All categories

2 years ago

QUIZ

Mathematics

2 answers:

marta [7]2 years ago

5 0

Answer:

B (5/8 - 6/8)

Step-by-step explanation:

Since we can just multiply the second fraction by two to get a similar denominator, we just need to multiply the second fraction.

Butoxors [25]2 years ago

3 0

Answer:

5/8 - 6/8 (the second one)

Step-by-step explanation:

If you look at the denominators with the goal of making them the same, you can see that 8 is a multiple of 4, meaning your job is easier than it could be. Multiply both the top and the bottom of -3/4 by 2 to make the bottom 8. You should get -6/8 while your 5/8 remains unchanged, meaning your final expression in 5/8 - 6/8.

You might be interested in

Solve the formula 27a + 1/3 * b = 4 for a.

Aneli [31]

Answer:

a = 4/27 - b/81

Step-by-step explanation:

27a + 1/3 * b = 4

Subtract the term with b from both sides.

27a = 4 - 1/3 *b

Divide both sides by 27.

a = 4/27 - 1/81 * b

a = 4/27 - b/81

4 0

2 years ago

Read 2 more answers

A book club offers a choice of 8 books from a list of 40. In how many can a member make a collection?

egoroff_w [7]

You have to divide 40 by 8
40/8=5

8 0

3 years ago

Help me pleas please can i get help

givi [52]

Answer:

I can't see the question

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Suppose the box you want to ship has a square base (meaning that the dimensions of the bottom of the box are equal). If you choo

JulsSmile [24]

Answer:

24 inches

Step-by-step explanation:

When dealing with 3-Dimensional objects such as a box, the depth of that object refers to the distance between the highest and lowest points of that object. Therefore since it the question it states that the length, width and depth is 24 inches. Then the height of the box (as well as maximum height) would be a total of 24 inches. This would be in the case that the entire box was cubed, if the box is triangular with a square base then this height would be shorter as all the points would meet in a shorter position.

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:

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

6 0

2 years ago