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ipn [44]
2 years ago
15

An experienced carpenter can frame a house twice as fast as an apprentice. Working together, it takes the carpenters 2 days. How

long would it take the apprentice working alone?
Mathematics
1 answer:
AleksAgata [21]2 years ago
4 0

Answer:

<em>6 days</em>

<em></em>

Step-by-step explanation:

Let the time taken by Carpenter working alone = C days

Then time taken by apprentice alone = Twice as that of taken by Carpenter = 2C days

Time taken working together = 2 days

Work done in one day working together = \frac{1}{2}

Work done in one day by Carpenter working alone = \frac{1}{C}

Work done in one day by apprentice working alone = \frac{1}{2C}

Work done in one day by Carpenter working alone + Work done in one day by Carpenter working alone =  \frac{1}{C}+\frac{1}{2C} = Work done in one day working together = \frac{1}{2}

\dfrac{1}{C}+\dfrac{1}{2C}=\dfrac{1}{2}\\\Rightarrow \dfrac{2+1}{2C}=\dfrac{1}{2}\\\Rightarrow C = 3\ days

Time taken by Carpenter alone to complete the work = 3 days

Time taken by Apprentice alone to complete the work = 3 \times 2= <em>6 days</em>

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A group of 40 people went to the theme park. While there, each person bought popcorn. Regular bags of popcorn sold for $6 per ba
Y_Kistochka [10]

Answer:

17 regular popcorn bags and 23 super size bags.

Step-by-step explanation:

A group of 40 people went to the theme park. While there, each person bought popcorn.

Let the regular popcorn bags be = x

Let the super size bags be = y

x+y=40        ................(1)

Next equation becomes:

6x+8y=286        ...............(2)

Multiplying (1) by 6 : 6x+6y=240

Now subtracting this from (2) we get,

2y=46

=> y = 23

And x = 40-23

=> x = 17

So, there were 17 regular popcorn bags and 23 super size bags that the group bought.

4 0
2 years ago
Suppose students' ages follow a skewed right distribution with a mean of 24 years old and a standard deviation of 3 years. If we
natta225 [31]

Answer: Option 'c' is correct.

Step-by-step explanation:

Since we have given that

Mean of students' age = 24 years

Standard deviation of students' age = 3 years

Sample size = number of students = 350

So, according to options,

a. The shape of the sampling distribution is approximately normal.

It is true as n >30, we will use normal.

b. The mean of the sampling distribution is approximately 24-years old.

It is true as it is given.

c. The standard deviation of the sampling distribution is equal to 5 years.

It is not true as it is given 3 years.

Hence, Option 'c' is correct.

8 0
2 years ago
How many ways are there to seat six different boys and six different girls along one side of a long table with 12 seats? how man
GaryK [48]
Q1: 479 001 600 ways

Explanation:
You have 12 people, 12 seats, ¹²P₁₂ ways in arranging them, or essentially 12! ways

Q2: 1 036 800 ways

Explanation:
Let's break it up into two cases.

Case 1: BGBGBGBGBGBG
Case 2:GBGBGBGBGBGB

Let's deal with case 1, because case 2 will pop out. There are ⁶P₆ ways in sorting out the boys and ⁶P₆ ways in sorting out the girls, so for case 1, we'd have (⁶P₆)² ways in sorting out boys and girls in case 1.

This is exactly the same in case 2, so it'd be (⁶P₆)² ways in sorting out girls and boys in case 2.

So, (⁶P₆)² + (⁶P₆)² ways = 1 036 800 ways if they alternate
3 0
2 years ago
Factorise 125x^3 -27y^3
telo118 [61]

Answer:

(5x - 3y)(25x² +15xy + 9y²)

Step-by-step explanation:

125x³ - 27y³ is a difference of cubes and factors in general as

a³ - b³ = (a - b)(a² + ab + b²)

Given

125x³ - 27y³

= (5x)³ - (3y)³

= (5x - 3y)((5x)² + (5x)(3y) + (3y)²)

= (5x - 3y)(25x² + 15xy + 9y²)

6 0
2 years ago
Assume that a fair die is rolled. The sample space is , 1, 2, 3, 4, 56, and all the outcomes are equally likely. Find P8. Expres
liraira [26]

Answer:

P\left(8\right)=0

Step-by-step explanation:

Assuming that a fair die is rolled.

  • The sample space is 1, 2, 3, 4, 5, 6 and all the outcomes are equally likely.

Let  X  be the set of all possible outcomes. Let  A  be an outcome.

So, the probability that  A  occurs is:

                                                        P\left(A\right)=\frac{|A|}{|X|}

As the set of all possible outcomes of the roll of a single die is:

X=\left\{1,2,3,4,5,6\right\}

Observe that

|X|=6

Here  

|A|=0   because 8 is not in the set sample space. So, the outcome of occurring the number 8 is not possible from the all possible outcomes.

So, the probability must be zero.

In other words,

                        P\left(A\right)=\frac{|A|}{|X|}

                         P\left(8\right)=\frac{0}{6}=0

Therefore,

                   P\left(8\right)=0

5 0
2 years ago
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