All categories

2 years ago

Ben can type 153 words in 3 minutes. at this rate, how many words can he type in 10 minutes?

Mathematics

2 answers:

Alexxandr [17]2 years ago

7 0

Answer:

Ben can type 510 words in 10 minutes.

Step-by-step explanation:

The relation between the number of words typed and the minutes is direcltly proportional, that is, with Ben type more minutes, he will end up typing more words.

Thus, we can use the rule of three:

words | minutes

153 --- 3

x --- 10

Now, we can just cross multiply them:

$153~ x~10 = x~ times~3$

$1530 = 3x\\3x= 1530\\x=\frac{1530}{3} \\= 510$

Thus, Ben can type 510 words in 10 minutes.

SpyIntel [72]2 years ago

4 0

Answer:

Ben can type 510 words in 10 min.

Solution:

153/3=51 words per min

51*10 min= 510 words per 10 min

You might be interested in

What is 3z + 2 when z = 2?

I am Lyosha [343]

Answer:

b 8

Step-by-step explanation:

just replace the z with a 2 and solve

3(2)+2

4 0

3 years ago

Read 2 more answers

A turtle swims 1 7/8 miles in 1/6 hour.

FrozenT [24]

Answer:

Speed of turtle = 11.25 miles/hour

Step-by-step explanation:

1 7/8 miles / 1/6 hour

= 15/8 / 1/6

= 15/8 * 6

= 11.25 miles/hour

3 0

2 years ago

Read 2 more answers

Como se realiza eso?

amm1812

Answer:

english

Step-by-step explanation:

3 0

3 years ago

Y=12x<br> Do x and y vary directly ?

77julia77 [94]

Answer:

Yes, they vary directly

Step-by-step explanation:

y=kx; k is the constant (12 in this case) and as x increases y increases, so they vary directly

6 0

3 years ago

Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca

Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

$\frac{dQ}{dt} = -rt$

Which has the following solution:

$Q(t) = Q(0)e^{-rt}$

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that $Q(6) = (1-0.03)Q(0) = 0.97Q(0)$ . We use this to find r.

$Q(t) = Q(0)e^{-rt}$

$0.97Q(0) = Q(0)e^{-6r}$

$e^{-6r} = 0.97$

$\ln{e^{-6r}} = \ln{0.97}$

$-6r = \ln{0.97}$

$r = -\frac{\ln{0.97}}{6}$

$r = 0.0051$

$Q(t) = Q(0)e^{-0.0051t}$

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

$Q(t) = Q(0)e^{-0.0051t}$

$0.5Q(0) = Q(0)e^{-0.0051t}$

$e^{-0.0051t} = 0.5$

$\ln{e^{-0.0051t}} = \ln{0.5}$

$-0.0051t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.0051}$

$t = 135.9$

The half-life of the radioactive substance is 135.9 hours.

6 0

2 years ago