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MA_775_DIABLO [31]
2 years ago
9

Thanks for all the help

Mathematics
1 answer:
uysha [10]2 years ago
7 0

No problem

That's what brainly is here for

when i first found out ab it i was overjoyed


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¿que son los numeros naturales?​
Inga [223]
Huh? Someone translate this for me plz
8 0
2 years ago
Read 2 more answers
During a promotional weekend, a state fair gives a free admission to every 179th person that enters the fair. On Saturday, there
V125BC [204]

Answer:

89

Step-by-step explanation:

Given that,

During a promotional weekend, a state fair gives a free admission to every 179th person that enters the fair.

No of people attending the fair on Saturday is 8,633 amd No of people attending the fair on Sunday is 7,400.

We need to find the no of people that received a free admission over the two days.

Dividing 8,633 by 179 gives 48 as quotient and 41 as remainder. It means on Saturday 48 people entered for free.

Dividing 7,400 by 179 gives 41 as quotient and 61 as remainder. It means on Sunday 41 people entered for free.

Total no of people,

T = 48 + 41

T = 89

Hence, there are 89 people for free entries.

6 0
3 years ago
A box contains different colored paper clips. The probability of drawing two red paper clips from the box without replacement is
ivann1987 [24]

Answer: 5/14 which is choice B

================================================

How I got this answer:

Define the following events

A = event of picking a red paper clip on the first selection

B = event of picking a red paper clip on the second drawing

Replacement is not made.

Now onto the probabilities for each

P(A) = 2/5 = 0.4 is given to us as this is simply the probability of picking red on the first try

P(A and B) = probability of both events A and B happeing simultaneously = 1/7

P(B|A) = probability event B occurs, given event A has occured

P(B|A) = probability of selecting red on second selection, given first selection is red (no replacement)

P(B|A) = P(A and B)/P(A)

P(B|A) = (1/7) / (2/5)

P(B|A) = (1/7) * (5/2)

P(B|A) = (1*5)/(7*2)

P(B|A) = 5/14

So if event A happens, then the chances of event B happening is 5/14

------------------

A more concrete example:

If we had 15 paperclips, and 6 of them were red, then

P(A) = (# of red)/(# total) = 6/15 = 2/5

P(B|A) = (# of red left)/(# total left) = (6-1)/(15-1) = 5/14

P(A and B) = P(A)*P(B|A) = (2/5)*(5/14) = 10/70 = 1/7

7 0
3 years ago
Read 2 more answers
if the cost price of 10 tables is equal to the selling price of 16 tables , find out the gain or loss percentage​
Dovator [93]

Hi,

To solve this problem, Let us take the LCM of 10 and 16 which will come 80.

Now suppose the cost price of 10 tables =₹n CP of 80 tables will be ₹ 8n

According to the question, CP of 10 tables is equal to the SP of 16 tables, then

the SP of 16 tables will also be ₹ n.

So, SP of 80 tables will be ₹ 5n

So, Loss = CP-SP

→ 8n - 5n = ₹ 3n

Loss%= (3n×100)/8n

Loss%= 37.5%.

Hence the correct answer will be a <u>loss of 37.5%.</u>

4 0
2 years ago
Peter rolls 2 fair dice and adds the results from each.
VikaD [51]

ANSWER: Well I say that your probability is low. 1/12

7 0
2 years ago
Read 2 more answers
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