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svp [43]
3 years ago
10

5. Which of the following is always true about

Mathematics
1 answer:
Lady bird [3.3K]3 years ago
5 0

Answer:

D. The slope of the graph will be zero.

Step-by-step explanation:

Slope of a line is given as \frac{rise}{run} = \frac{y_2 - y_1}{x_2 - x_1}.

For a horizontal line, the rise is always equal to zero since the line doesn't rise vertically. This means, y_2 - y_1 = 0.

No matter the value of the run, x_2 - x_1, when we divide zero by the value of the run, we would always have 0.

Therefore, the slope of an horizontal line would always be zero.

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6x - 7 y = -1 solve using elimination
Aleks [24]

Answer:

6x_7=-1

+7 +7

6x=6

÷6 ÷6

x=1

Step-by-step explanation:

shown above

5 0
3 years ago
Evaluate abc² if a= -3, b= 6, and c= -8​
Andreas93 [3]

Answer:

-1152

Step-by-step explanation:

abc² the expression can be rewritten using the given values for each letter

(-3)*6*(-8)^2 now we find the second power of (-8) by multiplying it with itself

(-8)*(-8) = 64

(-3)*6*64 = -1152

8 0
3 years ago
X^2+14x-15=0 solve by completing square<br>​
Softa [21]

Answer:

{x}^{2}  + 14x - 15 = 0 \\  {x}^{2}  - x + 15x -15 = 0 \\ x(x - 1) + 15(x - 1) = 0 \\ (x - 1)(x + 15) = 0 \\  \boxed{x = 1} \\  \boxed{x =  - 15}

<h3><u>x=1 and x=</u><u>-</u><u>15</u> is the right answer.</h3>
7 0
3 years ago
How do you solve c for a/b=c
asambeis [7]
The variable your solving for should be on the left side, so just flip it
c=a/b
5 0
3 years ago
Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65
erma4kov [3.2K]

Answer:

Probability that the sample average is at most 3.00 = 0.98030

Probability that the sample average is between 2.65 and 3.00 = 0.4803

Step-by-step explanation:

We are given that the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65 and standard deviation 0.85.

Also, a random sample of 25 specimens is selected.

Let X bar = Sample average sediment density

The z score probability distribution for sample average is given by;

               Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean = 2.65

           \sigma  = standard deviation = 0.85

            n = sample size = 25

(a) Probability that the sample average sediment density is at most 3.00 is given by = P( X bar <= 3.00)

    P(X bar <= 3) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{3-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z <= 2.06) = 0.98030

(b) Probability that sample average sediment density is between 2.65 and 3.00 is given by = P(2.65 < X bar < 3.00) = P(X bar < 3) - P(X bar <= 2.65)

P(X bar < 3) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{3-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z < 2.06) = 0.98030

 P(X bar <= 2.65) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{2.65-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z <= 0) = 0.5

Therefore, P(2.65 < X bar < 3)  = 0.98030 - 0.5 = 0.4803 .

                                                                             

8 0
4 years ago
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