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3 years ago

15

An elementary school class polled 198 people at a shopping center to determine how many read the Daily News and how many read th

e Sun Gazette. They found the following information: 171 read the Daily News, 40 read both, and 18 read neither. How many read the Sun Gazette?

1 answer:

lawyer [7]3 years ago

6 0

Answer: 49 people

Step-by-step explanation:

First you need to separate the people that read both from the people that read the daily news. Do this by subtracting 171 - 40 = 131.

Now you can subtract the total amount of people that only read the Daily Mews from the total amount of people polled (198-131=67)

From here you need to subtract the amount of people that read neither from the remaining total. (67-18=49)

this is where you get the 49 people out of 198 that read the Sun Gazette

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If the fixed cost is 9000 per year. Variable costs are estimated to be Tk. 60.75 / item. The firm wants to break even if 80 item

Mnenie [13.5K]

9514 1404 393

Answer:

Tk 173.25

Step-by-step explanation:

The firm will break even if its cost is equal to its revenue. That is, the price of each item sold must equal the cost of producing it. To cover the fixed cost, a share of it must be added to each of the items sold. Then the break-even price for 80 items is ...

price = variable cost + share of fixed cost

price = Tk 60.75 +9000/80 = Tk 60.75 +112.50 = Tk 173.25

6 0

3 years ago

On a recent trip to a local orchard, the Morgan family picked four different kinds of apples - Braeburn, Cortland, Fuji, and Rom

ivanzaharov [21]

Answer:

Morgan family picked 144 Braeburn apples, 96 Cortland apples, 72 Fuji apples and 48 Rome apples.

Step-by-step explanation:

Let B, C, F and R represent Braeburn apples, Cortland apples, Fuji apples, and Rome apples respectively.

We have been given that Morgan family picked a total of 360 apples. We can represent this information in an equation as:

$B+C+F+R=360...(1)$

We are told that Morgan family picked twice as many Braeburn as Fuji. We can represent this information in an equation as:

$B=2F...(2)$

We are told that Morgan family picked twice as many Cortland as Rome, that is:

$C=2R...(3)$

We are told that Morgan family picked 50% more Fuji than Rome, that is:

$F=1.5R...(4)$

Upon substituting equation (4) in equation (2), we will get:

$B=2(1.5)R$

$B=3R$

Now, each apple in in terms of Rome apples, so we will get:

$3R+2R+1.5R+R=360$

Let us solve for R.

$7.5R=360$

$\frac{7.5R}{7.5}=\frac{360}{7.5}$

$R=48$

Therefore, Morgan family picked 48 Rome apples.

Upon substituting $R=48$ in (3), we will get:

$C=2R\Rightarrow 2(48)=96$

Therefore, Morgan family picked 96 Cortland apples.

Upon substituting $R=48$ in (4), we will get:

$F=1.5R\Rightarrow 1.5(48)=72$

Therefore, Morgan family picked 72 Fuji apples.

Upon substituting $F=72$ in (2), we will get:

$B=2F\Rightarrow 2(72)=144$

Therefore, Morgan family picked 144 Braeburn apples.

3 0

3 years ago

If x is a rational number and y is the opposite of x, why do x and y have the same absolute value

grin007 [14]

Answer:

Absolute value is the distance away from zero, so if x is a rational number and y is opposite, they would both be the same distance away from zero, regardless of their sign.

Step-by-step explanation:

Absolute value is the distance of a number away from zero. For example, if x is equal to the rational number -2, the absolute value of -2 is just 2, given that both 2 and -2 are only a distance of 2 away from the number zero. Think of it as your walk or drive to school. If you live 4 miles from school, you drive a distance of 4 miles there and 4 miles back. That distance is not calculated as a positive 4 miles there and a negative 4 miles back, but rather just 4 miles there and 4 miles back, for a total distance of 8 miles. So, if x=4 and y=(-4), the absolute value of both would be 4.

8 0

2 years ago

One household is to be selected at random from a town. The probability that the household has a cat is 0.20.2 . The proba

dimulka [17.4K]

Answer:

There is a 50% probability that the household has a dog, given that the household has a cat.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a household has a cat.

B is the probability that a household has a dog.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a household has a cat but not a dog and $A \cap B$ is the probability that a household has both a cat and a dog.

By the same logic, we have that:

$B = b + (A \cap B)$

The probability that the household has a cat or a dog is 0.5

$a + b + (A \cap B) = 0.5$

The probability that the household has a dog is 0.4

$B = 0.4$

$B = b + (A \cap B)$

$b = 0.4 - (A \cap B)$

The probability that the household has a cat is 0.2.

$A = 0.2$

$A = a + (A \cap B)$

$a = 0.2 - (A \cap B)$

So

$a + b + (A \cap B) = 0.5$

$0.2 - (A \cap B) + 0.4 - (A \cap B) + (A \cap B) = 0.5$

$A \cap B = 0.1$

What is the probability that the household has a dog, given that the household has a cat?

20% of the households have a cat, and 10% have both a cat and a dog. So

$P = \frac{A \cap B}{A} = {0.1}{0.2} = 0.5$

There is a 50% probability that the household has a dog, given that the household has a cat.

4 0

2 years ago

11 times what give you 187

mr_godi [17]

17. 187 divided by 11 equals 17.

6 0

2 years ago

Read 2 more answers