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3 years ago

What are some possible drawbacks or consequences of calling a period of time a Golden Age?

Mathematics

1 answer:

pshichka [43]3 years ago

6 0

Answer:

Once we call one period that, we discount any future time that times will be better. What about later? Can things not get better than the golden age?

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What are the factors of m2 – 6m + 6m – 36? (m – 2)(m + 18) (m – 3)(m + 3) (m – 4)(m + 9) (m – 6)(m + 6)

77julia77 [94]

(m – 6)(m + 6) is the answer.
If you distribute them, you'll get m^2 - 36

3 0

3 years ago

Read 2 more answers

(x + 3)(x-8) Multiply and simplify

harkovskaia [24]

$(x+3)(x-8)=x\cdot x+x\cdot(-8)+3\cdot x+3\cdot(-8)=\\\\=x^2-8x+3x-24=\boxed{x^2-5x-24}$

6 0

3 years ago

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Willian and Amy are stuffing envelope for a charity today, William stuff a total of 70 envelope. This is 10 more than twice the

LiRa [457]

Answer:

because it is 10 major then we remove those ten and it gives us 60 that between 2 is 30 :))

7 0

3 years ago

HELP MATH ASAP! MARKING BRANLIST

qwelly [4]

Answer:

-4/3

Step-by-step explanation:

Use the slope formula:

$\frac{y2-y1}{x2-x2}$

-4 - 12/ 13 - 1

-16/12

Simplifies to -4/3

8 0

3 years ago

Any athlete who fails the Enormous State University's women's soccer fitness test is automatically dropped from the team. Last y

castortr0y [4]

Using conditional probability, it is found that there is a 0.7873 = 78.73% probability that Mona was justifiably dropped.

Conditional Probability

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.
$P(A \cap B)$ is the probability of both A and B happening.
P(A) is the probability of A happening.

In this problem:

Event A: Fail the test.
Event B: Unfit.

The probability of failing the test is composed by:

46% of 37%(are fit).
100% of 63%(not fit).

Hence:

$P(A) = 0.46(0.37) + 0.63 = 0.8002$

The probability of both failing the test and being unfit is:

$P(A \cap B) = 0.63$

Hence, the conditional probability is:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.63}{0.8002} = 0.7873$

0.7873 = 78.73% probability that Mona was justifiably dropped.

A similar problem is given at brainly.com/question/14398287

3 0

3 years ago