There are 9 students entering the school district's science project with all students having an equal chance

Match each statement with the right choice.

KengaRu [80]

Answer:

Your answer is:

1) 6 ( 5 + 8 ) = 78

2) 5 ( 10 + 3 ) = 65

3) 9 ( 2 + 4 ) = 54

4) 7 ( 3 + 6 ) = 63

Step-by-step explanation:

Hope this helped : )

4 0

4 years ago

Find the value of x.<br> 48<br> (7x + 13)<br> A. 7<br> B. 11<br> C. 3<br> D. 5

Lesechka [4]

The value of x is D. 5

3 0

3 years ago

I’m confused on this one

Lorico [155]

The question is asking which of the following does not have side that are parallel. So which of them have lines that would never touch if they were never ending lines.

4 0

3 years ago

ALGEBRA On the first Tuesday of every month, Roth's gives seniors a 15%

iVinArrow [24]

Answer:

Let selling price of three-piece suit = x. Applying 7% sales tax, total amount paid = x + xx7 - $152.79. 100 207 100 ०१ This is the $ 152.79 os, x- $142.7943 price after marked down of 40%, let the regular price of three-piece suit be = y. Therefore y [ 3 - 4m] $342.1943 1 - - $237.99 Regular price of 1 piece suit $ 237.99

3 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 $\displaystyle P(A_2\; \cap \; B) = \frac{1}{25}$ .

(c) $\displaystyle P(B) = \frac{9}{100}$ .

(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

3 years ago