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3 years ago

80 is 4/5 of what number

Mathematics

2 answers:

ahrayia [7]3 years ago

4 0

its simple, first make 4/5 a decimal; 0.8

then multiply it by 80 cause of means multiplication; 80 × 0.8 = 64

So your final answer would be; 64

MAVERICK [17]3 years ago

4 0

To find the answer to this, you first divide 80 by four to find what 1/5 is.
80 divided by four is twenty, so 1/5=20.

20x5=100, so 80 is 4/5 of 100.

Hope I could help!

You might be interested in

What are the factors of 5x2 + 39x − 8?

prisoha [69]

Answer:

C. (5x - 1)(x + 8)

Step-by-step explanation:

In a factoring problem by hand, the common steps to solve would be the "AM" method, or (Add-Multiply). Simply stated, finding what adds to the x-coeficient and what multiplies to the constant. In this case, what adds to 39(x) and what multiplies to 8. Just reading that, we can see that there aren't any numbers that can do that, so we have to improvise. Set up the problem equal to something that it can factor into:

<em>Multiply the 5 and -8</em> = -40: <em>this is what we will factor</em>
<em>The factors with add to 39 are 40 and -1</em>.
<em>Set up the problem:</em>

$5x^{2}+39x-8 = 5x^{2}+40x-x-8$

Now, we can fo the math:

$5x^{2} -8 = 5x^{2} -8$

8(5x - 1) + x(5x - 1)

<em>Factor (5x - 1) from this:</em>

(5x - 1)(x + 8)

And we have our answer.

4 0

3 years ago

The prior probabilities for events A1 and A2 are P(A1) = 0.20 and P(A2) = 0.80. It is also known that P(A1 ∩ A2) = 0. Suppose P(

Umnica [9.8K]

Answer:

(a) $A_1$ and $A_2$ are indeed mutually-exclusive.

(b) $\displaystyle P(A_1\; \cap \; B) = \frac{1}{20}$ , whereas $\displaystyle P(A_2\; \cap \; B) = \frac{1}{25}$ .

(d) $\displaystyle P(A_1 \; |\; B) \approx \frac{5}{9}$ , whereas $P(A_1 \; |\; B) = \displaystyle \frac{4}{9}$

Step-by-step explanation:

$P(A_1 \; \cap \; A_2) = 0$ means that it is impossible for events $A_1$ and $A_2$ to happen at the same time. Therefore, event $A_1$ and $A_2$ are mutually-exclusive.

By the definition of conditional probability:

$\displaystyle P(B \; | \; A_1) = \frac{P(B \; \cap \; A_1)}{P(B)} = \frac{P(A_1 \; \cap \; B)}{P(B)}$ .

Rearrange to obtain:

$\displaystyle P(A_1 \; \cap \; B) = P(B \; |\; A_1) \cdot P(A_1) = 0.25 \times 0.20 = \frac{1}{20}$ .

Similarly:

$\displaystyle P(A_2 \; \cap \; B) = P(B \; |\; A_2) \cdot P(A_2) = 0.80 \times 0.05 = \frac{1}{25}$ .

Note that:

$\begin{aligned}P(A_1 \; \cup \; A_2) &= P(A_1) + P(A_2) - P(A_1 \; \cap \; A_2) = 0.20 + 0.80 = 1\end{aligned}$ .

In other words, $A_1$ and $A_2$ are collectively-exhaustive. Since $A_1$ and $A_2$ are collectively-exhaustive and mutually-exclusive at the same time:

$\displaystyle P(B) = P(B \; \cap \; A_1) + P(B \; \cap \; A_2) = \frac{1}{20} + \frac{1}{25} = \frac{9}{100}$ .

By Bayes' Theorem:

$\begin{aligned} P(A_1 \; |\; B) &= \frac{P(B \; | \; A_1) \cdot P(A_1)}{P(B)} \\ &= \frac{0.25 \times 0.20}{9/100} = \frac{0.05 \times 100}{9} = \frac{5}{9}\end{aligned}$ .

Similarly:

$\begin{aligned} P(A_2 \; |\; B) &= \frac{P(B \; | \; A_2) \cdot P(A_2)}{P(B)} \\ &= \frac{0.05 \times 0.80}{9/100} = \frac{0.04 \times 100}{9} = \frac{4}{9}\end{aligned}$ .