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3 years ago

−16t2+75 Solve for t=2

Mathematics

2 answers:

MrMuchimi3 years ago

6 0

Answer:

Step-by-step explanation:

-16(2) + 75

-32 + 75

43 (assuming that -16t2 means -16(2) )

If what you meant was -16t^2, then the answer would be this:

-16(2^2) + 75

-16(4) + 75

-64 + 75

SVETLANKA909090 [29]3 years ago

5 0

Answer:

=−16t^2+75

Step-by-step explanation:

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Last two!!(ty for all the help) :

malfutka [58]

With the b2+b- 12 your gonna want to do this
b2+b-12
b2+4b-3b-12
which is the sum product
after doing that you wanna common the factors from the two pairs. Which is b2+4b-3b-12 then you do you parentheses around b(b+4)-3(b+4) just like that after you do that then you rewrite it in factored form b(b+4)-3(b+4) you then want to rewrite it to this (b-3)(b+4) after that your solution will be (b-3)(b+4) for the first one.

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2 years ago

Read 2 more answers

1.8.2

VLD [36.1K]

Answer:

4.66

-0.97

-3/2

-20

3 0

2 years ago

1) Graph y = f(x) = 3(2)%. Adjust the equation to make the y-intercept (0, -3). Which response is not correct?

Karolina [17]

Using the concept of y-intercept, it is found that the response that is not correct is given by:

B) Replace x with -x.

<h3>What is the y-intercept of a function?</h3>

It is the value of the function when x = 0, hence it is the point (0, f(0)).

In this problem, the function is:

$f(x) = 3(2)^x$

The y-intercept is given by:

$f(0) = 3(2)^0 = 3$

If we just replace x with -x, we still have the same expression as above, hence the y-intercept does not change and option B is the answer to this question.

More can be learned about y-intercepts at brainly.com/question/24737967

7 0

2 years ago

The point B lies on the segment AC.

svet-max [94.6K]

Answer:

(9, -5)

Step-by-step explanation:

The vector AC has coordinates (27, -18), because:

21 - (-6) = 27

-13 - 5 = -18

If the ratio of AB to BC is 5 to 4, it means that

AB = [5/(5+4)]AC = (5/9)AC

Therefore, we have

AB = (5/9)AC = (5/9) (27, -18) = (15, -10)

Which means that:

xB - xA = 15 <=> xB - (-6) = 15 <=> xB + 6 = 15 <=> xB = 9

yB - yA = -10 <=> yB - 5 = -10 <=> yB = -10 + 5 <=> yB = -5

So B has coordinates (9, -5).

4 0

2 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

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