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RideAnS [48]
3 years ago
15

Select the correct answer.

Mathematics
1 answer:
Kamila [148]3 years ago
7 0
That leaves C...because u have repeating x values (3 is ur repeating x value)...so C is not a function
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Which expression is equal to f(x) = g(x)? <br><br>2.1 + 54 <br>-2.r2 – 54r<br>-2x-54<br>or x+27
Sphinxa [80]

Answer:

Simplify radical expressions. … 1.04%. Squares and Square Roots. Find each square root. 26. √36. 27. √81. 28. … Step 2 (stage 1) Divide the square into four equal squares. … An exponential function has the form f (x) = a b x, where a ≠ 0, b ≠ 1, and b > 0. … There are 2.5 grams of fluorine-20 remaining after 44 seconds.

Step-by-step explanation:

4 0
3 years ago
Which is one one way to check 102 ÷ 6 = <br>17
sleet_krkn [62]

One way to check if 102 ÷ 6 = 17 is by focusing on the left side of the equation.


102 ÷ 6 can be converted into \frac{102}{6}


\frac{102}{6} can be simplified, lets start by dividing the numerator and denominator by 2.

\frac{102/2}{6/2} = \frac{51}{3}

\frac{51/3}{3/3} = \frac{17}{1}

\frac{17}{1} can be rewritten as 17.

5 0
4 years ago
Read 2 more answers
PLEASE HELP!!!! Complete the frequency table using the ages above and use the frequency table to make a graph
xxTIMURxx [149]
50-54= 2
55-59= 4
60-64= 3
65-69= 3

I hope this helps But what I did was I counted how many numbers were in between each of the two numbers.
7 0
4 years ago
Read 2 more answers
A company manufactures tires at a cost of $40 each. The following are probabilities of defective tires in a given production run
asambeis [7]

Answer:

The expected cost of the company for a 3000 tires batch is $120255

Step-by-step explanation:

Recall that given a probability of defective tires p, we can model the number of defective tires as a binomial random variable. For 3000 tires, if we have a probability p of having a defective tire, the expected number of defective tires  is 3000p.

Let X be the number of defective tires. We can use the total expectation theorem, as follows: if there are A_1,\dots, A_n events that partition the whole sample space, and we have a random variable X over the sample space, then

E(X) = p(A_1)E(X|A_1) + \dots + p(A_n)E(X|A_n).

So, in this case, we have the following

E(X) = P(p=0\%)E(X|p=0\%)+P(p=1\%)E(X|p=1\%)+P(p=2\%)E(X|p=2\%)+P(p=3\%)E(X|p=3\%) = 0.1\cdot 3000\cdot 0 + 0.3\cdot 3000\cdot 0.01+ 0.4\cdot 3000\cdot 0.02+ 0.2\cdot 3000\cdot 0.03 = 51.

Let Y be the number non defective tires. then X+Y = 3000. So Y = 3000-X. Then E(Y) = 3000-E(X). Then, E(Y) = 2949.

Finally, note that the cost of the batch would be 40Y+45X. Then

E(40Y+45X) = 40E(Y)+45E(X) = 40\cdot 2949+45\cdot 51=120255

3 0
3 years ago
Which of the following statements have the same result? Explain each step in solving each one.. . I f(3) when f(x) = 2x + 2. II
zhuklara [117]
F(x) = 2x + 2.....find f(3)
f(3) = 2(3) + 2
f(3) = 6 + 2
f(3) = 8

f(x) = (3x - 4)/5....find f^-1(4)
y = (3x - 4)/5...swap the variables
x = (3y - 4)/5 ...solve for y
5x = 3y - 4
5x + 4 = 3y
(5x + 4)/3 = y
f^-1(x) = (5x + 4)/3.....for f(4)
f(4) = (5(4) + 4)/3
f(4) = (20 + 4)/3
f(4) = 24/3
f(4) = 8

y + 10 = 2y + 1
y - 2y = 1 - 10
-y = -9
y = 9

so 1 and 2 have the same result...8

8 0
4 years ago
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