What is the lcm of 14 and 42

The population of the world on July 4, 2010 was more than 6,850,000,000. Write the number in scientific notation

Wewaii [24]

Answer:

6.85 × 10^9

Step-by-step explanation:

Move decimal twice to the right from were the first zero starts, count the number of zeros and that's your exponent.

5 0

2 years ago

If 'f(x)=x^2-18' and 'g(x)=(x-9)^(-1)(x+9)', find 'g(x)*f(x)

mezya [45]

The answer I would say would be B.

4 0

3 years ago

At a high school movie night, the refreshments stand sells popcorn and soft drinks. Of the 100 students who came to the movie, 6

Olenka [21]

Answer:

47% and 62%

Step-by-step explanation:

1. Drink

The probability that a student buys a drink is 0.47

Step-by-step explanation:

The probability that a student buys a drink will be given by;

( the number of students who bought a drink)/(the total number of students)

We are told that;

Of the 100 students who came to the movie, 62 bought popcorn and 47 bought a drink. Therefore, the required probability is;

47/100= 0.47

0.47 = 47%

2. Popcorn

For popcorn probability, it's basically the same.

The probabilty that a student buys popcorn is 0.62

The probability that a student buys popcorn will be given by;

( the number of students who bought popcorn)/(the total number of students)

So therefore,

62/100 = 0.62

0.62 = 62%

3 0

3 years ago

1. At a speed of 80km/h, it takes 10hrs to travel between Abuja and Lagos. How long will it takes to cover the same distance at

Softa [21]

Answer:

1. 8 hours 2. 12 pumps

Step-by-step explanation:

1. 80 ×10=800

there are 800 miles between abuja and lagos

800÷100=8

it will take 8 hours

2. with 3 pumps it takes 36 hours. you need to be 4 times quicker so you need 4 times more pumps.

4×3=12 pumps

3 0

3 years ago

Read 2 more answers

A small military base housing 1,000 troops, each of whom is susceptible to a certain virus infection. Assuming that during the c

slava [35]

Answer:

$I=\frac{1000}{exp^{0,806725*t-0.6906755}+1}$

Step-by-step explanation:

The rate of infection is jointly proportional to the number of infected troopers and the number of non-infected ones. It can be expressed as follows:

$\frac{dI}{dt}=a*I*(1000-I)$

Rearranging and integrating

$\frac{dI}{dt}=a*I*(1000-I)\\\\\frac{dI}{I*(1000-I)}=a*dt\\\\\int\frac{dI}{I*(1000-I)}=\int a*dt\\\\-\frac{ln(1000/I-1)}{1000}+C=a*t$

At the initial breakout (t=0) there was one trooper infected (I=1)

$-\frac{ln(1000/1-1)}{1000}+C=0\\\\-0,006906755+C=0\\\\C=0,006906755$

In two days (t=2) there were 5 troopers infected

$-\frac{ln(1000/5-1)}{1000}+0,006906755=a*2\\\\-0,005293305+0,006906755=2*a\\a = 0,00161345 / 2 = 0,000806725$

Rearranging, we can model the number of infected troops (I) as

$-\frac{ln(1000/I-1)}{1000}+0,006906755=0,000806725*t\\\\-\frac{ln(1000/I-1)}{1000}=0,000806725*t-0,006906755\\-ln(1000/I-1)=0,806725*t-0.6906755\\\\\frac{1000}{I}-1=exp^{0,806725*t-0.6906755} \\\\\frac{1000}{I}=exp^{0,806725*t-0.6906755}+1\\\\I=\frac{1000}{exp^{0,806725*t-0.6906755}+1}$

6 0

2 years ago