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zhenek [66]
3 years ago
7

The elevator at the state capital has a maximum weight capacity of 2000 pounds. If there are 3

Mathematics
1 answer:
Korolek [52]3 years ago
3 0

Answer:

Step-by-step explanation:

Maximum capacity of the elevator = 2000 pounds

Combined weight of the elevator = 428 pounds

A). Let the possible weights that can be added to the elevator = w pounds

B). Therefore, inequality representing this situation will be,

    428 + w < 2000

C). By solving the given inequality,

    (428 + w) - 428 < 2000 - 428

    w < 1572 pounds

    Now we draw this inequality on a number line.

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The sum of two positive numbers is 16 and their difference is 2 . the larger number is?​
Aleonysh [2.5K]

Answer:

9

Step-by-step explanation:

7 + 9 = 16

9 - 7 = 2

The larger number is 9

Hope that helps!

8 0
2 years ago
Read 2 more answers
A hot air balloon went from an elevation of 4,064 feet to an elevation of 3,322 feet in 28 minutes. What
ELEN [110]

Answer:

<em>The air balloon descends 26.5 feet per minute</em>

Step-by-step explanation:

<u>Rate of Change</u>

The rate of change is a measure that compares two quantities, usually to know how one variable changes in time.

The hot air balloon was 4,064 feet high and, 28 minutes later, it went to 3,322 feet. We can calculate the rate of descent R by dividing the change of elevation over time:

\displaystyle R=\frac{3,322-4,064}{28}=\frac{-742}{28}=-26.5

Thus the air balloon descends 26.5 feet per minute

3 0
3 years ago
a slitter assembly contains 48 blades five blades are selected at random and evaluated each day for sharpness if any dull blade
son4ous [18]

Answer:

P(at least 1 dull blade)=0.7068

Step-by-step explanation:

I hope this helps.

This is what it's called dependent event probability, with the added condition that at least 1 out of 5 blades picked is dull, because from your selection of 5, you only need one defective to decide on replacing all.

So if you look at this from another perspective, you have only one event that makes it so you don't change the blades: that 5 out 5 blades picked are sharp. You also know that the probability of changing the blades plus the probability of not changing them is equal to 100%, because that involves all the events possible.

P(at least 1 dull blade out of 5)+Probability(no dull blades out of 5)=1

P(at least 1 dull blade)=1-P(no dull blades)

But the event of picking one blade is dependent of the previous picking, meaning there is no chance of picking the same blade twice.

So you have 38/48 on getting a sharp one on your first pick, then 37/47 (since you remove 1 sharp from the possibilities, and 1 from the whole lot), and so on.

Also since are consecutive events, you need to multiply the events.

The probability that the assembly is replaced the first day is:

P(at least 1 dull blade)=1-P(no dull blades)

P(at least 1 dull blade)=1-(\frac{38}{48}* \frac{37}{47} *\frac{36}{46}*\frac{35}{45}*\frac{34}{44})

P(at least 1 dull blade)=1-0.2931

P(at least 1 dull blade)=0.7068

5 0
3 years ago
Please answer!!!! please!!!!!!
olga55 [171]

The problem here is:  Too many words.

If you boil the problem down to the bare facts, the question is:

             "$210 is 15% of what number ?"

I'm pretty sure you can take it from there,
but here's the rest of it anyway:

                                                $210  =  15% of (March amount)

                                                $210  =  0.15 x (March amount)

Divide each side by  0.15 :     $1400 = March amount .

Check:
1400 + 210 =  1610
1.15 x 1400 = 1610                                      yay!
5 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
ehidna [41]

Answer:  1) 0.10

               2) 0.60

               3) 0.20

               4) 0.10

<u>Step-by-step explanation:</u>

The total frequency is 20+120+40+20 = 200.  This means they ran the experiment 200 times.  The probability distribution is calculated by the satisfactory number of outcomes (frequency) divided by the total number of experiments/outcomes (total frequency):

\begin{array}{c|c||lc}\underline{x}&\underline{f}&\underline{f\div 200}&\underline{\text{Probability Distribution}}\\1&20&20\div200=&0.10\\2&120&120\div 200=&0.60\\3&40&40\div 200=&0.20\\4&20&20\div 200=&0.10\end{array}\right]

6 0
3 years ago
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