5. If 15 men can finish a work in 16 days in how many days will 8 men can finish the same work? a. 15 b. 20 c. 30 d. 42​

the sample space is:

{ (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (3, 1), (4, 1), (2, 2), (2, 3), (3, 2)}

Event A is rolling a sum less than 6.

Let's define the possible elements in this experiment as:

(outcome of dice 1, outcome of dice 2)

The outcomes where the sum is less than 6 are:

dice 1    dice 2     sum

  1               1           2

  1               2           3

  1               3          4

  1               4          5

  2               1          3

  3              1           4

  4               1          5

  2              2          4

  3              2          5

  2              3           5

So there are 10 outcomes, then the sample space is:

{ (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (3, 1), (4, 1), (2, 2), (2, 3), (3, 2)}

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Answer:

B) {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}

Step-by-step explanation:

If the question said less than 6 meaning you have to find all possible solution that are 5 or lower.

However, if the problem said equal or less than 6 then you have to find all possible solution that are 6 or lower.

B option is only option that don't have sum of 6. Therefore, option B is correct.

The equation of the function f(x) is f(x) = 2 sin(π/2(x + 6)) - 3

A sine function is represented as:

f(x) = A sin(B(x + C)) + D

Where

A = Amplitude

Period = 2π/B

C = Phase shift

D = Vertical shift

The requirements in the question are:

So, we can use the following assumptions

A = 2

Period = 4

C = 6

D = -3

So, we have:

f(x) = 2 sin(B(x + 6)) - 3

The value of B is

4 = 2π/B

This gives

B = π/2

So, we have:

f(x) = 2 sin(π/2(x + 6)) - 3

Using the representations in (a), we have:

See attachment for the graph of f(x)

Let θ = 3π

So, we have:

cos(3π)

This is calculated as:

cos(3π) = cos(2π + π)

Expand

cos(3π) = cos(2π) *cos(π) - sin(2π) *sin(π)

Evaluate

cos(3π) = -1

Let θ = 3π

So, we have:

sin(3π)

This is calculated as:

sin(3π) = sin(2π + π)

Expand

sin(3π) = sin(2π) *cos(π) + cos(2π) *sin(π)

Evaluate

sin(3π) = 0

Let θ = 3π

So, we have:

tan(2 * 3π)

tan(6π)

This is calculated as:

tan(6π) = tan(3π + 3π)

Evaluate

tan(3π) = 0

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Answer: x = -1 or x = 5/3

Step-by-step explanation:

|3x - 1| = 4

=> 3x - 1 = 4 or 3x - 1 = -4

=>x = 5/3 or x = -1

The domain of the functions are:

The domain of a function is the set of input values the function can take i.e. the set of values the independent variable can assume?

Function 1

The function is given as:

f(x) = √4x + 6

Set the radicand greater than 0

4x + 6 > 0

Subtract 6 from both sides

4x > -6

Divide by 4

x > -3/2

Express as interval notation

[-3/2, ∞)

Hence, the domain of f(x) = √4x + 6 is [-3/2, ∞) or x >-3/2

Function 2

The function is given as:

g(x) = -4√-20x - 6

Set the radicand greater than 0

-20x - 6 > 0

Add 6 to both sides

-20x > 6

Divide by -20

x < -6/20

Simplify

x < -3/10

Express as interval notation

(-∞, -3/10]

Hence, the domain of g(x) = -4√-20x - 6 is (-∞, -3/10] or x < -3/10

Function 3

The function is given as:

f(x) = 15 + √5x - 16

Set the radicand greater than 0

5x - 16 > 0

Add 16 to both sides

5x > 16

Divide by 5

x > 16/5

Express as interval notation

[16/5, ∞)

Hence, the domain of f(x) = 15 + √5x - 16 is [16/5, ∞) or x >16/5

Function 4

The function is given as:

p(x) = √20x + 6

Set the radicand greater than 0

20x + 6 > 0

Subtract 6 from both sides

20x > -6

Divide by 20

x > -6/20

Simplify

x > -3/10

Express as interval notation

(-3/10, ∞]

Hence, the domain of p(x) = √20x + 6 is (-3/10, ∞] or x > -3/10

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Answer:

x=25 and y=-10

Step-by-step explanation:

x+2y=5 ..........(1)

4x+12y=-20.........(2)

using elimination method.

multiply equ(1) by 4 and equ(2) by 1

so we have

x+2y=5..........*4

4x+12y=-20.........*1

4x+8y=20................(3)

4x+12y=-20.............(4)

subtract eq(4) from (3) we have

-4y=40

y=-10

substitute y=-10 in equation (1)

we have:

x+2(-10)=5

x-20=5

x=25

For a probability distribution to be represented, it is needed that P(X = 0) + P(X = 1) = 0.44. Hence one possible example is:

The sum of all the probabilities must be of 1, hence:

P(X = 0) + P(X = 1) + P(X = 3) + P(X = 4) + P(X = 5) = 1.

Then, considering the table:

P(X = 0) + P(X = 1) + 0.15 + 0.17 + 0.24 = 1

P(X = 0) + P(X = 1) + 0.56 = 1

P(X = 0) + P(X = 1) = 0.44.

Hence one possible example is:

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Answer: A

Step-by-step explanation:

The domain is the range of possible x that doesn't make y impossible to find, in this case, all real numbers work

The range is the range of possible y that doesn't make x impossible to find in the inverse function, in this case, all real numbers work

The answer is A.

As any number can be replaced in place of x for the function y, the domain of the function is All Real Numbers. Since any number can be inputted, the result will also vary accordingly. Hence, the range of the function is All Real  Numbers.

This is true for the listed function :

[tex]\boxed {y = \sqrt[3]{x-2} - 5}[/tex]

The chances of it happening more than once in 4 trials is 13%

From the information given, we have can deduce that;

Probability of 1 trial = 55%

= 55/ 100

Find the ratio

= 0. 55

We are to find the probability of it happening more than once in 4 different trials

If the probability of it happening in one trial is 555 which equals 0. 55

Then the probability of it happening in 1 in 4 trials is given as;

P(1/4 trials) = 1/ 4 × 55%

P(1/4 trials) = 1/ 4 × 0. 55

Put in decimal form

P(1/4 trials) = 0. 25 × 0. 55

P(1/4 trials) = 0. 138

But we have to know the percentage

= 0. 138 × 100

Multiply the values, we have

= 13. 8 %

Thus, the chances of it happening more than once in 4 trials is 13%

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Answer:

[tex](-2,0) \textsf{ and }\left(\dfrac{5}{2},0\right)[/tex]

Step-by-step explanation:

Given quadratic function:

[tex]f(x)=2x^2-x-10[/tex]

The x-intercepts of the graph are the points at which the curve crosses the x-axis, so when f(x) = 0.

To find these points, factor the quadratic equation, set it to zero, then solve for x.

To factor a quadratic in the form [tex]ax^2+bx+c[/tex], find two numbers that multiply to ac and sum to b.

[tex]\implies ac=2 \cdot -10 = -20[/tex]

[tex]\implies b=-1[/tex]

Therefore, the two numbers are -5 and 4.

Rewrite the middle term as the sum of these two numbers:

[tex]\implies 2x^2-5x+4x-10[/tex]

Factor the first two terms and the last two terms separately:

[tex]\implies x(2x-5)+2(2x-5)[/tex]

Factor out the common term (2x - 5):

[tex]\implies (x+2)(2x-5)[/tex]

Set the factored function to zero:

[tex]\implies (x+2)(2x-5)=0[/tex]

Apply the zero-product property:

[tex]\implies (x+2)=0 \implies x=-2[/tex]

[tex]\implies (2x-5)=0 \implies x=\dfrac{5}{2}[/tex]

Therefore, the x-intercepts of the graph are:

[tex](-2,0) \textsf{ and }\left(\dfrac{5}{2},0\right)[/tex]

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Part A - The average value of v(t) over the interval  (0, π/2) is 6/π

Part B -  The displacement of the yo-yo from time t = 0 to time t = π is 0 m

Part C - The total distance the yo-yo travels from time t = 0 to time t = π is 6 m.

The average value of a function f(t) over the interval (a,b) is

[tex]f(t)_{avg} = \frac{1}{b - a} \int\limits^b_a {f(t)} \, dx[/tex]

So, since  velocity at time t is given by v(t) = 3cos(t) for time t ≥ 0. Its average value over the interval  (0, π/2) is given by

[tex]v(t)_{avg} = \frac{1}{\frac{\pi }{2} - 0} \int\limits^{\frac{\pi }{2} }_0 {v(t)} \, dt[/tex]

Since v(t) = 3cost, we have

[tex]v(t)_{avg} = \frac{1}{\frac{\pi }{2} - 0} \int\limits^{\frac{\pi }{2} }_0 {3cos(t)} \, dt\\= \frac{3}{\frac{\pi }{2}} \int\limits^{\frac{\pi }{2} }_0 {cos(t)} \, dt\\= \frac{6}{{\pi}} [{sin(t)}]^{\frac{\pi }{2} }_{0} \\= \frac{6}{{\pi}} [{sin(\frac{\pi }{2})} - sin0]\\ = \frac{6}{{\pi}} [1 - 0]\\ = \frac{6}{{\pi}} [1]\\ = \frac{6}{{\pi}}[/tex]

So, the average value of v(t) over the interval  (0, π/2) is 6/π

To find the displacement of the yo-yo, we need to find its position.

So, its position x = ∫v(t)dt

= ∫3cos(t)dt

= 3∫cos(t)dt

= 3sint + C

Given that at t = 0, x = 3. so

x = 3sint + C

3 = 3sin0 + C

3 = 0 + C

C = 3

So, x(t) = 3sint + 3

So, its displacement from time t = 0 to time t = π is

Δx = x(π) - x(0)

= 3sinπ + 3 - (3sin0 + 3)

= 3 × 0 + 3 - 0 - 3

= 0 + 3 - 3

= 0 + 0

= 0 m

So, the displacement of the yo-yo from time t = 0 to time t = π is 0 m

The total distance the yo-yo travels from time t = 0 to time t = π is given by

[tex]x(t) = \int\limits^{\pi}_0 {v(t)} \, dt\\= \int\limits^{\pi }_0 {3cos(t)} \, dt\\= 3 \int\limits^{\pi }_0 {cos(t)} \, dt\\ = 3 \int\limits^{\frac{\pi }{2} }_0 {cos(t)} \, dt + 3\int\limits^{\pi }_{\frac{\pi }{2}} {cos(t)} \, dt\\= 3 \times 2\int\limits^{\frac{\pi }{2} }_0 {cos(t)} \, dt\\= 6 [{sin(t)}]^{\frac{\pi }{2} }_{0} \\= 6[{sin\frac{\pi }{2} - sin0]\\\\= 6[1 - 0]\\= 6(1)\\= 6[/tex]

So, the total distance the yo-yo travels from time t = 0 to time t = π is 6 m.

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So, the equation is sin(x + y)/sin(x - y) = (1 + cotxstany)/(1 + cotxtany)

The question has to to with trigonometric identities?

Trigonometric identities are equations that show the relationship between the trigonometric ratios.

Given the equation sin(x + y)/sin(x - y)

Using the trigonometric identities.

So, sin(x + y)/sin(x - y) = (sinxcosy + cosxsiny)/(sinxcosy + cosxsiny)

Dividing the rnumerator and denominator of ight hand side by sinx, we have

sin(x + y)/sin(x - y) = (sinxcosy + cosxsiny)/sinx/(sinxcosy + cosxsiny)/sinx

sin(x + y)/sin(x - y) = (sinxcosy/sinx + cosxsiny/sinx)/(sinxcosy/sinx + cosxsiny/sinx)

= (cosy + cotxsiny)/(cosy + cotxsiny) (since cosx/sinx = cotx)

Dividing the numerator and denominator of the right hand side by cosy, we have

= (cosy + cotxsiny)/cosy/(cosy + cotxsiny)/cosy

= (cosy/cosy + cotxsiny/cosy)/(cosy/cosy + cotxsiny/cosy)

= (1 + cotxstany)/(1 + cotxtany)   [since siny/cosy = tany]

So,  sin(x + y)/sin(x - y) = (1 + cotxstany)/(1 + cotxtany)

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Answer:

t = 1.75(s)

Step-by-step explanation:

I gotchu

Answer:

4:24 or also 1:6

Step-by-step explanation:

Answer:

event

An event can be described as an individual or set of outcomes of an experiment.

Events are a subset of the experiment's sample space.

The sample space of an experiment is the complete set of possible outcomes of that experiment.

Answer:

x = 1/3

Step-by-step explanation:

1/2 + X = 5/6

collecting the like terms

X = 5/6 - 1/2

making one fraction on the left side by taking the common denominator

X = (5 - 3)/6

X = 2/6

devide the numerator and denominator by 2, you get

X = 1/3

Seven less than the product of thirteen times a number, and two squared is as follows;

(13x × 2²) - 7

The expression says  seven less than the product of thirteen times a number, and two squared.

Let

the number  = x

Hence, the product of thirteen times a number, and two squared is as follows:

13(x)(2²)

Therefore,  seven less than the product of thirteen times a number, and two squared is as follows;

(13x × 2²) - 7

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Answer:

3

Step-by-step explanation:

6.01 has 3 sig fig

I think that's how I do it

Answer: Plant B produced 4000 panels

Step-by-step explanation: Let the number of panels produced by plant A = a and the number of panels produced by plant B = x. In total, we have 3000 + x.  Now for the defective panels. 10% of plant a + 3% of plant b = 6% of both a and b. 300 + 3/100(x) = 6/100(3000 + x). 300 + 0.03x = 180 + 0.06x. Now, we simplify. 300 - 180 = 0.06x = 0.03x. So, 120 = 0.03x and 40 = 1% of x. This means x = 4000. So, therefore, Plant B produced 4000 panels. We can check our work by using the value of x we got. 4000 + 3000 = 7000. 6% of 7000 = 420. 120 = 3% of 4000. 120 + 300 = 420 so the answer we got is correct. Hope this helped

Answer:

210

Step-by-step explanation:

[tex]512=2^9 \\[/tex] has 9 prime factors

Now, if you want to maximize the number of different prime factor, apply the following method: let [tex]p(n)[/tex]  be the nth prime number, in increasing order. Let [tex]v[/tex]  define as follow :  [tex]v(0)=1,v(n+1)=v(n)p(n+1)[/tex] , and find n  such that [tex]v(n) < 1000,v(n+1) > 1000[/tex]. n will be the number of factors and v(n) your integer.

[tex]v(1)=2v(0)=2v(2)=3v(1)=6v(3)=5v(2)=30v(4)=7v(3)=210v(5)=11v(4)=2310[/tex]

There you are,  210 , with only 4 different prime factors.

→Credits: Xavier Dectot←

Answer: C

Step-by-step explanation:

The line [tex]y=x+2[/tex] is shaded below and is solid.

This eliminates all the options except for C.

Answer:

  order: 2, 3, 1

Step-by-step explanation:

Reasonableness checks and your knowledge of integers and fractions will help you solve this. The offered questions are intended to help you think this through.

For an output of -31, the machine (x -2)² cannot possibly be last. Its output can only be positive.

  machine 2, (x-2)², cannot be last

Also, machine 3 cannot be last. For the output to be -31, the input to machine 3 must be -1/31. Neither of the other machines can produce a fraction with the inputs they might receive.

For an input of x=0, the machine 1/x cannot possibly be first. 1/0 is undefined.

The other two machines will give the following outputs for an input of 0:

  machine 1: 4(0) -32 = -32

  machine 2: (0 -2)² = 4

It is unlikely that machine 1 will be first, because the other two machines cannot do anything useful with -32 as an input.

  machine 3, 1/x, cannot be first

The reasoning of part (a) tells you the last machine must be machine 1. The reasoning of part (b) tells you the first machine cannot be 3, so must be 2. The order of the machines must be ...

Answer:

1/5 of it

Step-by-step explanation:

5/5 represents all of her homework

5/5-4/5= 1/5

4/5 represents the homework she did before dinner

1/5 represents the homework left to do

The excluded values are x = 4 and x - 1

The complete question is added as an attachment

The function is given as:

(2x^2 - 7x - 4)/(x^2 - 5x + 4)

Set the denominator to 0

x^2 - 5x + 4 = 0

Expand

x^2 - x - 4x + 4 = 0

Factorize the equation

x(x -1) - 4(x - 1) = 0

This gives

(x - 4)(x - 1) = 0

Solve for x

x = 4 and x - 1

Hence, the excluded values are x = 4 and x - 1

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Answer: Its D 1 and 4

Step-by-step explanation:

I took the assignment

Answer:

  (x, y) = (-√3/2, 1/2)

Step-by-step explanation:

The terminal point for that angle can be read from a unit circle chart.

On the attached chart, the point of interest is the first one above the -x axis on the left side. The chart tells you the coordinates are ...

  (x, y) = (-√3/2, 1/2)

The potential zeros of f(x)=6x^4+ 2x^3 - 4x^2 +2 are ±(1, 1/2, 1/3, 1/6, 2, 2/3)

The function is given as:

f(x)=6x^4+ 2x^3 - 4x^2 +2

For a function P(x) such that

P(x) = ax^n +...... + b

The rational roots of the function p(x) are

Rational roots = ± Possible factors of b/Possible factors of a

In the function f(x), we have:

a = 6

b = 2

The factors of 6 and 2 are

a = 1, 2, 3 and 6

b = 1 and 2

So, we have:

Rational roots = ±(1, 2)/(1, 2, 3, 6)

Split the expression

Rational roots = ±1/(1, 2, 3, 6)/ and ±2/(1, 2, 3, 6)

Evaluate the quotient

Rational roots = ±(1, 1/2, 1/3, 1/6, 2, 1, 2/3, 1/3)

Remove the repetition

Rational roots = ±(1, 1/2, 1/3, 1/6, 2, 2/3)

Hence, the potential zeros of f(x)=6x^4+ 2x^3 - 4x^2 +2 are ±(1, 1/2, 1/3, 1/6, 2, 2/3)

The complete parameters are:

The function is given as:

f(x) = 3x^3 + 2x^2 + 3x + 6

The potential zeros of f(x)=6x^4+ 2x^3 - 4x^2 +2 are ±(1, 1/2, 1/3, 1/6, 2, 2/3)

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Using the normal distribution, there is a 0.007 = 0.7% probability that the mean score for 10 randomly selected people who took the LSAT would be above 157.

The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

Researching this problem on the internet, the parameters are given as follows:

[tex]\mu = 150, \sigma = 9, n = 10, s = \frac{9}{\sqrt{10}} = 2.85[/tex]

The probability is one subtracted by the p-value of Z when X = 157, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

Z = (157 - 150)/2.85

Z = 2.46

Z = 2.46 has a p-value of 0.993.

1 - 0.993 = 0.007.

0.007 = 0.7% probability that the mean score for 10 randomly selected people who took the LSAT would be above 157.

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The cost of manufacturing the article bin 2004 will be 1875.

Percentage increase is the ratio of increase to the initial value in percentage.

Given that In 2001 the total cost of manufacturing an article was sh.1250 and this was divided between the cost of materials was doubled, labor cost increased by 30% and transport costs increased by 20%

Given that in 2001, the cost of manufacturing = 1250

If this cost was divided between the Material, labor, and transportation equally, then, cost for each sector will 1250/3 = 416.67

The cost of manufacturing = 833.33 + 541.67 + 500

The cost of manufacturing = 1875

Therefore, the cost of manufacturing the article bin 2004 will be 1875.

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The difference between Alok and Prabir share exists 8000.

Given: Invested by Alok = Rs 90, 000

Invested by Prabir = Rs 1,20000

The time period of Alok = 3 months

The time period of Prabir = 2 years

They earn a profit = Rs 96,000.

Profit exists directly proportional to the product of the amount invested and the time period of investment.

8000 = 90000 [tex]*[/tex] 24/120000 [tex]*[/tex] 21

= 5/7

5x + 7x = 96000

x = 8000

first = 40000

second = 48000

so the difference exists at 8000.

The difference between Alok and Prabir share exists 8000.

Therefore, the correct answer is 8000.

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The solution to the inequality x-13<=7+4x is x >= -20/3

The inequality expression is given as:

x-13<=7+4x

Add 13 to both sides of the inequality

x <= 20 + 4x

Subtract 4x from both sides of the inequality

-3x <= 20

Divide both sides of the inequality by -3

x >= -20/3

Hence, the solution to the inequality x-13<=7+4x is x >= -20/3

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Complete question

Solve the inequality for x

x-13<=7+4x