into -4
5-4(-4)
21
Answer: -11
Step-by-step explanation:
x=5
and
c=4
the equation written in numbers is:
5-4x4 which simplified equals 5-16
this equals 5-5-11 so it should be -11
The dotplots below display the scores for two classes on a 30-point statistics quiz. Class A has 26 students and Class B has 25 students.
2 dot plots. For Class A, the dots are spread out more. For Class B, the dots are more grouped together.
Which statement best compares the variability of the quiz scores for Class A and Class B?
The scores on the quiz for Class A have more variability than the scores for Class B.
The scores on the quiz for Class B have more variability than the scores for Class A.
The scores on the quiz for Class A have less variability than the scores for Class B.
The scores on the quiz for Class B have about the same variability as the scores for Class A.
I think its (A), The scores on the quiz for Class A have more variability than the scores for Class B. Can someone check me?
The statement that best compares the variability of the quiz scores for Class A and Class B is A; The scores on the quiz for Class A have more variability than the scores for Class B.
How does the dot plot work?Suppose we're measuring something whose values are numeric. For each value of that thing we observe, we plot a dot above that value in the number line. Thus, the total number of dots in the dot plot tells us the total number of observations of the values of that thing we did.
LEt suppose if we observed the value 'x', then we will make a dot above 'x'. If there is already a dot over 'x', then we will make a new dot over that dot.
The dot plot is skewed to the right due to majority of the data occur to the right of the dot plot.
An outlier is a number that is way smaller or way larger than that of other numbers in a data set. Outliers usually occur with a frequency of one.
For Class A the dots are spread out more.
For Class B, the dots are more grouped together.
Here, Looking at the dot plot, it can be seen that the numbers that range from 16 to 21 have only one dot.
The statement that best compares the variability of the quiz scores for Class A and Class B is A; The scores on the quiz for Class A have more variability than the scores for Class B.
To learn more about outliers, please check: brainly.com/question/27197311
SPJ2
Trevor is studying a polynomial function f(x). Three given roots of f(x) are -7, 2i, and 7. Trevor concludes that f(x) must
be polynomial with degree 3. Which statement is true?
Answer:
B Trevor isn't correct because -2i must also be a root
Step-by-step explanation:
A Trevor is correct.
B Trevor is not correct because –2i must also be a root.
C Trevor is not correct there cannot be an odd number of roots.
D Trevor is not correct because there cannot be both rational and complex roots.
statements^^^^^^
The statement that f(x) must be a polynomial with degree 3 is not necessarily true based on the given roots.
What is a function?A function has an input and an output.
A function can be one-to-one or onto one.
It simply indicated the relationships between the input and the output.
Example:
f(x) = 2x + 1
f(1) = 2 + 1 = 3
f(2) = 2 x 2 + 1 = 4 + 1 = 5
The outputs of the functions are 3 and 5
The inputs of the function are 1 and 2.
We have,
Since one of the roots is complex (2i), it follows that the coefficients of f(x) must be complex as well, and therefore f(x) must be a polynomial with complex coefficients.
However, it is possible for f(x) to have a higher degree than 3.
For example,
The polynomial (x + 7)(x - 2i)(x + 2i)(x - 7) has degree 4 and has the given roots of -7, 2i, and 7. Therefore, f(x) could be a polynomial with degree 4 or higher.
Thus,
The statement that f(x) must be a polynomial with degree 3 is not necessarily true based on the given roots.
Learn more about functions here:
https://brainly.com/question/28533782
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What is 9,000,000 + 8,000 + 90,000,000 + 100 + 2 + 90,000 + 90 in standard form?
Answer:
i thank its 4000
Step-by-step explanation:
please mark this answer as brainlist
:
The width of a rectangle is 5 cm more than triple its length. The perimeter of the
rectangle is 240 cm. What is the length and width of the rectangle?
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Answer:
length: 28.75 cmwidth: 91.25 cmStep-by-step explanation:
Let L represent the length of the rectangle. Then the width is W=5+3L, and the perimeter is ...
P = 2(L+W)
240 = 2(L +(5 +3L))
120 = 5 +4L
115 = 4L
115/4 = L = 28.75 . . . . cm
W = 5+3L = 5 +3(28.75) = 91.25 . . . . cm
The length and width of the rectangle are 28.75 cm and 91.25 cm.
A manufacturer of computer memory chips produces chips in lots of 1000. If nothing has gone wrong in the manufacturing process, at most 7 chips each lot would be defective, but if something does go wrong, there could be far more defective chips. If something goes wrong with a given lot, they discard the entire lot. It would be prohibitively expensive to test every chip in every lot, so they want to make the decision of whether or not to discard a given lot on the basis of the number of defective chips in a simple random sample. They decide they can afford to test 100 chips from each lot. You are hired as their statistician.
There is a tradeoff between the cost of eroneously discarding a good lot, and the cost of warranty claims if a bad lot is sold. The next few problems refer to this scenario.
Problem 8. (Continues previous problem.) A type I error occurs if (Q12)
Problem 9. (Continues previous problem.) A type II error occurs if (Q13)
Problem 10. (Continues previous problem.) Under the null hypothesis, the number of defective chips in a simple random sample of size 100 has a (Q14) distribution, with parameters (Q15)
Problem 11. (Continues previous problem.) To have a chance of at most 2% of discarding a lot given that the lot is good, the test should reject if the number of defectives in the sample of size 100 is greater than or equal to (Q16)
Problem 12. (Continues previous problem.) In that case, the chance of rejecting the lot if it really has 50 defective chips is (Q17)
Problem 13. (Continues previous problem.) In the long run, the fraction of lots with 7 defectives that will get discarded erroneously by this test is (Q18)
Problem 14. (Continues previous problem.) The smallest number of defectives in the lot for which this test has at least a 98% chance of correctly detecting that the lot was bad is (Q19)
(Continues previous problem.) Suppose that whether or not a lot is good is random, that the long-run fraction of lots that are good is 95%, and that whether each lot is good is independent of whether any other lot or lots are good. Assume that the sample drawn from a lot is independent of whether the lot is good or bad. To simplify the problem even more, assume that good lots contain exactly 7 defective chips, and that bad lots contain exactly 50 defective chips.
Problem 15. (Continues previous problem.) The number of lots the manufacturer has to produce to get one good lot that is not rejected by the test has a (Q20) distribution, with parameters (Q21)
Problem 16. (Continues previous problem.) The expected number of lots the manufacturer must make to get one good lot that is not rejected by the test is (Q22)
Problem 17. (Continues previous problem.) With this test and this mix of good and bad lots, among the lots that pass the test, the long-run fraction of lots that are actually bad is (Q23)
Step-by-step explanation:
A manufacturer of computer memory chips produces chips in lots of 1000. If nothing has gone wrong in the manufacturing process, at most 7 chips each lot would be defective, but if something does go wrong, there could be far more defective chips. If something goes wrong with a given lot, they discard the entire lot. It would be prohibitively expensive to test every chip in every lot, so they want to make the decision of whether or not to discard a given lot on the basis of the number of defective chips in a simple random sample. They decide they can afford to test 100 chips from each lot. You are hired as their statistician.
There is a tradeoff between the cost of eroneously discarding a good lot, and the cost of warranty claims if a bad lot is sold. The next few problems refer to this scenario.
Problem 8. (Continues previous problem.) A type I error occurs if (Q12)
Problem 9. (Continues previous problem.) A type II error occurs if (Q13)
Problem 10. (Continues previous problem.) Under the null hypothesis, the number of defective chips in a simple random sample of size 100 has a (Q14) distribution, with parameters (Q15)
Problem 11. (Continues previous problem.) To have a chance of at most 2% of discarding a lot given that the lot is good, the test should reject if the number of defectives in the sample of size 100 is greater than or equal to (Q16)
Problem 12. (Continues previous problem.) In that case, the chance of rejecting the lot if it really has 50 defective chips is (Q17)
Problem 13. (Continues previous problem.) In the long run, the fraction of lots with 7 defectives that will get discarded erroneously by this test is (Q18)
Problem 14. (Continues previous problem.) The smallest number of defectives in the lot for which this test has at least a 98% chance of correctly detecting that the lot was bad is (Q19)
(Continues previous problem.) Suppose that whether or not a lot is good is random, that the long-run fraction of lots that are good is 95%, and that whether each lot is good is independent of whether any other lot or lots are good. Assume that the sample drawn from a lot is independent of whether the lot is good or bad. To simplify the problem even more, assume that good lots contain exactly 7 defective chips, and that bad lots contain exactly 50 defective chips.
Problem 15. (Continues previous problem.) The number of lots the manufacturer has to produce to get one good lot that is not rejected by the test has a (Q20) distribution, with parameters (Q21)
Problem 16. (Continues previous problem.) The expected number of lots the manufacturer must make to get one good lot that is not rejected by the test is (Q22)
Problem 17. (Continues previous problem.) With this test and this mix of good and bad lots, among the lots that pass the test, the long-run fraction of lots that are actually bad is (Q23)
HELP PLEASE ASAPPPP!!
Answer:
12
Step-by-step explanation:
(10/y + 13) -3
Let y=5
(10/5 + 13) -3
PEMDAS says parentheses first
(2 +13) -3
15 -3
12
Marissa constructed a figure with these views.
HELP ASAP EXTRA POINTS
Answer:
a triangular pyramid
Simplify: (-2)^-3
a) 8
b) 1/8
c) -8
d) -1/8
Answer:
-1/8
Step-by-step explanation:
(-2)^-3
We know that a^-b = 1/a^b
1/(-2)^3
We know (-2)^3 = -8
1/(-8)
-1/8
What 2/3 as a percentage
Answer:
66 2/3 %
Step-by-step explanation:
2/3
Take 2 and divide by 3
.666666(repeating)
Multiply by 100
66.6666(repeating)%
But .6666666(repeating) is 2/3
66 2/3 %
Answer:
[tex] 66.66\%[/tex]
Step-by-step explanation:
[tex] \frac{2}{3} \times 100\% \\ \frac{200}{3} \% \\ 66.66\%[/tex]
Can’t find answers online to check mine.
Answer:
3. 100% = 1
3/4 = 0.75
Now, 0.75 is halfway between 0.5 and 1, so Chris is correct.
4. 10% = 10/100 = 0.1
3/5 = 0.6
Now, 0.2 is not halfway between 0.1 and 0.6, so Emily is wrong.
Answer:
3 đúng 4 wrong
Step-by-step explanation:
100%=1
giữa 0, 5 và 1 =(0,5+1)/2=3/4
10%= 0,1
giữa 0,1 và 3/5 =(0,1+3/5)/2= 0,35 #0,2
The lines shown below are parallel. If the green line has a slope of -2, what is
the slope of the red line?
Answer:
Hi! There's no picture, but we don't need that to find the answer. Parallel lines always have the same slope. I suppose you're saying that the green and red line are parallel -- so, the red line's slope is also -2.
-2 <--
Hope this helps!! Have a nice day & please mark brainliest if you don't mind!
write fifty and two hundreds eight thousandths as a mixed decimal
Answer:
Pretty sure it's 0.528
g Jack is polling New Yorkers to determine what proportion of them want to legalize recreational marijuana. How many New Yorkers should the sample to ens
Answer:
1006
Step-by-step explanation:
Take proportion, p = 62%
n = (Z² * pq) / M.E²
Margin error = 0.03
p = 62% = 0.62
q = 1 - p = 1 - 0.62 = 0.38
Zcritical at 95% = 1.96
n = (1.96² * (0.62*0.38)) / 0.03²
n = 0.90508096 / 0.0009
n = 1005.6455
n = 1006
Sample size = 1006
Find the value of x.
Answer: x = 6
Concept:
From the given graph, we can see that it is an isosceles triangle since the two base angles are congruent.
In geometry, an isosceles triangle is a triangle that has two sides of equal length.
If you are still confused, you may refer to the attachment below for a graphical explanation or tell me.
Solve:
Given information
One side = 6
Second side = x
Given expression deducted from the definition of an isosceles triangle
Second side = First side
Substitute values into the expression
[tex]\boxed {x=6}[/tex]
Hope this helps!! :)
Please let me know if you have any questions
Help with question b please
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Answer:
(a) 5.82 cm (correctly shown)
(b) 10.53 cm
Step-by-step explanation:
a) The length BC can be found from the law of sines:
BD/sin(C) = BC/sin(D)
BC = BC·sin(C)/sin(D) = (6 cm)sin(48°)/sin(50°) ≈ 5.82 cm
__
b) The angle ABD is the sum of the angles shown:
angle ABD = 50° +48° = 98°
We know the lengths BA and BD and the included angle ABD, so we can use the law of cosines to find AD.
AD² = BA² +BD² -2·BA·BD·cos(98°)
AD² ≈ 8² +5.82² -2(8)(5.82)(-0.139173) ≈ 110.8411
AD ≈ √110.8411 ≈ 10.53 . . . . cm
Onetta goes to the food court to get a salad and sandwich for lunch. The Daily Deli has 8 varieties of sandwiches and 3 salads. Better Bites has 2 varieties of sandwiches and 7 salads. The Lunch Spot has 5 varieties of sandwiches and 8 salads. Determine the number of ways Onetta can select a sandwich and a salad.
Answer:
Onetta can salect a sandwich and a salad in 78 different ways.
Step-by-step explanation:
Since Onetta goes to the food court to get a salad and sandwich for lunch, and the Daily Deli has 8 varieties of sandwiches and 3 salads, while Better Bites has 2 varieties of sandwiches and 7 salads, and the Lunch Spot has 5 varieties of sandwiches and 8 salads, to determine the number of ways Onetta can select a sandwich and a salad, the following calculation must be performed:
8 x 3 + 2 x 7 + 5 x 8 = X
24 + 14 + 40 = X
78 = X
Therefore, Onetta can salect a sandwich and a salad in 78 different ways.
What is the domain of the function shown on the graph?
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Answer:
all real numbers
Step-by-step explanation:
The arrows on the ends of the curve indicate that the graph extends to infinity horizontally. The domain is the horizontal extent, so is all real numbers.
__
Additional comment
Apparently, y=-7 is a horizontal asymptote, so the range is y > -7.
please help i am stuck on this assignment
Answer:
answer
x = -13/ 15, 0
Step-by-step explanation:
15x^2 + 13 x = 0
or, x(15x + 13) = 0
either, x = 0
or, 15x + 13 = 0
x = -13/15
Answer:
The answer should be C...............
imma sorry if I'm wrong
Which point is collinear to the point (2, 1)?
OA) (1,0)
OB) (3,2)
OC) (4,1)
OD (1,3)
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Answer:
all of them (A, B, C, D)
Step-by-step explanation:
Two unique points define a line. Any pair of unique points will always be collinear (reside on the same line).
Each of the points listed, together with (2, 1), will define a line. The two points will be collinear.
$17,818 is invested, part at 11% and the rest at 6%. If the interest earned from the amount invested at 11% exceeds the interest earned from the amount invested at 6% by $490.33, how much is invested at each rate? (Round to two decimal places if necessary.)
Answer:We know the total amount of money invested. $17818
x+y=17818,
We know that the difference in interest earned by the two accounts is $490.33
0.11*x-0.06*y=490.33
x=17818-y
We substitute for x
0.11*(17818-y)-0.06*y=490.33
We multiply out
1959.98-0.11y-0.06*y=490.33
We combine like terms.
1469.65=0.17*y
Isolate y
y=1469.65/0.17
y=8645 at 6%
Calculate x
x=17818-8645
x=9173 at 11%
Check
0.11*9173-0.06*8645=490.33
interest earned at 11%=1009.03
interest earned at 6%=518.70
1009.03-518.7=490.33
490.33=490.33
Since this statement is TRUE and neither amount is negative then all is well.We know the total amount of money invested. $17818
x+y=17818,
We know that the difference in interest earned by the two accounts is $490.33
0.11*x-0.06*y=490.33
x=17818-y
We substitute for x
0.11*(17818-y)-0.06*y=490.33
We multiply out
1959.98-0.11y-0.06*y=490.33
We combine like terms.
1469.65=0.17*y
Isolate y
y=1469.65/0.17
y=8645 at 6%
Calculate x
x=17818-8645
x=9173 at 11%
Check
0.11*9173-0.06*8645=490.33
interest earned at 11%=1009.03
interest earned at 6%=518.70
1009.03-518.7=490.33
490.33=490.33
Since this statement is TRUE and neither amount is negative then all is well.
Evaluate 3x ^ 2 + 3x - 9 , when x = 2
A=-3
B=3
C=9
D=27
Answer:
C. 9
Step-by-step explanation:
Start plugging in the number 2
3(2)^2+3(2)-9
6^2+6-9
12+6-9
18-9
9
5404 buttons are produced by a factory in Jebel Ali in a week. If the factory produced same number of buttons every day of the week, buttons produced in a day is _________________
We need to find the number of buttons the company produced per day
The company produced 772 buttons per day
Total bottons produced in a week = 5404
There are 7 days in a week
If the factory produced same number of buttons every day of the week
Then,
Buttons produced per day = Total bottons produced in a week / Total number of days in a week
= 5404 / 7
= 772 buttons
The company produced 772 buttons per day
Read more: https://brainly.com/question/24368335
