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stellarik [79]
2 years ago
7

Good luck with answering the question. :)

Mathematics
2 answers:
VladimirAG [237]2 years ago
5 0

Answer:

Step-by-step explanation:

JERRODHASCBASEBALLCARDS.LINDAHAS38BASEBALLCARDS,WHICHIS5MORETHANONE-THIRDTHENUMBERJERRODHAS.WHICHEQUATIONSHOWSANEQUALITYBETWEENTWODIFFERENTWAYSOFEXPRESSINGHOWMANYBASEBALLCARDSLINDAHAS?A.)C538B.)C538C.)3C538D.)1/3C538

✨✨HOPE IT HELPS✨✨

NemiM [27]2 years ago
5 0

Answer:

D. \frac{1}{3}C + 5 = 38

Step-by-step explanation:

Given: Jerrod has C cards.

Linda has 38 cards.

The problem tells us that the amount of cards Linda has is 5 more than one-third of Jerrod's cards. This can be represented as an equation:

38 = \frac{1}{3}C+5

This is the same equation as the one in answer choice D. So, answer choice D. is the correct answer.

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Amir has saved $1200, so he has enough money to purchase the computer.
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I took the test.

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work for a publishing company. The company wants to send two employees to a statistics conference. To be​ fair, the company deci
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Answer:

Step-by-step explanation:

Here is the complete question.

Dominique, Marco, Roberto , and John work for a publishing company. The company wants to send two employees to a statistics conference. To be fair, the company decides that the two individuals who get to attend will have their names drawn from a hat. This is like obtaining a simple random sample of size 2. (a) Determine the sample space of the experiment. That is, list all possible simple random samples of size n = 2. (b) What is the probability that Dominique and Marco attend the conference? (c) What is the probability that John attends the conference?  ​(d) What is the probability that John stays​ home?

Since there are four employees to select the two to send from. then for us to get the sample space for the experiment we need to merge all possible two employees and represent them as set.

Let D = Dominique, M = Marco, R = Roberto and J = John

a) The sample space for the experiment is the total number of possible outcomes that we can have. It is as given below

S = 4C2 = 4!/(4-2)!2! (Selecting 2 out of 4 employees)

Total sample space = 4!/2!2!

Total sample space = 4*3*2!/2!2

Total sample space = 12/2 = 6

The sample space are S = {DM, DR, DJ, MR, MJ, RJ}

b) Probability is the ratio of number of event to the sample space.

P = n(E)/n(S)

Given n(S) = 6

n(E) is the event of Dominique and Marco attending the conference.

E = {DM}

n(E) = 1

P(D and M) = 1/6

Hence  the probability that Dominique and Marco attend the conference is 1/6

c) For John to attend the conference, the event outcome will be given as;

E = {DJ, MJ, RJ}

n(E) = 3

n(S) = 6

Probability for John to attend the conference is 3/6 = 1/2

d) Probability that John stays at home = 1 - Prob (John attends the conference)

Probability that John stays at home = 1 - 1/2

Probability that John stays at home = 1/2

7 0
3 years ago
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