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4 years ago

11

The following graph shows the consumer price index (CPI) for a fictional country from 1990 to 2000. During which of these time p

eriods was there a period of deflation?

2 answers:

netineya [11]4 years ago

6 0

the first graph is the answer

IgorC [24]4 years ago

4 0

Answer:

The answer for apex is A. 1974 - 1976 hope this help also have a great day

Step-by-step explanation:

You might be interested in

Which polynomial identity will prove that 19 = 27 − 8?

3241004551 [841]

The answer is Difference of Cubes.

Difference of cubes is a³ - b³.
27 can be written as 3³ (= 3 × 3 × 3 = 27). So, a = 3.
8 can be written as 2³ (= 2 × 2 × 2 = 8). So, b = 2.

Difference of cubes can be expressed as:
a³ - b³ = (a - b)(a² + ab + b²)
⇒ 3³ - 2³ = (3 - 2)(3² + 3×2 + 2²) = 1 × (9 + 6 + 4) = 1 × 19 = 19
⇒ 27 - 8 = 19

5 0

3 years ago

Drone INC. owns four 3D printers (D1, D2, D3, D4) that print all their Drone parts. Sometimes errors in printing occur. We know

USPshnik [31]

Answer:

Step-by-step explanation:

Hello!

There are 4 3D printers available to print drone parts, then be "Di" the event that the printer i printed the drone part (∀ i= 1,2,3,4), and the probability of a randomly selected par being print by one of them is:

D1 ⇒ P(D1)= 0.15

D2 ⇒ P(D2)= 0.25

D3 ⇒ P(D3)= 0.40

D4 ⇒ P(D4)= 0.20

Additionally, there is a chance that these printers will print defective parts. Be "Ei" represent the error rate of each print (∀ i= 1,2,3,4) then:

P(E1)= 0.01

P(E2)= 0.02

P(E3)= 0.01

P(E4)= 0.02

Ei is then the event that "the piece was printed by Di" and "the piece is defective".

You need to determine the probability of randomly selecting a defective part printed by each one of these printers, i.e. you need to find the probability of the part being printed by the i printer given that is defective, symbolically: P(DiIE)

Where "E" represents the event "the piece is defective" and its probability represents the total error rate of the production line:

P(E)= P(E1)+P(E2)+P(E3)+P(E4)= 0.01+0.02+0.01+0.02= 0.06

This is a conditional probability and you can calculate it as:

$P(A/B)= \frac{P(AnB)}{P(B)}$

To reach the asked probability, first, you need to calculate the probability of the intersection between the two events, that is, the probability of the piece being printed by the Di printer and being defective Ei.

P(D1∩E)= P(E1)= 0.01

P(D2∩E)= P(E2)= 0.02

P(D3∩E)= P(E3)= 0.01

P(D4∩E)= P(E4)= 0.02

Now you can calculate the probability of the piece bein printed by each printer given that it is defective:

$P(D1/E)= \frac{P(E1)}{P(E)} = \frac{0.01}{0.06}= 0.17$

$P(D2/E)= \frac{P(E2)}{P(E)} = \frac{0.02}{0.06}= 0.33$

$P(D3/E)= \frac{P(E3)}{P(E)} = \frac{0.01}{0.06}= 0.17$

$P(D4/E)= \frac{P(E4)}{P(E)} = \frac{0.02}{0.06}= 0.33$

P(D2)= 0.25 and P(D2/E)= 0.33 ⇒ The prior probability of D2 is smaller than the posterior probability.

The fact that P(D2) ≠ P(D2/E) means that both events are nor independent and the occurrence of the piece bein defective modifies the probability of it being printed by the second printer (D2)

I hope this helps!

8 0

3 years ago

Can anyone explain how to do this?

Aliun [14]

<h3>Answer: 85 units ^2</h3>

Step-by-step explanation:

Using the equation $a^{2} + b^{2} = c^{2}$

so $a^{2} = 35$ and $b^{2} =50$

$a^{2} + b^{2} = 85^{2}$

5 0

3 years ago

Read 2 more answers

I travel 1/10 mile in 1/2 second. how many can i travel in one second

Lorico [155]

Answer:

2/10 miles

Step-by-step explanation:

2 x 1/10 = 2/10

3 0

3 years ago

Help please please help

Artemon [7]

Answer:

could you maybe give me a better picture in the light or something? it is hard for me to see it

Step-by-step explanation:

7 0

3 years ago