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lianna [129]
3 years ago
13

I need help on number 8​

Mathematics
1 answer:
tester [92]3 years ago
5 0

Answer:

(-3,5) (-3,4)

5-4/-3--3=1/0

It is a vertical line so it's y∈R

Step-by-step explanation:

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X = 42 and y = - 23 are the two numbers
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Question is shown in the image
astraxan [27]
Your average rate of change for the interval is 3/2 or 1.5, so that should be your answer.
6 0
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Given the graph below, which of the following statements is true?
nirvana33 [79]

Answer:

D) The graph does not represent a one-to-one function because the y-values between 0 to 2 are paired with multiple x-values.

Step-by-step explanation:

Let's find the ordered pairs.

(-4, 4), (-3, 2), (-2, 0), (-1, 0.5), (0, 1), (1, 1.5), (2, 2), (3, 0)

The graph passed through above points.

In the x-coordinates -3, 2 gives the same output. Therefore, the given function is not one-to-one.

Therefore, the answer D)

The graph does not represent a one-to-one function because the y-values between 0 to 2 are paired with multiple x-values.

Hope you will understand the concept.

Thank you.

7 0
3 years ago
Solve using the quadratic formula: <br><img src="https://tex.z-dn.net/?f=%20%7B3x%7D%5E%7B2%7D%20%20%2B%202%20%3D%206" id="TexFo
satela [25.4K]

Answer:

x = 2/3, -2/3

Step-by-step explanation:

Solve the equation for  x  by finding  a ,  b , and  c  of the quadratic then apply the quadratic formula.

7 0
3 years ago
Suppose the sediment density (g/cm3 ) of a randomly selected specimen from a certain region is known to have a mean of 2.80 and
Irina18 [472]

Answer:

0.918 is the probability that the sample average sediment density is at most 3.00

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 2.80

Standard Deviation, σ = 0.85

Sample size,n = 35

We are given that the distribution of sediment density is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

Standard error due to sampling:

=\dfrac{\sigma}{\sqrt{n}} = \dfrac{0.85}{\sqrt{35}} = 0.1437

P(sample average sediment density is at most 3.00)

P( x \leq 3.00) = P( z \leq \displaystyle\frac{3.00 - 2.80}{0.1437}) = P(z \leq 1.3917)

Calculation the value from standard normal z table, we have,  

P(x \leq 3.00) = 0.918

0.918 is the probability that the sample average sediment density is at most 3.00

4 0
3 years ago
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