Answer:
-7/25
Step-by-step explanation:
1/5 ÷ -5/7
Copy dot flip
1/5 * -7/5
-7/25
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.
PLEASE HELP ASAP !!!! THANKS
Answer:
Its the first one
$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.
what shape is the 1st one? I already know the second one
the first one is a quadrilateral
Step-by-step explanation:
a quadrilateral have four side
Solve the following quadratic function
by utilizing the square root method.
y = 16 – x2
Answer:
The solutions are x = -4 and x = 4.
Step-by-step explanation:
Solving a quadratic equation:
We have to find x for which [tex]y = 0[/tex].
In this question:
[tex]y = 16 - x^2[/tex]
So
[tex]16 - x^2 = 0[/tex]
[tex]x^2 = 16[/tex]
[tex]x = \pm \sqrt{16}[/tex]
[tex]x = \pm 4[/tex]
The solutions are x = -4 and x = 4.
What is AE?
Enter your answer in the box.
units
Answer:
AE = 18
Step-by-step explanation:
The triangles must be similar so we can use ratios
2x+4 12
-------- = ---------
x+8 10
Using cross products
(2x+4)10 = 12 (x+8)
20x +40 = 12x+96
Subtract 12x from each side
20x-12x +40 = 12x-12x +96
8x +40 = 96
Subtract 40 from each side
8x = 96-40
8x = 56
Divide by 8
x = 56/8
x = 7
AE = 2x+4 = 2(7) +4 = 14+4 = 18
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)
Distance between two points
What is the length of the line?
Answer:
c
Step-by-step explanation:
that is the procedure above
write fifty and two hundreds eight thousandths as a mixed decimal
Answer:
Pretty sure it's 0.528
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
:
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.
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.
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
PAIesung
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Reading list
- Blake bought a motorcycle for $550 last year and sold it for $330 this year. What is his sale
price as a percentage of his purchase price?
Answer:
The sale price was 60% of the purchase price.
Step-by-step explanation:
Given that Blake bought a motorcycle for $ 550 last year and sold it for $ 330 this year, to determine what is his sale price as a percentage of his purchase price, the following calculation must be performed:
550 = 100
330 = X
330 x 100/550 = X
33000/550 = X
60 = X
Therefore, the sale price was 60% of the purchase price.
©/17
Correct
Question 1 of 17, Step 1 of 1
Write a mixed number to describe the length of the ribbon shown in the figure below.
Please enter your answer in the box below.
Inches
Answer
How to enter your answer
If your answer is a whole number, enter it in the left most box and leave the numerator and denominator boxes blank.
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Answer:
3 3/8 inches
Step-by-step explanation:
If you spend a little time looking at the marks on the ruler, you see that the smallest marks divide each inch into 8 parts. The ribbon comes to the 3rd small mark* after the 3-inch mark, so the length of the ribbon is 3 3/8 inches.
_____
* Technically, it is the second small mark, as the marks are small, medium, small. It marks the end of the third space, where each space is 1/8 inch.
Can I please get help it’s urgent . Find the lateral surface area and volume of the solid object.
Find the value of x.
A. 15
B. 10
C. 6
D. 60
Answer:
6x+24=10
4x=24
x=6
OAmalOHopeO
Answer:
Step-by-step explanation:
OK, Since angle B and angle K are equal and the line segment CE is a bisector,So the shape is divided into two equal triangular parts.
therefore angle E divide into two equal part(angles) ;then: 6x+24=10x--->4x=24--->x=6
A random number is selected from the interval [6.35, 10]. Find the probability that the number is within a distance of 0.25 from an even integer. (Answer as a decimal number, and round to 4 decimal places).
Let X be a random number selected from the interval. Then the probability density for the random variable X is
[tex]f_X(x)=\begin{cases}\dfrac1{10-6.35}=\dfrac1{3.65}\approx0.2740&\text{if }6.35\le x\le 10\\0&\text{otherwise}\end{cases}[/tex]
8 and 10 are the only even integers that fit the given criterion (6 is more than 0.25 away from 6.35), so that we're looking to compute
P(|X - 8| < 0.25) + P(|X - 10| < 0.25)
… = P(7.75 < X < 8.25) + P(9.75 < X < 10.25)
… = P(7.75 < X < 8.25) + P(9.75 < X < 10)
(since P(X > 10) = 0)
… = 0.2740 (8.25 - 7.75) + 0.2740 (10 - 9.75)
… = 0.2055
The room numbers of two adjacent classrooms are two consecutive odd numbers. If their sum is 860, find the classroom numbers
Answer:
429 & 431
Step-by-step explanation:
Consecutive numbers are numbers that come right after the other.
Let x = first room number
[tex]x+x+2=860\\2x+2=860\\2x=858\\x=429\\\\(429) + 2 = 431[/tex]
Therefore, the two room numbers are 429 and 431.
Plz show steps for this
Answer: Choice D. 3 : r=3
Step-by-step explanation:
Easiest method and probably only method given the graph without knowing exact points besides an asymptote at x=-3.
Since we know there is an asymptote at x=-3, we just solve for the denominator and find r, when x=-3.
We are setting equation equal to 0, because when the denominator is 0, the graph has an asymptote at that point.
x+r=0
-3+r=0
r=3
Answer:
r=3
Step-by-step explanation:
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
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
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).
A runner increases his velocity from 0 m/s to 20 m/s in 2.0 s. What was his average acceleration?
Answer:
[tex]a = \frac{dv}{dt } = \frac{20 - 0}{2} = 10[/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
please help me with this on the image
Answer:
6ab
Step-by-step explanation:
write your answer in simplest radical form
Answer:
[tex]5 \sqrt{2 } [/tex]
Step-by-step explanation:
In a 45 45 90 triangle, the hypotenuse is equal to
[tex]x \sqrt{2} [/tex]
where x is the length of a leg of the triangle
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
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.
Find m/c.
A
18 in
12 in
C
B
28 in