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2 years ago

-Write 0.875 as a reduced fraction

2 answers:

7 0

We can reduce this fraction, 875 / 1000 to lowest
terms by dividing both the numerator and denominator by 125 ;
Why divide by 125? 125 Greatest Common Factor (GCF) of the numbers 875 and 1000.
So, this fraction reduced to lowest terms is 7 / 8 ;
In the same way, 35 / 100 is reduced to lowest terms as 7 / 20 ;

Zolol [24]2 years ago

5 0

7/8 and 7/20. those are the answers.

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Answer:

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2 years ago

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3 years ago

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PLS HELP!!! factor: a. 16b^2c^12-0.25 b. 81x^6y^2-0.36a^2

mafiozo [28]

Answer:

Step-by-step explanation:

a. 16b^2c^12-0.25 can be rewritten as follows because it is the difference of two squares:

(4bc^6 - .5)(4bc^6 + .5)

b. 81x^6y^2-0.36a^2

is the difference of two squares, just like (a) (above); its factors are:

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7 0

3 years ago

55g as a percentage of 2.2kg

Evgesh-ka [11]

Answer:

2.5%

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2 years ago

A random variable x follows a normal distribution with mean d and standard deviation o=2. It is known that x is less than 5 abou

Vaselesa [24]

Answer:

The mean of this distribution is approximately 3.96.

Step-by-step explanation:

Here's how to solve this problem using a normal distribution table.

Let $z$ be the

$\displaystyle z = \frac{x - \mu}{\sigma}$ .

In this question, $x = 5$ and $\sigma = 2$ . The equation becomes

$\displaystyle z = \frac{5 - \mu}{2}$ .

To solve for $\mu$ , the mean of this distribution, the only thing that needs to be found is the value of $z$ . Since

The problem stated that $P(X \le 5) = 69.85\% = 0.6985$ . Hence, $P(Z \le z) = 0.6985$ .

The problem is that the normal distribution tables list only the value of $P(0 \le Z \le z)$ for $z \ge 0$ . To estimate $z$ from $P(Z \le z) = 0.6985$ , it would be necessary to find the appropriate

Since $P(Z \le z) = 0.6985$ and is greater than $P(Z \le 0) = 0.50$ , $z > 0$ . As a result, $P(Z \le z)$ can be written as the sum of $P(Z < 0)$ and $P(0 \le Z \le z)$ . Besides, $P(Z < 0) = P(Z \le 0) = 0.50$ . As a result: