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Harman [31]
2 years ago
8

Cómo se resuelven las ecuaciones químicas?

Mathematics
1 answer:
prohojiy [21]2 years ago
4 0

Answer:

La ecuación química necesita ser equilibrada para que siga la ley de conservación de la masa.

Step-by-step explanation:

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Identify the solution set of the given inequality using the given replacement set.
scoundrel [369]
Another user previously answered this same question and had a good rating, his answer was as follows:

"{-9; -6.5; -5.2}" 

Which I suppose would be answer D.
- Hope it helps.

3 0
3 years ago
Mrs. Ross needs to buy dish soap. There are four differently sized containers. Sort the brands from least to the greatest unit p
r-ruslan [8.4K]

Answer: Spotless soap < Bright wash < Lemon bright < lot of suds

Step-by-step explanation:

Given that 8 oz of Lots of suds cost = $0.98

Then unit price of 1 lot of suds = 0.98/8 = $0.1225

Given that 12 oz of Bright wash cost = $1.29

Then unit price of 1 Bright wash = 1.29/12 = $0.1075

Given that 30 oz of Spotless soap cost = $3.14

Then unit price of 1 Spotless soap = 3.14/30 = $0.104666666667

Given that 32 oz of Lemon bright cost = $3.50Then unit price of 1 Lemon bright = 3.50/32 = $0.109375

We also need to round the values to the nearest thousandth means three digits after decimal

Then unit price of 1 lot of suds = $0.123

unit price of 1 Bright wash = $0.108

unit price of 1 Spotless soap = $0.105

unit price of 1 Lemon bright = $0.109

Now we can easily sort them

0.105 < 0.108 < 0.109 < 0.123

respective names in sorted order are:

Spotless soap < Bright wash < Lemon bright < lot of suds

8 0
3 years ago
The probability that a randomly selected 2 2​-year-old male garter snake garter snake will live to be 3 3 years old is 0.98861 0
Mnenie [13.5K]

Answer:

a. Probability = 0.97735

b. Probability = 0.92294

c. P(At\ Least\ One) = 1

No, it is not unusual if at least 1 lives up to 3.

Step-by-step explanation:

Given

Represent the probability that a 2 year old snake will live to 3 with P(Live);

P(Live) = 0.98861

Solving (a): Probability that two selected will live to 3 years.

Both snakes have a chance of 0.98861 to live up to 3 years.

So, the required probability is:

Probability = P(Live)\ and\ P(Live)

Probability = 0.98861 * 0.98861

Probability = 0.9773497321

Probability = 0.97735 <em>--- Approximated</em>

Solving (b): Probability that seven selected will live to 3 years.

All 7 snakes have a chance of 0.98861 to live up to 3 years.

So, the required probability is:

Probability = P(Live)^n

Where n = 7

Probability = 0.98861^7

Probability = 0.92294324145

Probability = 0.92294 <em>--- Approximated</em>

Solving (c): Probability that at least one of seven selected will not live to 3 years.

In probabilities, the following relationship exist:

P(At\ Least\ One) = 1 - P(None).

So, first we need to calculate the probability that none of the 7 lived up to 3.

If the probability that one lived up to 3 years is 0.98861, then the probability than one do not live up to 3 years is 1 - 0.98861

This gives:

P(Not\ Live) = 0.01139

The probability that none of the 7 lives up to 3 is:

P(None) = P(Not\ Live)^7

P(None) = 0.01139^7

Substitute this value for P(None) in

P(At\ Least\ One) = 1 - P(None).

P(At\ Least\ One) = 1 - 0.01139^7

P(At\ Least\ One) = 0.99999999999997513055642436060443621

P(At\ Least\ One) = 1 ---- Approximated

No, it is not unusual if at least 1 lives up to 3.

This is so because the above results, which is 1 shows that it is very likely for at least one of the seven to live up to 3 years

7 0
3 years ago
A=5x+20
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