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3 years ago

What is 20/10 as a mixed number

Mathematics

2 answers:

Lostsunrise [7]3 years ago

5 0

It will just be 2 since 10 goes into 20 2 times

inna [77]3 years ago

5 0

Let's see, we have to find out how many times does 10 fit into 20, and we can see that 10 fits to times into 20 because 10 + 10 equals 20, so 20/10 as a mixed number is 2.

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Can any one give me a fast answer please I need help

icang [17]

Answer:

C = n + 2

Step-by-step explanation:

Well looking at the line on the graph we can see that the y intercept is 2 because the y intercept is the point in the line that touches the y axis.

And the slope is how fat away each points are from each other on a line so we can find the slope by using two points on the line, we can use (1,3) and (2,4).

So we set up the formula like this $\frac{y^2-y^1}{x^2-x^1}$ .

And now we gotta plug in the numbers and solve so the answer is 4-3 = 1 and 2-1 = 1 so the slope is 1.

And we can’t write that as just n.

So the answer is C = n + 2.

For proof look at the image below.

5 0

3 years ago

How could you put that is a mathematic form??

Vladimir [108]

Where's the picture???? I'm confusedd

7 0

2 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:

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