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4 years ago

8

Help me with this please show work please help!!!!!!!!!!!!!!!!!!!!!!...

2 answers:

Paul [167]4 years ago

8 0

The total cost to buy 2 bananas and 2 oranges is 1.70$
To find the cost of the bananas you would divide 1.50 by 6 and you arrive at 0.25 then each banana costs 25 cents, so if you want 2 it would cost you 50 cents.
To find the price of the oranges you would divide 3.00 dollars buy 5 and you arrive at 0.60 cents, so that means each orange costs .60 cents. If you want 2 oranges then you would add 60+60 and get 1.20$
Then you would add 50 cents and 1.20$ and arrive at 1.70$

Ulleksa [173]4 years ago

3 0

Correct me if im wrong but i think the answer is 4.5

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Helpppp me plz i dont get it

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3 years ago

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Step-by-step explanation:

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3 years ago

Suppose on a certain planet that a rare substance known as Raritanium can be found in some of the rocks. A raritanium-detector i

aleksandrvk [35]

Answer:

0.967 = 96.7% probability the rock sample actually contains raritanium

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

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

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 question:

Event A: Positive reading

Event B: Contains raritanium

Probability of a positive reading:

98% of 13%(positive when there is raritanium).

0.5% of 100-13 = 87%(false positive, positive when there is no raritanium). So

$P(A) = 0.98*0.13 + 0.005*0.87 = 0.13175$

Positive when there is raritanium:

98% of 13%

$P(A) = 0.98*0.13 = 0.1274$

What is the probability the rock sample actually contains raritanium?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1274}{0.13175} = 0.967$

0.967 = 96.7% probability the rock sample actually contains raritanium

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