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3 years ago

What are all of the possible classifications for a 3 × 3 system? Check all of the boxes that apply.

Mathematics

2 answers:

Degger [83]3 years ago

7 0

Consistent and Dependent
Consistent and Independent
Inconsistent and Independent

Alex_Xolod [135]3 years ago

6 0

I believe the answer would be inconsistent and dependent, and also inconsistent and independent. System of equations with three variables are either independent, dependent or inconsistent , each case can be established algebraically and represented graphically.

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What is the length of side a? Round to the nearest tenth of an inch. Please and thank youuu

lord [1]

Answer:

using Pythagoras rule, 16^2 -7^2 =a^2

so 256-49 = 297.

√207 =14.38 to the nearest tenth of an inch is 14inch.which is your answer

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3 years ago

Alex must read at least 500 pages of his book this month. He has already read 260 pages. If he reads 30 pages a day, what is the

slamgirl [31]

So he has to read 500 and he already read 260 500 - 260 = 240. 240 divided by 30 equals 8. So the answer will be 8 days

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3 years ago

Find the length of a diagonal of a square with sides of 14 feet long. [?][feet /

inessss [21]

Answer:

14√2 feet

Step-by-step explanation:

The length of a diagonal of a square is $L=S\sqrt{2}$ where $S$ is the side length of the square. Therefore, $L=14\sqrt{2}$ .

4 0

3 years ago

Solve for z. z/12=6 3/4

Tatiana [17]

6 3/4=6*4+3/4=24+3/4=27/4

z/12=27/4
4z=27*12
4z=324
z=81

3 0

3 years ago

Read 2 more answers

(a) Let {A1, A2} be a partition of a sample space and let B be any event. State and prove the Law of Total Probability as it app

Nataliya [291]

Answer:

0.625

Step-by-step explanation:

Given that {A1, A2} be a partition of a sample space and let B be any event. State and prove the Law of Total Probability as it applies to the partition {A1, A2} and the event B.

Since A1 and A2 are mutually exclusive and exhaustive, we can say

b) P(B) = P(A1B)+P(A2B)

Selecting any one coin is having probability 0.50. and A1, A2 are events that the coins show heads. $P(B/A1) = 0.50 \\P(B/A2) = 0.75\\P(A1B) = 0.5(0.5) = 0.25 \\P(A2B) = 0.75(0.5) = 0.375\\P(B) = 0.625$

c) Using Bayes theorem

conditional probability that it wasthe biased coin

= $\frac{0.375}{0.625} \\=\frac{3}{5}$

d) Given that the chosen coin flips tails,the conditional probability that it was the biased coin= $\frac{0.25*0.5}{0.25*0.5+0.5*0.5} \\=\frac{1}{3}$

7 0

3 years ago