All categories

3 years ago

Nate solve the equation 3 times 5 equals 15 by writing and solving 15 divided by 5 is 3 x plane why Nate's method works

Mathematics

1 answer:

Vsevolod [243]3 years ago

8 0

Division is the opposite from multiplication

You might be interested in

{0.23,19%1/5} Order each set of numbers from least to greatest show work

Liono4ka [1.6K]

First you have to change them all to the same form, so you have to pick one out of them, so for instance you could pick percent which would be; 23%, 19%, and 20%, so from least to greatest it would be; 19%, 20%, and 23%. Hope this helps!

4 0

3 years ago

In a survey of 1,400 people who owned a certain type of car, 630 said they would buy that type of car again. What percent of the

Shalnov [3]

Answer:

45%

Step-by-step explanation:

(630 / 1400) * 100 = 45%

3 0

2 years ago

The number of requests for assistance received by a towing service is a Poisson process with rate θ = 4 per hour.a. Compute the

aliya0001 [1]

Answer:

a) 9.93% probability that exactly ten requests are received during a particular 2-hour period

b) 13.53% probability that they do not miss any calls for assistance

c) 2

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

Poisson process with rate θ = 4 per hour.

This means that $\mu = 4n$ , in which n is the number of hours.

a. Compute the probability that exactly ten requests are received during a particular 2-hour period.

n = 2, so $\mu = 4*2 = 8$

This is P(X = 10). So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 10) = \frac{e^{-8}*8^{10}}{(10)!} = 0.0993$

9.93% probability that exactly ten requests are received during a particular 2-hour period

b. If the operators of the towing service take a 30-min break for lunch, what is the probability that they do not miss any calls for assistance?

n = 0.5, so $\mu = 4*0.5 = 2$

This is P(X = 0). So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353$

13.53% probability that they do not miss any calls for assistance

c. How many calls would you expect during their break?

n = 0.5, so $\mu = 4*0.5 = 2$

4 0

3 years ago

Four less than the quotient of a number and 4 is 14. What is the number? Let n be the unknown number.

Maru [420]

So the equation is n/4-4=14
then you +4 to both sides
n/4-4=14 so now the equation is n/4=18 then you multiply 18 by 4 and get 72
+4 +4
then to check your work
72/4-4=14
so N=72
Hope I helped :)

8 0

2 years ago

Please help me this is too hard.

REY [17]

Answer:a

Step-by-step explanation:did it

4 0

2 years ago