Ask question

Ask question

All categories

3 years ago

13

A certain plastic cup is manufactured by a robot. If 5 such robots, working at the same individual rate, can manufacture a combi

ned total of 5 cups in 5 seconds, how many seconds will it take 25 robots to manufacture a total of 25 cups?A) 25B) 20C) 15D) 10E) 5

1 answer:

iragen [17]3 years ago

8 0

Answer:

E) 5 seconds

Step-by-step explanation:

We know that 5 robots can manufacture 5 cups in 5 seconds.

R1,R2,R3,R4,R5: 5 cups in 5 seconds

If we have 25 robots we can divide these 25 robots in 5 groups having 5 robots in each group. And each group will manufacture 5 cups in 5 seconds.

Team 1 - R1,R2,R3,R4,R5 : 5 cups in 5 seconds

Team 2 - R6,R7,R8,R9,R10: 5 cups in 5 seconds

Team 3 - R11,R12,R13,R14,R15: 5 cups in 5 seconds

Team 4 - R16,R17,R18,R19,R20: 5 cups in 5 seconds

Team 5 - R21,R22,R23,R24,R25: 5 cups in 5 secods

Then we have:

<h3> total of robots: 25 total of cups: 25 total of seconds: 5</h3>

Meaning that if we have the 25 robots <u>working simultaneously</u>, it will take 5 seconds to each of the 5 teams (being a total of 25 robtos) to manufacture a total of 25 cups.

You might be interested in

Use induction to prove: For every integer n > 1, the number n5 - n is a multiple of 5.

nignag [31]

Answer:

we need to prove : for every integer n>1, the number $n^{5}-n$ is a multiple of 5.

1) check divisibility for n=1, $f(1)=(1)^{5}-1=0$ (divisible)

2) Assume that $f(k)$ is divisible by 5, $f(k)=(k)^{5}-k$

3) Induction,

$f(k+1)=(k+1)^{5}-(k+1)$

$=(k^{5}+5k^{4}+10k^{3}+10k^{2}+5k+1)-k-1$

$=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k$

Now, $f(k+1)-f(k)$

$f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-(k^{5}-k)$

$f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-k^{5}+k$

$f(k+1)-f(k)=5k^{4}+10k^{3}+10k^{2}+5k$

Take out the common factor,

$f(k+1)-f(k)=5(k^{4}+2k^{3}+2k^{2}+k)$ (divisible by 5)

add both the sides by f(k)

$f(k+1)=f(k)+5(k^{4}+2k^{3}+2k^{2}+k)$

We have proved that difference between $f(k+1)$ and $f(k)$ is divisible by 5.

so, our assumption in step 2 is correct.

Since $f(k)$ is divisible by 5, then $f(k+1)$ must be divisible by 5 since we are taking the sum of 2 terms that are divisible by 5.

Therefore, for every integer n>1, the number $n^{5}-n$ is a multiple of 5.

3 0

3 years ago

Find the value of the variable (n)

alisha [4.7K]

Answer:

D. 65

Step-by-step explanation:

180-50=130

130 divided by 2 =65

hope this helps:)

5 0

3 years ago

Read 2 more answers

A chocolate shop sold 1 1/4 ounces of white chocolate. It sold 8 1/4 times as much milk chocolate as white chocolate. How many o

rjkz [21]

Answer:

10.3125 (NOT ROUNDED)

Step-by-step explanation:

1. I changed it to decimals. 1/4 can easily be changed into a decimal.

2. I set 1/4 = x/100. Then cross multiplied. That would give you 4x = 100.

3. Now you would simplify that. x would be 25. So its 25/100 wich then turns to .25.

4. Now you are going to replace them so it would be 1.25 and 8.25.

5. To find out how much milk chocolate there was you would multiply 1.25 and 8.25.

6. That's your final answer

6 0

3 years ago

Help me with this one problem please :( <br> Find the following arc measures.

r-ruslan [8.4K]

Answer:

Arc KL $=23$

Arc OM $=113$

Step-by-step explanation:

We know that Angle NPM is equal to 90 degrees.

Therefore, Angle KPM is also equal to 90 degrees.

We are told that Arc LM is equal to 67.

Thus, the measure for Arc KL is equal to $90-67=23$

We also know that Angle OPN is equal to Angle KPL.

Therefore, the measure of Arc ON $=23$

Using that information, we know that Arc OM is equal to $90+23=113$

Hope I helped!

3 0

3 years ago

On one day a pet store sells 2 birds 6 gerbils 3 fish and 3 hamsters whats the data and relative frequency as a percent?

bonufazy [111]

Answer:

Birds $F_b=14.3\%$

Gerbils $F_g=42.9\%$

Fish $F_f=21.4\%$

Hamsters $F_h=21.4\%$

Step-by-step explanation:

From the question we are told that:

Sales

Number of birds $n_b=2$

Number of gerbils $n_g=6$

Number of fish $n_f=3$

Number of hamsters $n_h=3$

Generally the frequency for birds $F_b$ is mathematically given b

$F_b=\frac{n_b}{n}*100$

$F_b=\frac{2}{14}*100$

$F_b=14.3\%$

Generally the frequency for Gerbils $F_g$ is mathematically given b

$F_g=\frac{n_g}{n}*100$

$F_g=\frac{6}{14}*100$

$F_g=42.9\%$

Generally the frequency for fish $F_f$ is mathematically given b

$F_f=\frac{n_f}{n}*100$

$F_f=\frac{3}{14}*100$

$F_f=21.4\%$

Generally the frequency for fish $F_h$ is mathematically given b

$F_h=\frac{n_h}{n}*100$

$F_h=\frac{3}{14}*100$

$F_h=21.4\%$

5 0

3 years ago