Answer: D
Step-by-step explanation:
D
Answer:
D
Step-by-step explanation:
it is between 0 and 1
- 18 = -3x + 6
Plz help
Answer:
8 =x
Step-by-step explanation:
- 18 = -3x + 6
Subtract 6 from each side
-18-6 = -3x+6-6
-24 = -3x
Divide each side by -3
-24/-3 = -3x/-3
8 =x
Answer:
x= 8
Step-by-step explanation:
[tex]\sf{}[/tex]
=> -3x+6 = -18
=> -3x+6-6= -8-6
=> -3x= -24
=> x= 8
1. Find the length of the side indicated.
6
9.1
?
Answer:
10.9
Step-by-step explanation:
[tex]a^2+b^2=c^2[/tex]
[tex]9.1^2+6^2=c^2[/tex]
[tex]82.81+36=c^2[/tex]
[tex]c^2=118.81[/tex]
[tex]c=\sqrt{118.81} =10.9[/tex]
2(3+ ‐ 4)(7‐3)÷(2‐ ‐2)
Simplify i to the 31st power
Answer:
?
How did you simplify
(20/2 + 4)/2
^^^ I NEED A EQUATION LIKE THAT BUT FOR IT TO EQUAL 16
Answer:
Here are a few examples.
(30/2 + 1)/2
(26/2 + 3)/2
(28/2 + 2)/2
Consider the following information. 1 hour = 3.6 · 10^3 seconds 1 day = 24 hours 1 year = 3.65 · 10^2 days Use scientific notation to calculate the number of seconds in 3 days.
Answer:
2.592 x 10^5
Step-by-step explanation:
1st step: Calculate for 1 day.
if 1 hour = 3.6 X 10³
24 hrs = (3.6 x 10³) X (2.4 x 10¹)
= 3.6 x 2.4 X 10⁴
= 8.64 x 10⁴
2nd Step: Calculate for 3 days;
if 1 day = 8.64 x 10⁴
3 days = (8.64 x 10⁴) X (0.3 x 10¹)
= 8.64 x 0.3 x 10^5
= 2.592 x 10^5
find the derivative of e power ax divide by log bx
Answer:
Step-by-step explanation:
Select the correct answer.plz answer fast due at 11:59
What is the solution for x in the equation?
-4 + 5x − 7 = 10 + 3x − 2x
A. x=4/13
B. x=13/4
C. x=4/21
D. x=21/4
Question: -4 + 5x -7 = 10 + 3x -2x
⇒ 5x -11 = 10 + x
⇒ 5x - x = 10+11
⇒ 4x = 21
⇒ x = 21/4
Answer is Option D
x = 21/4
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llana has 70¢ in her purse. All the coins have the same value. What could describe all the coins she has in her
purse?
71 pennies
8 dimes
1 0 0
14 nickels
3 quarters
Answer:
14 nickels
Step-by-step explanation:
One nickel is worth 5 cents, so find the value of 14 nickels by multiplying 14 by 5:
14(5)
= 70
So, 14 nickels is equal to the 70 cents that she has in her purse.
The correct answer is 14 nickels.
please help me!
simplify
Answer:
sec^2(x)
Tell me if I'm correct or wrong. If I'm correct, plz mark me brainliest!
Step-by-step explanation:
Recall that
[tex]\sin (\frac{\pi}{2} - x) = \cos x[/tex]
[tex]\cos (\frac{\pi}{2} - x) = \sin x[/tex]
So we can rewrite the given expression as
[tex]\dfrac{\cos^2 x}{\sin^2 x} + (\sin^2 x + \cos^2 x)[/tex]
[tex]\Rightarrow \dfrac{\cos^2 x}{\sin^2 x} + 1[/tex]
or
[tex]\dfrac{\cos^2 x + \sin^2 x}{\sin^2 x} = \dfrac{1}{\sin^2 x} = \csc^2 x[/tex]
One of the numbers 0 through 9 is drawn at random. Match the following events with their probabilities.
1. drawing 4
2. drawing an odd number
3. not drawing 3
4. drawing 5 or 6
5. drawing a number greater than 9
1/2
1/5
1/10
0
9/10
Answer:
Drawing 4 - 1/10
Drawing odd number - 1/2
Not drawing 3- 9/10
Drawing 5 or 6 - 1/5
Drawing a number greater than 9 - 0
Let me know if this helps!
How do I solve for y:
x²y² = 12/z²
Answer:
Y=Srt(12xs^2/z^2)
Step-by-step explanation:
Firstly
We multiply both sides with 1/x^2
We get
Y^2=12/z^2*1/x^2
Y^2=12x^2/z^2
Next: introduce a srt root
We have
Y=srt(12x^2/z^2)
Which of the following factorizations is correct?
O x3 + 27 = (x - 3)(x2 + 3x – 9)
O x - 27 = (x - 3)(x2 + 3x +9)
O x3 - 27 = (x - 3)(x2 +9)
O x3 + 27 = (x - 3)(x2 - 9)
Answer:
first one is correct bro
The correct factorization of x³ - 27 is (x - 3)(x² + 3x + 9).
What are the formulas for a³ + b³ and a³ - b³?a³ + b³ = (a + b)(a² -ab + b²) and a³ - b³ = (a - b)(a² + ab + b²).
If we know the formula of these two we can conclude by observing that
the second option x³ - 27 = (x - 3)(x² + 3x +9) is correct.
As x³ - 27
= x³ - 3³.
= (x - 3)(x² + 3x + 9).
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Amy works 40 hours a week and gets paid 15.50 an hour. Tom works 40 hours a week and gets paid 10 an hour. How much does Amy earn per week? How much does Tom get pay for week?
9514 1404 393
Answer:
Amy: $620
Tom: $400
Step-by-step explanation:
The pay for 40 hours (1 week) is 40 times the pay for 1 hour:
Amy: 40 × $15.50 = $620 . . . . Amy's pay for 1 week
Tom: 40 × $10.00 = $400 . . . . Tom's pay for 1 week
a game is played with a fishpond containing 100 fish; 90 white, 9 red, and 1 blue. a contestant randomly catches a fish and receives payments as follows: $0.30 for white, $1.00 for red, and $10.00 for blue. If it cists $0.60 to play this game, how much (on average) does a contestant win on each play
Answer:
loses 14 cents
- $0.14
Step-by-step explanation:
90% $0.30 $(0.30) $(0.27)
9% $1.00 $0.40 $0.04
1% $10.00 $9.40 $0.09
$(0.14)
Help me with this question plz
9514 1404 393
Answer:
17
Step-by-step explanation:
The points at the ends of the interval are ...
(0, f(0)) = (0, 0)
(7, f(7)) = (7, 119)
The average rate of change is given by the slope formula:
m = (y2 -y1)/(x2 -x1)
m = (119 -0)/(7 -0) = 119/7 = 17
How many more cherry pies have 60 to 89 cherries than 90 to 119 cherries?
Answer:
6 pies
Step-by-step explanation:
For 60 - 89 cherries there are 10 pies
For 90 -119 cherries there are 4 pies
10 -4 = 6
There are 6 more pies in the 60 -89 cherries category than in the 90-119 category
A cupboard costing Rs.16800 is depreciated at the rate of 15% per year. What will be the cost of the cupboard after 2 years.
Answer:
Rs. 11132
Step-by-step explanation:
formula for depriciation= P(1-R/100)^T
p= principal, r=rate, t=time
Answer:
15162
Step-by-step explanation:
after 1 year= 16800*95%= 15960
after 2 year= 15960*95%= 15162
(b) An economy has an agricultural industry and a textile industry. Each unit of agricultural output requires 0.4 unit of agricultural input and 0.1 unit of textiles input. Each unit of textiles output requires 0.1 unit of agricultural input and 0.2 unit of textiles input.
(i) Write the technology matrix for this economy. [2 marks]
(ii) If surpluses of 5 units of agricultural products and 195 units of textiles are desired, find the gross production of each industry
Leontief input output model (technology matrix) is an economic model that shows the quantitative relationship and sectorial interdependency in a national economy
The responses with regards to the question are;
(i) The technology matrix for the economy is presented as follows;
[tex]\mathbf{ A} =\left[\begin{array}{ccc}Agric&&Textile\\0.4&&0.1\\&&\\0.1&&0.2\end{array}\right] \begin{array}{ccc}\mathbf{Per \ Unit}\\Agriculture\\\\Textile\end{array}\right][/tex]
(ii) The required gross production of each industry to meet the desired surplus are;
50 units of agriculture and 250 units of textile
The reason the above values are correct is as follows:
(i) The given parameters are;
The industries in the economy = Agricultural industry and textile industry
Units of agricultural input required per unit of agricultural output = 0.4
Units of textile input required per unit of agricultural output = 0.1
Units of agricultural input required per unit of textile output = 0.1
Units of textile input required per unit of textile output = 0.2
Let X represent agriculture, and let Y represent textile, we have;
[tex]Agric \ for \ agric = \dfrac{0.4 \ units \ of \ agriculture}{1\ unit \ of \ agric \ produced} \times X \ Agric \ produced= 0.4 \cdot X[/tex]
[tex]Agric \ for \ textile = \dfrac{0.1 \ units \ of \ agriculture}{1\ unit \ of \ textile \ produced} \times Y \ textile \ produced= 0.1 \cdot Y[/tex]
We also have;
Textile for agriculture = 0.1·X
Textile for textile = 0.2·Y
Therefore;
X = 0.4·X + 0.1·Y
Y = 0.1·X + 0.2·Y
Therefore;
The technology matrix for the economy is presented as follows;
[tex]\mathbf{Technology \ matrix, A} =\left[\begin{array}{ccc}Agric&&Textile\\0.4&&0.1\\&&\\0.1&&0.2\end{array}\right] \begin{array}{ccc}\mathbf{Per \ Unit}\\Agriculture\\\\Textile\end{array}\right][/tex]
(ii) Let P represent the production vector, and let d represent the demand vector, we have;
[tex]P = \left[\begin{array}{c}X \\Y\end{array}\right][/tex], [tex]d = \left[\begin{array}{c}5 \\195\end{array}\right][/tex]
P = A·P + d
∴ P - A·P = d
Therefore;
[tex]P = \mathbf{ \dfrac{d}{(I - A)}}[/tex]
Where I = The 2 by 2 identity matrix
We get;
[tex]I - A =\left[\begin{array}{ccc}1&&0\\&&\\0&&1\end{array}\right] - \left[\begin{array}{ccc}0.4&&0.1\\&&\\0.1&&0.2\end{array}\right] = \mathbf{\left[\begin{array}{ccc}0.6&&-0.1\\&&\\-0.1&&0.8\end{array}\right]}[/tex]
With the use of a graphing calculator, we have;
[tex]P =\left[\begin{array}{c}X \\Y\end{array}\right] = \dfrac{\left[\begin{array}{c}5 \\195\end{array}\right]}{\left[\begin{array}{ccc}0.6&&-0.1\\&&\\-0.1&&0.8\end{array}\right]} = \left[\begin{array}{ccc}50\\\\\ 250\end{array}\right][/tex]
The required gross product of agriculture, X = 50 units
The required gross product of textile, Y = 250 units
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We have that he technology matrix for this economy and the the gross production of each industry are
a) [tex]X= \begin{vmatrix}0.4 & 0.1 \\0.1 & 0.2\end{vmatrix}[/tex]
b) [tex]\begin{vmatrix}A\\T\end{vmatrix}=\begin{vmatrix}50\\250\end{vmatrix}[/tex]
From the Question we have told that
Each unit of agricultural output requires 0.4 unit of agricultural input
Each unit of agricultural output requires 0.1 unit of textiles input.
Each unit of textiles output requires 0.1 unit of agricultural input
Each unit of textiles output requires 0.2 unit of textiles input.
Generally the technology matrix for this economy is given below
With
X =Agricultural industry Gross output
Y= Textile industry Gross Output
Therefore
[tex]X= \begin{vmatrix}0.4 & 0.1 \\0.1 & 0.2\end{vmatrix}[/tex]
b)
From the Question we are told that
Surpluses of 5 units of agricultural products and 195 units of textiles are desired.
Therefore, we have Desired surplus matrix of
[tex]D= \begin{vmatrix}5\\195\end{vmatrix}[/tex]
Generally the Technology equation is mathematically given as
[tex](I-X)\phi=D[/tex]
Where
X =Agricultural industry Gross output
I=A Unit matrix
\phi=Matrix of gross production
Therefore
[tex]\begin{vmatrix}1 & 0\\0 & 1\end{vmatrix}-(\begin{vmatrix}0.4 & 0.1 \\0.1 & 0.2\end{vmatrix}))\begin{vmatrix}A\\T\end{vmatrix}=\begin{vmatrix}5\\195\end{vmatrix}[/tex]
[tex]\begin{vmatrix}A\\T\end{vmatrix}=\begin{vmatrix}50\\250\end{vmatrix}[/tex]
In conclusion
The technology matrix for this economy and the the gross production of each industry are
[tex]X= \begin{vmatrix}0.4 & 0.1 \\0.1 & 0.2\end{vmatrix}[/tex]
[tex]\begin{vmatrix}A\\T\end{vmatrix}=\begin{vmatrix}50\\250\end{vmatrix}[/tex] Respectively
In conclusion
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(4x - 3) × (x + 2)=0
Hi,
AxB = 0 means A=0. or B=0
so 2 solutions :
4x-3= 0
4x=3
x = 3/4
and x+2 = 0.
x= -2
solutions are : -2 +and 3/4
2 answers
1) -2
2) -3/4
At a sale this week, a desk is being sold for $312. This is a 35% discount from the original price.
What is the original price?
Answer:
$480
Step-by-step explanation:
Let original price be x
{Original price - discount amount = sale price}
So, x - (x * 35/100) = 312
x - 35x/100 = 312
100x - 35x / 100 = 312
65x/100 = 312
65x = 31200
x = 31200/65
x = $480
So I think the original price of desk was $480
The original price of the desk is $480.
What is discount?A discount is the reduction of either the monetary amount or a percentage of the normal selling price of a product or service.
Given that, a desk is being sold for $312.
Let the original price of the desk be x.
Here, x-35% of x =312
x- 35/100 ×x=312
x-0.35x=312
0.65x=312
x=312/0.65
x=480
Therefore, the original price of the desk is $480.
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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)
8 millions =
how many hundred thousands
Answer:
100000
Step-by-step explanation:
100000x80=8m
Which compound inequality could this graph be the solution of?
-4 ≤ x ≤ 6
x ≥ -4
3x ≥ -12
3x + 12 ≥ 0
3x + 14 ≥ 2
6 ≥ x
6 - x ≥ 0
1 - x ≥ -5
A. 3x + 14 ≥ 2 and 1 - x ≥ -5
hat is the measure of the unknown angle?
Image of a straight angle divided into two angles. One angle is ninety seven degrees and the other is unknown
pls
Let on of the angle be x
ATQ
97 + x = 180
x = 180 - 97 (Angles on same line adds upto 180°) (Linear pair cuz 2 angles)
x = 83
The unknown angle is 83°
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I need help to fine the statement that is true
Answer:
option A
Step-by-step explanation:
wx and zy making 90 angle with each other therefore they are perpendicular.
wx and ab making 0 angle with each other therefore they are parallel
What is the equation of a line that passes through points A
(1, 2) and B (4, 3)?
Answer:
3y=x+5
Step-by-step explanation:
The slope of the line is (1/3). The equation of the line is y=(1/3)x+5/3 or 3y=x+5
n(AnB)=3 and n(AuB)=10, then find (p(A∆B))?
I assume A ∆ B denotes the symmetric difference of A and B, i.e.
A ∆ B = (B - A) U (A - B)
where - denotes the set difference or relative complement, e.g.
B - A = {b ∈ B : b ∉ A}
It can be established that
A ∆ B = (A U B) - (A ∩ B)
so that
n(A ∆ B) = n(A U B) - n(A ∩ B) = 10 - 3 = 7
Not sure what you mean by p(A ∆ B), though... Probability?
2067 Supp Q.No. 2a Find the sum of all the natural numbers between 1 and 100 which are divisible by 5. Ans: 1050
5
Answer:
1050
Step-by-step explanation:
Natural Numbers are positive whole numbers. They aren't negative, decimals, fractions. We can just divide 5 into 100 to find how many natural numbers go up to 100 and just add them but that is just to much.
There is a easier method.
E.g: Natural Numbers that are divisible by a Nth Number. is the same as adding the Nth Numbers to a multiple of that Nth Term. For example, let say we need to find numbers divisible by 2. We know that 4 is divisible by 2 because 4/2=2. We can add the Nth numbers which is 2 to 4. 4+2=6. And 6 is divisible by 2 because 6/2=3. We can call this a arithmetic series. A series which has a pattern of adding a common difference
Back to the problem, we can use the sum of arithmetic series formula,
[tex]y = x( \frac{z {}^{1} + {z}^{n} }{2} )[/tex]
Where x is the number of terms in our sequence. Z1 is the fist term of our series. ZN is our last term. And y is the sum of all of the terms
The first term is 5, the numbers of terms being added is 20 because 100/5=20. The last term is 100.
[tex]y = 20( \frac{5 + 100}{2} )[/tex]
[tex]y = 20( \frac{105}{2} )[/tex]
[tex]y = 1050[/tex]
Find the area of the figure
A =
Answer:
207m
Step-by-step explanation:
(17+5)*6=132
(15*5)=75
132+75=207