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swat32
3 years ago
5

Who made math? Explain??

Mathematics
2 answers:
Ne4ueva [31]3 years ago
5 0

Answer:

Beginning in the 6th century BC with the Pythagoreans, with Greek mathematics the Ancient Greeks began a systematic study of mathematics as a subject in its own right. Around 300 BC, Euclid introduced the axiomatic method still used in mathematics today, consisting of definition, axiom, theorem, and proof

Step-by-step explanation:

dsp733 years ago
3 0

Answer:

, Step-by-step explanation:

so it was the greeks along with the pythagoreans  during the 6th century the greeks began a study of mathematics. around 300bc euclid intoduced the axiomatic method still used today consisting of definition, axiom, theorem, and proof so it was really the greeks and pythagoreans. your welcome

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Evaluate 5! + 2!.<br> Idk what is it
Sedbober [7]

Answer:

122

Step-by-step explanation:

If a number has an explanation point on the side of it, you should multiply all numbers that are equal to are less than that number.

5!=5*4*3*2*1=120\\2!=2*1=2\\120+2=122

4 0
3 years ago
Check my answer?
Effectus [21]
D = 2r => 12.6 = 2*r => r=6.3
A = π*r^2 = (3.14) * (6.3)^2 = (3.14)*(39.69) = 124.6266 which is approximately 124.63

Hope this helps!
6 0
3 years ago
Read 2 more answers
4(x−0.8)=3.8x−5.8 PLZZZ ANSWER FAST!
Hunter-Best [27]
To find the answer:
4(x - 0.8) = 3.8x - 5.8 \\ 4x - 3.2 = 3.8x - 5.8 \\ 4x - 3.8x =  - 5.8 + 3.2 \\ 0.2x =  - 2.6 \\ x =   - \frac{ 2.6}{0.2}  \\ x =  - 13

Therefore, x=-13.

Hope it helps!
5 0
3 years ago
Read 2 more answers
X−3y=1x+y=9
Brrunno [24]

I got this same question on a test, the answer is A.

7 0
3 years ago
Read 2 more answers
The prior probabilities for events A1 and A2 are P(A1) = 0.35 and P(A2) = 0.50. It is also known that P(A1 ∩ A2) = 0. Suppose P(
forsale [732]

Answer:

Step-by-step explanation:

Hello!

Given the probabilities:

P(A₁)= 0.35

P(A₂)= 0.50

P(A₁∩A₂)= 0

P(BIA₁)= 0.20

P(BIA₂)= 0.05

a)

Two events are mutually exclusive when the occurrence of one of them prevents the occurrence of the other in one repetition of the trial, this means that both events cannot occur at the same time and therefore they'll intersection is void (and its probability zero)

Considering that P(A₁∩A₂)= 0, we can assume that both events are mutually exclusive.

b)

Considering that P(BIA)= \frac{P(AnB)}{P(A)} you can clear the intersection from the formula P(AnB)= P(B/A)*P(A) and apply it for the given events:

P(A_1nB)= P(B/A_1) * P(A_1)= 0.20*0.35= 0.07

P(A_2nB)= P(B/A_2)*P(A_2)= 0.05*0.50= 0.025

c)

The probability of "B" is marginal, to calculate it you have to add all intersections where it occurs:

P(B)= (A₁∩B) + P(A₂∩B)=  0.07 + 0.025= 0.095

d)

The Bayes' theorem states that:

P(Ai/B)= \frac{P(B/Ai)*P(A)}{P(B)}

Then:

P(A_1/B)= \frac{P(B/A_1)*P(A_1)}{P(B)}= \frac{0.20*0.35}{0.095}= 0.737 = 0.74

P(A_2/B)= \frac{P(B/A_2)*P(A_2)}{P(B)} = \frac{0.05*0.50}{0.095} = 0.26

I hope it helps!

5 0
3 years ago
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