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3 years ago

Simplest form and answer

Mathematics

1 answer:

Alinara [238K]3 years ago

4 0

Answer: 4 / 9

Solution:

Step 1: Simplify the possible terms in the equation

2/5 * 10/9

= 2 * 2 / 9 ( 10 would be cancelled out by 5 )

Step 2: multiply the remaining terms

= 4 / 9 Answer

Hope this solution will help you understand the question!

Tip: you can recheck the answer by a calculator, until you get a good grasp on the method. Always double check the answer or it will may cost you marks

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hichkok12 [17]

Answer:

Step-by-step explanation:

1. Cross Multiply. $\frac{5}{6} =\frac{30}{x}$

2. Solve. $\frac{5}{6} =\frac{30}{x} =36$

3. Your answer is 36

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3 years ago

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Explain how you know that in ?×6,273=6,273, the ? Will be 1

grigory [225]

Anything more or less than 1 would result in a different outcome, and anything times 1 results in itself.

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3 years ago

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Exercise #4: Each of the following fractions can be converted to a whole number using division.

solong [7]

A) 6 because 12 divided by 2 is 6

B) 9 because 45 divided by 5 is 9

C) 23 because 207 divided by 9 is 23

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3 years ago

If Mitzi has two dogs, a puppy that weighs 25 pounds and an adult that weighs 80 pounds, how much will it cost to bathe them bot

Sidana [21]

Answer:

The correct answer is A) $105

Step-by-step explanation:

In this real-world problem, the cost to groom both dogs is (the puppy) + (the adult dog) = x.

For the first step, y = 40, if x <= 25.

y = 50, if 25 < x < 50.

if x >= 50, then y = 0.50x + 25

So therefore, substitute the puppy + adult dog = x

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So, the final answer is $105.

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3 years ago

Suppose that a box contains one fair coin (Heads and Tails are equally likely) and one coin with Heads on each side. Suppose fur

stealth61 [152]

Using conditional probability, it is found that there is a 0.1 = 10% probability that the chosen coin was the fair coin.

Conditional Probability

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.
$P(A \cap B)$ is the probability of both A and B happening.
P(A) is the probability of A happening.

In this problem:

Event A: Three heads.
Event B: Fair coin.

The probability associated with 3 heads are:

$0.5^3 = 0.125$ out of 0.5(fair coin).
1 out of 0.5(biased).

Hence:

$P(A) = 0.125 + 0.5 = 0.625$

The probability of 3 heads and the fair coin is:

$P(A \cap B) = 0.5(0.125) = 0.0625$

Then, the conditional probability is:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0625}{0.625} = 0.1$

0.1 = 10% probability that the chosen coin was the fair coin.

A similar problem is given at brainly.com/question/14398287

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3 years ago