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2 years ago

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fred can complete a work in 6 days and Paul can complete a work in 3 days. if fred and Paul work together, how many days can the

work be completed?

1 answer:

defon2 years ago

3 0

Answer:

I think its 3 or 2 days bc if paul can get two works done by the time fred can get one done then the days should be divided.

Step-by-step explanation:

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Please Help need this ASAAP!! The First and correct answer will be marked brainlest!!

Alex_Xolod [135]

Answer:

y=1.5x+2

Step-by-step explanation:

Step 1, find slope

(20-8)/(12-4)

12/8 = 1.5

slope = 1.5

Step 2, find the y-intercept

y = 1.5x+b

use point (4,8)

8 = 1.5(4)+b

8 = 6+b

-6 -6

2 = b

Step 3, put it all in slope-intercept form (y = mx+b)

y = 1.5x+2

Answer: y = 1.5x+2

5 0

2 years ago

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If the polynomials p(x)=2x³ +bx² +3x-5 and q(x)= x³+x² -4x-b leave the same remainder, when divided by x-2, prove that b = -13/5

Tju [1.3M]

Step-by-step explanation:

Lets first find the remainder when p(x) is divided by x-2

p(2)=2(2)³+b(2)²+3(2)-5

p(2)=16+4b+6-5

p(2)=4b+17

Now let's find the remainder when q(x) is divided by x-2

q(2)=(2)³+(2)²-4(2)-b

q(2)=8+4-8-b

q(2)=4-b

Now let's equate both of them

4b+17=4-b

4b+b=4-17

5b=-13

b=-13/5 as required

5 0

2 years ago

Help pls!!!!!! im giving brainliest if you help.

devlian [24]

A is FALSE
B is FALSE
C is FALSE
D is TRUE

8 0

2 years ago

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A bookstore had 60 copies of a magazine. Yesterday, it sold 1/3 of them. Today, it sold 1/4 of what remained. How many copies do

lara [203]

Answer:

30

Step-by-step explanation:

1/3 of 60 is 20

40 would be left

1/4 of 40 is 10 so 30 would be left

7 0

3 years ago

An Airliner has a capacity for 300 passengers. If the company overbook a flight with 320 passengers, What is the probability tha

TEA [102]

Answer:

The probability is $P(X >300 ) = 0.97219$

Step-by-step explanation:

From the question we are told that

The capacity of an Airliner is k = 300 passengers

The sample size n = 320 passengers

The probability the a randomly selected passenger shows up on to the airport

$p = 0.96$

Generally the mean is mathematically represented as

$\mu = n* p$

=> $\mu = 320 * 0.96$

=> $\mu = 307.2$

Generally the standard deviation is

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

=> $\sigma = \sqrt{320 * 0.96 * (1 -0.96 ) }$

=> $\sigma =3.50$

Applying Normal approximation of binomial distribution

Generally the probability that there will not be enough seats to accommodate all passengers is mathematically represented as

$P(X > k ) = P( \frac{ X -\mu }{\sigma } > \frac{k - \mu}{\sigma } )$

Here $\frac{ X -\mu }{\sigma } =Z (The \ standardized \ value \ of \ X )$

=> $P(X >300 ) = P(Z > \frac{300 - 307.2}{3.50} )$

Now applying continuity correction we have

$P(X >300 ) = P(Z > \frac{[300+0.5] - 307.2}{3.50} )$

=> $P(X >300 ) = P(Z > \frac{[300.5] - 307.2}{3.50} )$

=> $P(X >300 ) = P(Z > -1.914 )$

From the z-table

$P(Z > -1.914 ) = 0.97219$

So

$P(X >300 ) = 0.97219$

8 0

3 years ago