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3 years ago

12

Stuffed animals each weighing 1 over 4 pound are on a scale. The scale shows 1 1 over 2 pounds. How many stuffed animals are on

the scale?

1 answer:

Anna35 [415]3 years ago

4 0

Answer:

6 stuffed animals

Step-by-step explanation:

if 1/4= 1 then 1x6= 1 1/2

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A weather station on the top of a mountain reports that the temperature is currently 0°C and has been falling at a constant rate

Nastasia [14]

Answer:

Step-by-step explanation:

Absolute zero, the lowest temperature possible, is defined as being exactly 0 K and −273.15 °C. Until 19 May 2019, the temperature of the triple point of water was defined as exactly 273.16 k

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3 years ago

If i am making $4 billion / sec then how much money would i have in 10 minutes? if i am making $6 billion / sec then how much mo

Ghella [55]

If you are making 4 billion a second do 4,000,000,000 x 60 = 240,000,000,000
Then times that by 10 which equals. 2,400,000,000,000

Is the answer for the second one
6,000,000,000 x 60 x 60 x 2 = 1,296,000,000,000,000

3 0

3 years ago

Which expression can be used to determine 50% of 42?<br> 42-2 ,42÷2,42÷10,42-10<br>

evablogger [386]

Answer:

42÷2

Step-by-step explanation:

3 0

3 years ago

Anybody have the answer?

Mariulka [41]

80 inches squared, I would assume based off of the other one.

6 0

2 years ago

Read 2 more answers

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