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olga55 [171]
2 years ago
7

Could 18 in, 6 in, 13 in form a triangle

Mathematics
1 answer:
Tanya [424]2 years ago
7 0

Answer: Yes

Step-by-step explanation:

The reason on why it could form a triangle because 18 is the highest side in the triangle and if you have 2 of the smaller sides that are greater than the larger size then you could form a triangle and 13 and 6 equals to 19 so you could make a triangle hopes this helps and have a good day.

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Ignore the blue and 13. But I don’t understand dividing the polynomials. It says to complete the problem.
Virty [35]
OK the next step is to multiply 3r -1 by -7r
this gives us -21r^2 + 7r  and we write it below -21r^2 + 25r
and subtract 
This gives us  18r and we bring down the -6   to give
18r - 6

Finally we divide  18r by  -3r   to give  -6 then  add - 6 to the r^2 - 7r on the top 
then multiply -3r - 1 by -6 to give 18r - 6
subtraction then leaves nothing.

so the answer is r^2 - 7r - 6

4 0
3 years ago
How do you do this problem?
erastovalidia [21]

Step-by-step explanation:

∫₀² x f(x²) dx

If u = x², then du = 2x dx, and ½ du = x dx.

When x = 0, u = 0.  When x = 2, u = 4.

∫₀⁴ ½ f(u) du

½ (16)

8

7 0
3 years ago
If 75% is 60 pounds what is the full cost
otez555 [7]
OK,  60 pounds is weight.... cost is money.... what the what?
3 0
3 years ago
What is the probability that the card drawn is a diamond and a jack​
SpyIntel [72]

Answer:

1/26=4/104

Step-by-step explanation:

The probability of pulling a diamond if completely shuffled properly, is a 1/4. Because there are 4 suits. This is only if, there are no jokers.

And the probability of pulling a jack, is a 4/51. Because there are 4 jacks out of the 51 cards, but you already pulled out a diamond card, so you take 1 out.

Multiply them together, and you will get 4/104 = 1/26

There is a 1/26 chance of pulling out a diamond card, then a jack.

4 0
3 years ago
An e-mail filter is planned to separate valid e-mails from spam. The word free occurs in 60% of the spam messages and only 4% of
ANEK [815]

Answer:

(a) 0.152

(b) 0.789

(c) 0.906

Step-by-step explanation:

Let's denote the events as follows:

<em>F</em> = The word free occurs in an email

<em>S</em> = The email is spam

<em>V</em> = The email is valid.

The information provided to us are:

  • The probability of the word free occurring in a spam message is,             P(F|S)=0.60
  • The probability of the word free occurring in a valid message is,             P(F|V)=0.04
  • The probability of spam messages is,

        P(S)=0.20

First let's compute the probability of valid messages:

P (V) = 1 - P(S)\\=1-0.20\\=0.80

(a)

To compute the probability of messages that contains the word free use the rule of total probability.

The rule of total probability is:

P(A)=P(A|B)P(B)+P(A|B^{c})P(B^{c})

The probability that a message contains the word free is:

P(F)=P(F|S)P(S)+P(F|V)P(V)\\=(0.60*0.20)+(0.04*0.80)\\=0.152\\

The probability of a message containing the word free is 0.152

(b)

To compute the probability of messages that are spam given that they contain the word free use the Bayes' Theorem.

The Bayes' theorem is used to determine the probability of an event based on the fact that another event has already occurred. That is,

P(A|B)=\frac{P(B|A)P(A)}{P(B)}

The probability that a message is spam provided that it contains free is:

P(S|F)=\frac{P(F|S)P(S)}{P(F)}\\=\frac{0.60*0.20}{0.152} \\=0.78947\\

The probability that a message is spam provided that it contains free is approximately 0.789.

(c)

To compute the probability of messages that are valid given that they do not contain the word free use the Bayes' Theorem. That is,

P(A|B)=\frac{P(B|A)P(A)}{P(B)}

The probability that a message is valid provided that it does not contain free is:

P(V|F^{c})=\frac{P(F^{c}|V)P(V)}{P(F^{c})} \\=\frac{(1-P(F|V))P(V)}{1-P(F)}\\=\frac{(1-0.04)*0.80}{1-0.152} \\=0.90566

The probability that a message is valid provided that it does not contain free is approximately 0.906.

4 0
3 years ago
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