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eduard
3 years ago
10

Frank solved an equation and got the result x = x. Sarah solved the same equation and got 12 = 12. Frank says that one of them i

s incorrect because you cannot get different results for the same equation. What would you say to Frank? If both results are indeed correct, explain how this happened.
Mathematics
1 answer:
jek_recluse [69]3 years ago
4 0
Frank is incorrect. 
They could have solved the same equations and got what seems like different answers, however both the answers of x = x, and 12 = 12 mean that the equation has an infinite amount of solutions, or that the answer is all real numbers.

This could have happened because Sarah and Frank may have approached the same equation in different ways, meaning that one simplified or solved the equation with a different, but still correct, strategy.

Hope this helps! :)

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Find f(-3) if f(x) = x 2.
vlabodo [156]

Answer:

f(-3) = 9

Step-by-step explanation:

Lets define f(a) first where a is any integer,

f(a) is found out by putting the integer in the function in place of the variable

So for the given question

f(x) = x^{2}

In order to find f(-3) we will put -3 in place of x

So putting -3,

f(-3) = (-3)^{2}

f(-3) = 9

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4 years ago
Read 2 more answers
4/5 divided by -3/5 any answers because right now I’m really stuck
MatroZZZ [7]

Answer:

-1 1/3 as a mixed number (if they ask for it in simplest form, choose this one)

-4/3 as an improper fraction

Step-by-step explanation:

1. Keep, change, flip

4/5 x -5/3

2. Cross cancel the fives

4/1 x -1/3

3. Simplify

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3 years ago
Hailey got a grade of 93 on her last math test. However, because she turned it in late her teacher deducted 15 points. What will
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Answer:

i believe 78

7 0
3 years ago
(Ross 5.15) If X is a normal random variable with parameters µ " 10 and σ 2 " 36, compute (a) PpX ą 5q (b) Pp4 ă X ă 16q (c) PpX
pochemuha

Answer:

(a) 0.7967

(b) 0.6826

(c) 0.3707

(d) 0.9525

(e) 0.1587

Step-by-step explanation:

The random variable <em>X</em> follows a Normal distribution with mean <em>μ</em> = 10 and  variance <em>σ</em>² = 36.

(a)

Compute the value of P (X > 5) as follows:

P(X>5)=P(\frac{x-\mu}{\sigma}>\frac{5-10}{\sqrt{36}})\\=P(Z>-0.833)\\=P(Z

Thus, the value of P (X > 5) is 0.7967.

(b)

Compute the value of P (4 < X < 16) as follows:

P(4

Thus, the value of P (4 < X < 16) is 0.6826.

(c)

Compute the value of P (X < 8) as follows:

P(X

Thus, the value of P (X < 8) is 0.3707.

(d)

Compute the value of P (X < 20) as follows:

P(X

Thus, the value of P (X < 20) is 0.9525.

(e)

Compute the value of P (X > 16) as follows:

P(X>16)=P(\frac{x-\mu}{\sigma}>\frac{16-10}{\sqrt{36}})\\=P(Z>1)\\=1-P(Z

Thus, the value of P (X > 16) is 0.1587.

**Use a <em>z</em>-table for the probabilities.

8 0
3 years ago
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