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2 years ago

Tyrone is comparing mobile phone plans. The plans are as follows:

Mathematics

2 answers:

Mashutka [201]2 years ago

8 0

Answer:

Tyrone is comparing mobile phone plans. The plans are as follows:

Plan A: $95 for unlimited talk time and text messaging

Plan B: 10 cents per minute for talk and 5 cents per text message

Plan C: $20 monthly fee plus 5 cents per minute for talk and 5 cents per text message

Plan D: $45 monthly fee with the first 500 minutes of talk free, then 10 cents per minute after that, along with unlimited text messages.

If Tyrone talks for 800 minutes each month and uses 1000 text messages, which plan is least expensive for him?

Plan A

Plan B

Plan C

Plan D

Step-by-step explanation:

The cheapest plan for Tyrone is the D: because it has unlimited free messages, and 500 minutes of free conversation, as 800 speaks, minus 500, there are 300 for which you would have to pay, 10 cent per minute, which would be $ 30 , plus the $ 45 that the plan costs, they would make a total of $ 75.

800 - 500 = 300 x 0.10 = 30 + 45 = 75 $

steposvetlana [31]2 years ago

6 0

Plan D is the least expensive for him

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Step-by-step explanation:

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2 years ago

Assume that x and y are both differentiable functions of t and find the required values of dy/dt and dx/dt. x2 + y2 = 25 (a) Fin

Artemon [7]

Answer:

(a) $\frac{dy}{dt}=-3\frac{3}{4}$

(b) $\frac{dx}{dt}=3\frac{3}{4}$

Step-by-step explanation:

$x^{2} +y^{2}=25$

Take $\frac{d}{dt}$ of of each term.

$\frac{d}{dt}(x^{2})+\frac{d}{dt}(y^{2})=\frac{d}{dt}(25)\\\\(\frac{d}{dx}(x^{2})*\frac{dx}{dt}) +(\frac{d}{dy}(y^{2})*\frac{dy}{dt})=\frac{d}{dt}(25)\\\\2x\frac{dx}{dt} +2y\frac{dy}{dt} = 0\\\\$

For Question a

$2y\frac{dy}{dt}=-2x\frac{dx}{dt}\\\\\frac{dy}{dt}=\frac{-2x\frac{dx}{dt}}{2y} \\\\\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$

Given that x = 3, y = 4, and dx/dt = 5.

$\frac{dy}{dt}=-\frac{3}{4}*5=-\frac{15}{4}\\ \\\frac{dy}{dt}=-3\frac{3}{4}$

For Question b

$2x\frac{dx}{dt}=-2y\frac{dy}{dt}\\\\\frac{dx}{dt}=\frac{-2y\frac{dy}{dt}}{2x} \\\\\frac{dx}{dt}=-\frac{y}{x}\frac{dy}{dt}$

Given that x = 4, y = 3, and dx/dt = -5.

$\frac{dx}{dt}=-\frac{3}{4}*-5=\frac{15}{4}\\ \\\frac{dx}{dt}=3\frac{3}{4}$

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2 years ago

If 8 identical blackboards are to be divided among 4 schools,how many divisions are possible? How many, if each school mustrecei

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Answer:

There are 165 ways to distribute the blackboards between the schools. If at least 1 blackboard goes to each school, then we only have 35 ways.

Step-by-step explanation:

Essentially, this is a problem of balls and sticks. The 8 identical blackboards can be represented as 8 balls, and you assign them to each school by using 3 sticks. Basically each school receives an amount of blackboards equivalent to the amount of balls between 2 sticks: The first school gets all the balls before the first stick, the second school gets all the balls between stick 1 and stick 2, the third school gets the balls between sticks 2 and 3 and the last school gets all remaining balls.

The problem reduces to take 11 consecutive spots which we will use to localize the balls and the sticks and select 3 places to put the sticks. The amount of ways to do this is ${11 \choose 3} = 165 .$ As a result, we have 165 ways to distribute the blackboards.

If each school needs at least 1 blackboard you can give 1 blackbooard to each of them first and distribute the remaining 4 the same way we did before. This time there will be 4 balls and 3 sticks, so we have to put 3 sticks in 7 spaces (if a school takes what it is between 2 sticks that doesnt have balls between, then that school only gets the first blackboard we assigned to it previously). The amount of ways to localize the sticks is ${7 \choose 3} = 35$ . Thus, there are only 35 ways to distribute the blackboards in this case.

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2 years ago

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