Answer:
7.5
Step-by-step explanation:
it is feometeic progression
r=0.5
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
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.
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]
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
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
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.
A sample of 24 observations is taken from a population that has 150 elements. The sampling distribution of is _____. an. approximately normal because is always approximately normally distributed b. approximately normal because the sample size is large in comparison to the population size c. approximately normal because of the central limit theorem d. normal if the population is normally distributed
Answer:
d. normal if the population is normally distributed
Step-by-step explanation:
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this question:
Sample size less than 30, so only will be normal if the population is normally distribution, and thus the correct answer is given by option d.
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
$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.
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%
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!
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.
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
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! :)
write fifty and two hundreds eight thousandths as a mixed decimal
Answer:
Pretty sure it's 0.528
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
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
HELP HELP HELP MATH⚠️⚠️⚠️⚠️⚠️
Find four consecutive integers with the sum of 2021
Answer:
This problem has not solution
Step-by-step explanation:
lets the integers be:
x
x+1
x+2
x+3
so:
x+(x+1)+(x+2)+(x+3)=2021
x+x+x+x+1+2+3=2021
4x+6=2021
4x=2021-6=2015
x=2015/4=503.75
x is not a integer
help with this please !
Answer:
c
Step-by-step explanation:
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
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
Marissa constructed a figure with these views.
HELP ASAP EXTRA POINTS
Answer:
a triangular pyramid
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]
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).
:
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.
Unit: Decimals
Progress
The movement of the progress bar may be uneven becouse questions can be worth more or less (including zero) depending on your answer.
Which digit is in the thousandths place in the number 48.5302?
O 3
O 5
O 2
O 0
Submit
Pass
Don't know answer
Answer:
0
Step-by-step explanation:
48.5302
The number just to the left of the decimal point is the ones place. The 8 is in the ones place. Every place to the left is 10 times greater than the previous one, and every place to the right is 10 times smaller than the previous one.
48.5302
4 - tens place - 4 tens means 40
48.5302
8 - ones place - 8 ones means 8
48.5302
5 - tenths place - 5 tenths means 0.5
48.5302
3 - hundredths place - 3 hundredths means 0.03
48.5302
0 - thousandths place - 0 thousandths means 0.000
48.5302
2 - ten-thousandths place - 2 ten-thousandths means 0.0002
Answer: 0
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
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
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)
Tìm GTLN và GTNN của
M=x+2y biết x^2 + 4y^2
Answer:
saya tidak tahu !)!(!)!(!)!(!)!(!)!(!)!(!)!(!)!(!)!(!)!(!)!(!)!(!qjjjjbugg7yhhuhhhuhgy fundes per grallahegeuebevhehehnjķßùèò395oìùýþŕèwqàďhĺ
