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Free_Kalibri [48]
3 years ago
9

Toysrus is having a 38% off sale. If a backpack normally costs $22, how much will it cost on sale

Mathematics
1 answer:
andrezito [222]3 years ago
6 0
It will be 13.64. it is bc 38% of 22 is 8.36 and 22-8.36=13.64
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Find the area of the figure<br> 40 yd<br> 36 yd<br> 20 yd<br> 10 yd<br> 10 yd
Ksenya-84 [330]

Answer:

1640yd

Step-by-step explanation:

10*10*2=200

36*40=1440

200+1440=1640 yd

4 0
3 years ago
Explain how you can use reasoning to compare two fractions with the same denominator
Talja [164]
When two fractions have the same denominator, you can compare which fraction is greater by just comparing the numerators.
For example, if you have 2/5 and 3/5, the denominators are the same. So you can just look at the numerators. 3 is greater than 5, so 3/5 is greater than 2/5.

The reasoning to this is that the two numbers are divided by the same number (5), so you can compare the original numbers (numerators).
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4 years ago
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Answer:

Step-by-step explanation:

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3 years ago
What is the probability of getting a factor of 6 when you have 1 2 3 4 5 6
Marrrta [24]
Hello,

I suppose you mean using all those digits one time.

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6 0
3 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
3 years ago
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