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3 years ago

There may be multiple answers here. please help

Mathematics

1 answer:

gogolik [260]3 years ago

6 0

I'm not positive I know the answer, but I have a feeling that because it's asking that you select all that apply, there is more than just one answer.

x + 3 = 0
-3 + 3 = 0

So yeah, just basically solve them as individual problems. Just do additive inverse for each expression.

You might be interested in

5.1 kilometers to meters and centimeters

Snezhnost [94]

Hey there

1 km = 1000 meters
So 5 km will equal 5000 meters
.1 will equal 100 meters

5.1 km = 5100 meters

<span>100,000 centimeters = 1 kilometer
</span>5.1 km = 510000 centimeters

Hope this helps

7 0

3 years ago

Read 2 more answers

The equation t^3=a^2 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A

ankoles [38]

Answer:

2√2

Step-by-step explanation:

We can find the relationship of interest by solving the given equation for A, the mean distance.

<h3>Solve for A</h3>

$T^3=A^2\\\\A=\sqrt{T^3}=T\sqrt{T}\qquad\text{take the square root}$

<h3>Substitute values</h3>

The mean distance of planet X is found in terms of its period to be ...

$D_x=T_x\sqrt{T_x}$

The mean distance of planet Y can be found using the given relation ...

$T_y=2T_x\\\\D_y=T_y\sqrt{T_y}=2T_x\sqrt{2T_x}=(2\sqrt{2})T_x\sqrt{T_x}\\\\D_y=2\sqrt{2}\cdot D_x$

The mean distance of planet Y is increased from that of planet X by the factor ...

2√2

8 0

2 years ago

The staff at the Arcola Country Club prepares tables for a wedding reception. They are able to prepare 6 tables in 45 minutes. H

TiliK225 [7]

24

Multiply 6 and 4 to get 24

They will be able to prepare 24 tables

5 0

3 years ago

Use the pattern below for questions 6 - 8.

Yuliya22 [10]

Answer:

See below

Step-by-step explanation:

<u>The sequence given</u>

6, 19, 58, 175

<u>We see the pattern: triple the previous term plus 1</u>

2) 19 = 6*3 + 1
3) 58 = 19*3 + 1
4) 175 = 58*3 + 1

<u>Next two terms</u>

5) 175*3 + 1 = 526
6) 526*3 + 1 = 1579

<u>Following two terms</u>

7) 1579*3 + 1 = 4738
8) 4738*3 + 1 = 14215

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:

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

6 0

3 years ago