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3 years ago

Please help and show work

Mathematics

1 answer:

Musya8 [376]3 years ago

4 0

Answer:

2 is x = 8

Step-by-step explanation:

I’m still figuring out 1), please don’t report this

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Drag each tile to the correct box. not all tiles will be used.Chris plans to mulch the border surrounding a rectangular garden i

amm1812

Length (L): 2w + 3

width (w): w

border (b): $\frac{1}{4} w$

Area (A) = (L + b) * (w + b) <em>NOTE: This is assuming the width also has a border</em>

= (2w + 3 + $\frac{1}{4} w$ ) * (w + $\frac{1}{4} w$ )

= $(2\frac{1}{4} w + 3)$ $(1\frac{1}{4} w)$

= $(\frac{9}{4} w + 3)$ $(\frac{5}{4} w)$

= $\frac{45}{16} w^2 + \frac{15}{4}w$

5 0

3 years ago

Please help i dont get it and its for a grade! :(

Tju [1.3M]

Answer:

y=-4x+5

Step-by-step explanation:

7 0

3 years ago

(x - 3)2 + (y + 2)2 = 4

vazorg [7]

As a function this is x=3 y=3

4 0

3 years ago

Read 2 more answers

Solve. 4 1/4−1/3x=−3/4 Enter your answer in the box.

romanna [79]

Answer:

x=33

Step-by-step explanation:

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4 0

3 years ago

Read 2 more answers

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

3 years ago