A tram moved downward 12 meters in 4 seconds at a constant rate. What was the change in the tram's elevation each second?

Answers

Answer 1

Therefore , the solution of the given problem of unitary method comes out to be during the 4-second period, the tram's elevation changed by 3 metres every second.

What is an unitary method?

To complete the assignment, use the iii . -and-true basic technique, the real variables, and any pertinent details gathered from basic and specialised questions. In response, customers might be given another opportunity to sample expression the products. If these changes don't take place, we will miss out on important gains in our knowledge of programmes.

Here,

By dividing the overall elevation change (12 metres) by the total time required (4 seconds),

it is possible to determine the change in the tram's elevation every second. We would then have the average rate of elevation change per second.

=> Elevation change equals 12 metres

=> Total duration: 4 seconds

=>  12 meters / 4 seconds

=> 3 meters/second

As a result, during the 4-second period, the tram's elevation changed by 3 metres every second.

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Related Questions

what is the 4th term/number of (a+b)^9, pascal’s triangle?

Answers

Step-by-step explanation:

hope this will help you Thanks

A tile is selected from seven tiles, each labeled with a different letter from the first seven letters of the alphabet. The letter selected will be recorded as the outcome. Consider the following events. Event X: The letter selected comes before "D". Event Y: The letter selected is found in the word "CAGE". Give the outcomes for each of the following events. If there is more than one element in the set, separate them with commas.

(a) Event "X or Y":

(b) Event "X and Y":

(c) The complement of the event X:

EXPLANATION/ANSWER

The sample space is the set of all possible outcomes.

In this case, the sample space is , A, B, C, D, E, F, G.

The event X is "The letter selected is found in the word "BEAD". "

The outcomes in this event are A, B, D, and E.

The event Y is "The letter selected comes after "D". "

The outcomes in this event are E, F, and G.

(a) Event "X or Y"

Outcomes in the event "

X or Y" are any outcomes from event X along with any outcomes from event Y.

So the outcomes in the event "X or Y " are A, B, D, E, F, and G. Event "X or Y": , A, B, D, E, F, G

(b) Event "X and Y"

The outcomes in the event "X and Y" are the outcomes from event X that also occur in event Y.

So the outcome in the event "X and Y" is E. Event "X and Y": E

(c) The complement of the event X

The complement of the event X is the event consisting of all possible outcomes not in the event X.

So the outcomes in the complement of the event X are C, F, and G

The complement of the event X: , C, F, G

(a) Event "X or Y": , A, B, D, E, F, G

(b) Event "X and Y": E

(c) The complement of the event X: , C, F, G

My problem that I am having trouble with:

A number cube with faces labeled 1 to 6 is rolled once.

The number rolled will be recorded as the outcome.

Consider the following events.

Event A: The number rolled is odd.

Event B: The number rolled is less than 4

Give the outcomes for each of the following events.

If there is more than one element in the set, separate them with commas.

(a) Event"A or B":

(b) Event"A and B":

(c) The complement of the event B:

Answers

Event "X or Y": A, B, C, E, G. Event "X and Y": C. The complement of event X: D, E, F, G.

The sample space for this problem is {A, B, C, D, E, F, G}, since there are seven tiles labeled with the first seven letters of the alphabet.

Event X: The letter selected comes before "D". Outcomes in this event are A, B, and C.

Event Y: The letter selected is found in the word "CAGE". Outcomes in this event are A, C, E, and G.

Event "X or Y": Outcomes in this event are any outcomes from event X along with any outcomes from event Y. So the outcomes in the event "X or Y" are A, B, C, E, and G.

Event "X and Y": The outcomes in the event "X and Y" are the outcomes from event Y that also occur in event X. So the only outcome in the event "X and Y" is C.

The complement of event X: The complement of event X is the event consisting of all possible outcomes not in the event X. So the outcomes in the complement of event X are D, E, F, and G.

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--The given question is incomplete, the complete question is given

" A tile is selected from seven tiles, each labeled with a different letter from the first seven letters of the alphabet. The letter selected will be recorded as the outcome. Consider the following events. Event X: The letter selected comes before "D". Event Y: The letter selected is found in the word "CAGE". Give the outcomes for each of the following events. If there is more than one element in the set, separate them with commas.

(a) Event "X or Y":

(b) Event "X and Y":

(c) The complement of the event X: "--

Consider a sequence whose first five terms are:-1.75, -0.5, 0.75, 2, 3.25
Which explicit function (with domain all integers n ≥ 1) could be used to define and continue this sequence?

Answers

Step-by-step explanation:

+ 1.25

every new term is the previous term + 1.25.

with starting value -1.75

f(n) = 1.25n - 1.75

brainlist
show all steps nd i will make u brainlist

Answers

Step-by-step explanation:

Again, using similar triangle ratios

7.2 m  is to 2.4 m  

     as   AB is to  12.0 m

7.2 / 2.4  = AB/12.0    Multiply both sides of the equation by 12

12 *   7.2 / 2.4   = AB = 36.0 meters

of the cartons produced by a company, 3% have a puncture, 6% have a smashed corner, and 1.4% have both a puncture and a smashed corner. find the probability that a randomly selected carton has a puncture or a smashed corner.

Answers

The probability that a randomly selected carton has a puncture or a smashed corner is 0.076, or 7.6%.

What is probability?

Probability is a measure of the likelihood of an event occurring. It is a number between 0 and 1, where 0 means the event is impossible and 1 means the event is certain to happen.

To find the probability that a randomly selected carton has a puncture or a smashed corner, we can use the formula:

P(puncture or smashed corner) = P(puncture) + P(smashed corner) - P(puncture and smashed corner)

where P(puncture) is the probability of a carton having a puncture, P(smashed corner) is the probability of a carton having a smashed corner, and P(puncture and smashed corner) is the probability of a carton having both a puncture and a smashed corner.

Substituting the given probabilities into the formula, we get:

P(puncture or smashed corner) = 0.03 + 0.06 - 0.014

P(puncture or smashed corner) = 0.076

Therefore, the probability that a randomly selected carton has a puncture or a smashed corner is 0.076, or 7.6%.

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