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3 years ago

Frank can mow lawns at a constant rate of 6/5 lawns per hour. what is the ratio of lawns to hours?

Mathematics

1 answer:

Igoryamba3 years ago

6 0

Answer:

6 : 5

Step-by-step explanation:

Rate = $\frac{6}{5}$

But Rate is also equal to number of lawns mowed in a certain amount of time.

Since ratio should always be in terms of simple whole numbers.

If 6/5 lawns are mowed in 1 hour, then,

6 lawns will be mowed in:

= $\frac{6 \times 1}{\frac{6}{5} }$

= 5 hours.

therefore, 6 laws will be mowed in 5 hours.

Ratio of lawns to hours in 6:5

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A hacker is trying to guess someone's password. The hacker knows (somehow) that the password is 10 characters long, and that eac

Gnoma [55]

The probability that the hacker guesses the password on his first try is:

P = 1/(62^10) = 1.19*10^(-18)

We know that the password is 10 characters long.

In each one of these, we can put.

One lower case letter (26 of these)

One upper case letter (26 of these)

one numerical digit (10 of these)

So, for every single digit, we have a total of:

26 + 26 + 10 = 62 options

Now we can find the total number of different passwords, which will be equal to the product between the number of options for each one of the characters.

We know that for each character we have 62 different options.

And we have 10 characters.

Then the product between the numbers of options is:

C = 62^10

Then if the hacker does a random guess, the probability that the random guess is correct is one over the total number of possible combinations.

P = 1/C = 1/(62^10)

The probability that the hacker guesses the password on his first try is:

P = 1/(62^10) = 1.19*10^(-18)

If you want to read more about probability, you can read:

brainly.com/question/427252

4 0

3 years ago

Lim <br> x → 1 <br> ln(x)/<br> sin(7πx)

mote1985 [20]

$\displaystyle\lim_{x\to1}\frac{\ln x}{\sin7\pi x}$

Note that both the numerator and denominator approach 0 as $x\to1$ , so we can try using L'Hopital's rule.

$\displaystyle\lim_{x\to1}\frac{\lnx }{\sin7\pi x}=\lim_{x\to1}\frac{\frac1x}{7\pi\cos7\pi x}=\lim_{x\to1}\frac1{7\pi x\cos7\pi x}$

The denominator is nonzero at $x=1$ , so the limit is equivalent to

$\displaystyle\frac1{\lim\limits_{x\to1}7\pi x\cos7\pi x}=\frac1{7\pi\cos7\pi}=-\frac1{7\pi}$

7 0

3 years ago

As Sales Manager for ISeeYou Productions, you are planning to review the prices you charge clients for television advertisement

aksik [14]

The development fee p can be modeled by the equation p=-400q+9600 where p is the price the company charges and q is the number of contracts they get. The total revenue R ISeeYou obtains by signing q contracts is found by using the function: $R(q)=-400q^{2}+9600q$ . The cost ISeeYou Productions can be found by using the following function: C(q)=800q+16000. The monthlhy profit of Express ISeeYou Productions if found with the next function: $P(q)=-400p^{2}+8800q-16000$ . ISeeYou should sign 2 or 20 contracts to break even.

a) In order to solve part a of the problem, we will suppose the price will have a linear relation with the amount of contracts they signed, so we need to find two points to build this function.

We know that when signing 6 contracts, the company charges $7,200, so our first point will be:

(6, 7200)

we also know that the company signs 14 contracts when they charge $4,000. So our second point is:

(14, 4000)

so we can use these points to find the slope of the line by using the

slope formula:

$m=\frac{p_{2}-p_{1}}{q_{2}-q_{1}}$

$m=\frac{4000-7200}{14-6}$

so the slope is:

m=-400

We can now use one of those points and the slope to find the function for the price, so we get:

$y-y_{1}=m(x-x_{1})$

$p-p_{1}=m(q-q_{1})$

if we used the point: (6, 7200) we get:

$p-7200=-400(q-6)$

which can now be simplified:

p=-400q+2400+7200

so the price function is:

p=-400q+9600

b) In order to find the revenue equation we need to multiply the price equation by the amount of contracts they signed in a month, so we get:

R(q)=pq

$R(q)=(-400q+9600)q$

$R(q)=-400q^{2}+9600q$

c) The cost can be found by adding the fixed cost and the variable cost, so we get:

C(q)=800q+16000

d) The profit is found by subtracting the montly cost from the monthly revenue, so we get:

P(q)=R(q)-C(q)

$P(q)=-400q^{2}+9600q-(800q+16000)$

so we can now simplify our function to get:

$P(q)=-400q^{2}+9600q-800q-16000$

$P(q)=-400q^{2}+8800q-16000$

e) In order to find the number of contracts for the company to break even, we need to find the q contracts that will give a profit of $0 so we set the function of the previous part of the problem equal to zero so we get:

$P(q)=-400q^{2}+8800q-16000$

$-400q^{2}+8800q-16000=0$