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3 years ago

Only C please:) Thanks

Mathematics

1 answer:

kirill115 [55]3 years ago

4 0

Answer:

Step-by-step explanation:

The f^-1 notation indicates you need to find the inverse of the function

y=2x+1

x=2y+1

Exchange the x & y variables. Note this is what accomplishes that reflection across the line y = x

7=2y+1

-2y=1-7

-2y=-6

Divide both side of the equation by -2

negative divide negative is positive so

y=3

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Rewrite 1/2 * 1/2 * 1/2 as an exponential expression with a base of 2.

Bumek [7]

Answer:

2^-3

Step-by-step explanation:

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3 years ago

Help please What is the answer I’m struggling

yaroslaw [1]

I believe the answer is x=120

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3 years ago

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In a binomial distribution, n = 8 and π=0.36. Find the probabilities of the following events. (Round your answers to 4 decimal p

skelet666 [1.2K]

Answer:

$\mathbf{P(X=5) =0.0888}$

P(x ≤ 5 ) = 0.9707

P ( x ≥ 6) = 0.0293

Step-by-step explanation:

The probability of a binomial mass distribution can be expressed with the formula:

$\mathtt{P(X=x) =(^{n}_{x} ) \ \pi^x \ (1-\pi)^{n-x}}$

$\mathtt{P(X=x) =(\dfrac{n!}{x!(n-x)!} ) \ \pi^x \ (1-\pi)^{n-x}}$

where;

n = 8 and π = 0.36

For x = 5

The probability $\mathtt{P(X=5) =(\dfrac{8!}{5!(8-5)!} ) \ 0.36^5 \ (1-0.36)^{8-5}}$

$\mathtt{P(X=5) =(\dfrac{8!}{5!(3)!} ) \ 0.36^5 \ (0.64)^{3}}$

$\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 \times 5!}{5!(3)!} ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 }{3 \times 2 \times 1} ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =({8 \times 7 } ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =0.0887645}$

$\mathbf{P(X=5) =0.0888}$ to 4 decimal places

b. x ≤ 5

The probability of P ( x ≤ 5) $\mathtt{P(x \leq 5) = P(x = 0)+ P(x = 1)+ P(x = 2)+ P(x = 3)+ P(x = 4)+ P(x = 5})$

${P(x \leq 5) = ( \dfrac{8!}{0!(8!)} \times (0.36)^0 \times (1-0.36)^8 \ ) + \dfrac{8!}{1!(7!)} \times (0.36)^1 \times (1-0.36)^7 \ +$ $\dfrac{8!}{2!(6!)} \times (0.36)^2 \times (1-0.36)^6 \ + \dfrac{8!}{3!(5!)} \times (0.36)^3 \times (1-0.36)^5 + \dfrac{8!}{4!(4!)} \times (0.36)^4 \times (1-0.36)^4 \ + \dfrac{8!}{5!(3!)} \times (0.36)^5 \times (1-0.36)^3 \ )$

P(x ≤ 5 ) = 0.0281+0.1267+0.2494+0.2805+0.1972+0.0888

P(x ≤ 5 ) = 0.9707

c. x ≥ 6

The probability of P ( x ≥ 6) = 1 - P( x ≤ 5 )

P ( x ≥ 6) = 1 - 0.9707

P ( x ≥ 6) = 0.0293

4 0

3 years ago

41. Assuming that a man can complete the work alone in x days, his work in four days would be: a) b) X X C d) 4x x 42. If a man

Schach [20]

Percentage and ratio word problems require understanding of the relationship between variables from which the question is formed

The options that give the correct values of the duration of the work are;

$41. \ c) \ \dfrac{4}{x}$

$42. \ d) \ \dfrac{4}{x} + \dfrac{6}{y} = \dfrac{1}{5}$

43. a) 35 days

44. c) 21·a + 28·b = 1

45. c) (42, 56)

Reasons:

41. Number of days it takes a man to complete the work alone = x days

Therefore;

$The \ work \ done \ by \ the \ man \ in \ one \ day = \dfrac{1}{x}$

$The \ work \ done \ in \ four \ days \ by\ the \ man = 4 \times \dfrac{1}{x} = \dfrac{4}{x}$

The correct option is $c) \ \dfrac{4}{x}$

42. Number of days it takes a man to complete the work alone = x days

$Work \ done \ by \ a\ man \ in \ one \ day = \dfrac{1}{x}$

$Work \ done \ by \ four \ men \ in \ one \ day = \dfrac{4}{x}$

Number of days it takes a boy to complete the work alone = y days

$Work \ done \ by \ a \ boy \ in \ one \ day = \dfrac{1}{x}$

$Work \ done \ by \ six \ boys \ in \ one \ day = \dfrac{6}{y}$

4 men and 6 boys work for 5 days to complete the work

Therefore, work done by 4 men and 6 boys in 1 day is therefore;

$\dfrac{4}{x} + \dfrac{6}{y} = \dfrac{1}{5}$

The correct option is therefore;

$d) \ \dfrac{4}{x} + \dfrac{6}{y} = \dfrac{1}{5}$

43. As per the case study, we have;

Case 1

$\dfrac{4}{x} + \dfrac{6}{y} = \dfrac{1}{5}$

Which gives;

$\dfrac{6\cdot x + 4\cdot y}{y \cdot x} = \dfrac{1}{5}$

30·x + 20·y = y·x

Case 2

$\dfrac{3}{x} + \dfrac{4}{y} = \dfrac{1}{7}$

Which gives;

$\dfrac{4\cdot x + 3\cdot y}{y \cdot x} = \dfrac{1}{7}$

28·x + 21·y = y·x

Therefore;

30·x + 20·y = 28·x + 21·y

∴ 2·x = y

Plugging in the value of y = 2·x, in Case 1 gives;

$\dfrac{4}{x} + \dfrac{6}{2 \cdot x} = \dfrac{1}{5}$

$\dfrac{2 \times 4 + 6}{2 \times x} = \dfrac{14}{2 \times x} =\dfrac{7}{x} = \dfrac{1}{5}$

7 × 5 = x

x = 7 × 5 = 35

The number of days, x, it takes a man to complete the work alone, is given by option; a) 35 days

44. For the equation $\dfrac{3}{x} + \dfrac{4}{y} = \dfrac{1}{7}$ , if $a = \dfrac{1}{x}$ , and $b = \dfrac{1}{y}$ , we have;

$3 \cdot a+ 4\cdot y = \dfrac{1}{7}$

21·a + 28·y = 1

The correct option is option C. 21·a + 28·b = 1

45. A solution to the equation $\dfrac{3}{x} + \dfrac{4}{y} = \dfrac{1}{7}$ , is given by the values of x, and y, that gives;

$\dfrac{1}{14} + \dfrac{1}{14} = \dfrac{1}{7}$

We have;

3 × 14 = 42

4 × 14 = 56

Therefore, a solution to the equation is (42, 56)

The correct option is $c) \ \underline{ (42, \ 56)}$

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2 years ago

Simplify 2x-3(y-2x) 8x-3y

olchik [2.2K]

Answer:

16x - 6y

Step-by-step explanation:

Hope that helps you

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3 years ago

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