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ycow [4]
3 years ago
11

Elevators have safety limits for how much weight they can lift. The elevator at the Terrace Hotel can hold up to 2,100 pounds. w

rite an equation or inequality that represents the allowable weight on the elevator at the Terrace Hotel. Graph the solution to the equation or inequality you wrote on the number line. Explain your answer.

Mathematics
2 answers:
hram777 [196]3 years ago
8 0
To represent this answer you need to find out what possible numbers of pounds would make this true. In this case, the elevator can hold any number of pounds up to and including 2100 pounds. Attached is a picture of the solutions to your any quality and the actual inequality that represents all of the possible solutions. The idea behind the number line is to show all numbers that are less than or equal to 2100 pounds. The arrow points to the left because numbers to the left are less than The numbers to the right, and we fill in the circle because it includes 2100 as a possible solution.

jok3333 [9.3K]3 years ago
4 0
If the elevator can hold up to 2,100 pounds, that means 2,100 is the maximum. So we have an inequality, not an equation because the elevator can also hold less than 2,100 pounds. Our inequality would represent "less than or equal to".

To write this out, we start with a variable, let's say "w" for "weight".

w

Then, we add our inequality symbol "less than or equal to". 

w \leq

Last, we insert our weight limit which is 2,100 pounds. 

w 
\leq 2,100

That represents the inequality for his specific problem. To graph this on a number line, we simply use the symbol that matches "less than or equal to", which is a closed circle. And since we are talking about the weight being LESS THAN, our arrow is going to the left. Attached is an image of how to graph this on a number line. 

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4 years ago
Perform the indicated operation and simplify the result.
harina [27]

\frac{b^{2}-3b-10 }{(b-2)^{2} }·\frac{b-2}{b-5}=

\frac{(b-5)(b+2)}{(b-2)^{2} }·\frac{b-2}{b-5}=

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3 years ago
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What is 100,203 in expanded form
alexgriva [62]
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6 0
3 years ago
A: {71,73,79,83,87} B:{57,59,61,67}
Jobisdone [24]

Answer:

\frac{3}{5}.

Step-by-step explanation:

We have been given two sets as A: {71,73,79,83,87} B:{57,59,61,67}. We are asked to find the probability that both numbers are prime, if one number is selected at random from set A, and one number is selected at random from set B.

We can see that in set A, there is only one non-prime number that is 87 as it is divisible by 3.

So there are 4 prime number in set A and total numbers are 5.

P(\text{Prime number from A})=\frac{4}{5}

We can see that in set B, there is only one non-prime number that is 57 as it is divisible by 3.

So there are 3 prime number in set B and total numbers are 4.

P(\text{Prime number from B})=\frac{3}{4}

Now, we will multiply both probabilities to find the probability that both numbers are prime. We are multiplying probabilities because both events are independent.

P(\text{Prime number from A and B})=\frac{4}{5}\times \frac{3}{4}

P(\text{Prime number from A and B})=\frac{1}{5}\times \frac{3}{1}

P(\text{Prime number from A and B})=\frac{3}{5}

Therefore, the probability that both numbers are prime would be \frac{3}{5}.

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3 years ago
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