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Mashcka [7]
2 years ago
11

A and b together can do a piece of work in 8 days. If a alone can do the same work in 12 days, then b alone can do the same work

in?.
Mathematics
1 answer:
Nostrana [21]2 years ago
6 0

B can finish the job alone in 24 days.

In work-rate problems, a certain person can complete a fraction of the job for a specific timeframe. As stated in the problem, A and B can finish the job by working together for 8 days. It must be noted that the fraction of the job they can finish individually per day is additive.

\frac{1}{A} + \frac{1}{B} = \frac{1}{8}

A can complete the work in 12 days and finish 1/12 fraction of a job per day. This can be substituted directly to the equation to solve for the time B needed to finish the work.

\frac{1}{12} + \frac{1}{B} = \frac{1}{8}

\frac{1}{B}= \frac{1}{8}- \frac{1}{12}

\frac{1}{B} = \frac{1}{24}

This means that B can do 1/24 fraction of the job per day and can completely finish the job in 24 days.

For more rate work-rate problems, please refer to the link brainly.com/question/14464103.

#SPJ4

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Solve for x and y <br> x+y=1<br> 2y-x=8
lara31 [8.8K]

Answer:

(- 2, 3 )

Step-by-step explanation:

Given the 2 equations

x + y = 1 → (1)

2y - x = 8 → (2)

Adding the 2 equations term by term will eliminate the x- term

(x - x) + (y + 2y) = (1 + 8), that is

3y = 9 ( divide both sides by 3 )

y = 3

Substitute y = 3 into either of the 2 equations and solve for y

Substituting y = 3 in (1)

x + 3 = 1 ( subtract 3 from both sides )

x = - 2

Solution is (- 2, 3 )

7 0
3 years ago
Suppose 52% of the population has a college degree. If a random sample of size 563563 is selected, what is the probability that
amm1812

Answer:

The value is  P(| \^ p -  p| < 0.05 ) = 0.9822

Step-by-step explanation:

From the question we are told that

    The population proportion is  p =  0.52

     The sample size is  n  =  563      

Generally the population mean of the sampling distribution is mathematically  represented as

           \mu_{x} =  p =  0.52

Generally the standard deviation of the sampling distribution is mathematically  evaluated as

       \sigma  =  \sqrt{\frac{ p(1- p)}{n} }

=>      \sigma  =  \sqrt{\frac{ 0.52 (1- 0.52 )}{563} }

=>      \sigma  =   0.02106

Generally the  probability that the proportion of persons with a college degree will differ from the population proportion by less than 5% is mathematically represented as

            P(| \^ p -  p| < 0.05 ) =  P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 ))

  Here  \^ p is the sample proportion  of persons with a college degree.

So

 P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = P(\frac{[[0.05 -0.52]]- 0.52}{0.02106} < \frac{[\^p - p] - p}{\sigma }  < \frac{[[0.05 -0.52]] + 0.52}{0.02106} )

Here  

    \frac{[\^p - p] - p}{\sigma }  = Z (The\ standardized \  value \  of\  (\^ p - p))

=> P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = P[\frac{-0.47 - 0.52}{0.02106 }  <  Z  < \frac{-0.47 + 0.52}{0.02106 }]

=> P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = P[ -2.37 <  Z  < 2.37 ]

=>  P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = P(Z <  2.37 ) - P(Z < -2.37 )

From the z-table  the probability of  (Z <  2.37 ) and  (Z < -2.37 ) is

  P(Z <  2.37 ) = 0.9911

and

  P(Z <  - 2.37 ) = 0.0089

So

=>P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) =0.9911-0.0089

=>P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = 0.9822

=> P(| \^ p -  p| < 0.05 ) = 0.9822

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kupik [55]

Step-by-step explanation:

answer is B

use unitary method to solve it

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