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dangina [55]
3 years ago
6

11. Evaluate: f(x) = x^2 - 2x + 1,

Mathematics
1 answer:
svetlana [45]3 years ago
6 0

Answer:

0

Step-by-step explanation:

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I need help PLEASE!!!!!! I need an answer for the 4th question!!!!
tester [92]

Answer:

The figures from POMN to ZYWX are similar because the scale factor is 1/3 from the biggest shape to the smallest shape. <u>!</u><u> </u><u>T</u><u>H</u><u>I</u><u>S</u><u> </u><u>I</u><u>S</u><u> </u><u>J</u><u>U</u><u>S</u><u>T</u><u> </u><u>A</u><u> </u><u>S</u><u>A</u><u>M</u><u>P</u><u>L</u><u>E</u><u>,</u><u> </u><u>Y</u><u>O</u><u>U</u><u> </u><u>C</u><u>A</u><u>N</u><u> </u><u>W</u><u>R</u><u>I</u><u>T</u><u>E</u><u> </u><u>I</u><u>T</u><u> </u><u>A</u><u>N</u><u>Y</u><u> </u><u>O</u><u>T</u><u>H</u><u>E</u><u>R</u><u> </u><u>W</u><u>A</u><u>Y</u><u>,</u><u> </u><u>I</u><u> </u><u>S</u><u>U</u><u>G</u><u>G</u><u>E</u><u>S</u><u>T</u><u> </u><u>Y</u><u>O</u><u>U</u><u> </u><u>T</u><u>R</u><u>Y</u><u> </u><u>A</u><u>N</u><u>D</u><u> </u><u>M</u><u>A</u><u>K</u><u>E</u><u> </u><u>I</u><u>T</u><u> </u><u>M</u><u>O</u><u>R</u><u>E</u><u> </u><u>I</u><u>M</u><u>P</u><u>R</u><u>O</u><u>V</u><u>E</u><u>D</u><u> </u><u>:</u><u>)</u><u>)</u><u>)</u><u> </u><u>!</u><u> </u>

3 0
3 years ago
What is the percent increase of 40 to 65
Darya [45]

Answer:

percentage increase calculator, formula

Step-by-step explanation:

calculations to find what percent increase from, 40 to 66, 65%

4 0
3 years ago
Andrea has a savings account with $80 that earns 2% interest per
lianna [129]

Answer:

After 10 years, she will has $96 in her bank.

Step-by-step explanation:

It is given that Andrea's saving account is $80 and earns 2% interest per year as a <em>S</em><em>i</em><em>m</em><em>p</em><em>l</em><em>e</em><em> </em><em>I</em><em>n</em><em>t</em><em>e</em><em>r</em><em>e</em><em>s</em><em>t</em><em> </em>(Not Compounded). Using simple interest formula, Interest = (P×R×T)/100 where <em>P</em> is the <em>principal</em>, <em>R</em> is the <em>interest rate</em> and <em>T</em> is <em>number of years</em><em> </em>:

I =  \frac{p \times r \times t}{100}

P = $80

R = 2%

T = 10 years

I =  \frac{80 \times 2 \times 10}{100}

I =  \frac{1600}{100}

I = 16

It is given that the interest amount is $16. So the total amount she has after 10 years in the bank is $96 :

interest amount = $16

principal = $80

total = $16 + $80

= $96

8 0
3 years ago
Suppose that bugs are present in 1% of all computer programs. A computer de-bugging program detects an actual bug with probabili
lawyer [7]

Answer:

(i) The probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

Step-by-step explanation:

Denote the events as follows:

<em>B</em> = bugs are present in a computer program.

<em>D</em> = a de-bugging program detects the bug.

The information provided is:

P(B) =0.01\\P(D|B)=0.99\\P(D|B^{c})=0.02

(i)

The probability that there is a bug in the program given that the de-bugging program has detected the bug is, P (B | D).

The Bayes' theorem states that the conditional probability of an event <em>E </em>given that another event <em>X</em> has already occurred is:

P(E|X)=\frac{P(X|E)P(E)}{P(X|E)P(E)+P(X|E^{c})P(E^{c})}

Use the Bayes' theorem to compute the value of P (B | D) as follows:

P(B|D)=\frac{P(D|B)P(B)}{P(D|B)P(B)+P(D|B^{c})P(B^{c})}=\frac{(0.99\times 0.01)}{(0.99\times 0.01)+(0.02\times (1-0.01))}=0.3333

Thus, the probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii)

The probability that a bug is actually present given that the de-bugging program claims that bug is present is:

P (B|D) = 0.3333

Now it is provided that two tests are performed on the program A.

Both the test are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is:

P (Bugs are actually present | Detects on both test) = P (B|D) × P (B|D)

                                                                                     =0.3333\times 0.3333\\=0.11108889\\\approx 0.1111

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii)

Now it is provided that three tests are performed on the program A.

All the three tests are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is:

P (Bugs are actually present | Detects on all 3 test)

= P (B|D) × P (B|D) × P (B|D)

=0.3333\times 0.3333\times 0.3333\\=0.037025927037\\\approx 0.037

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

4 0
3 years ago
Solve 6x + 2 = 5x + 17 for x if there is a solution.
labwork [276]

Answer: x=15


Step-by-step explanation:

To solve problems like this you need to simplify the equations on both sides and then isolate your variable x.


3 0
3 years ago
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