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Naily [24]
4 years ago
14

Which of these tables represents a linear function? A 2-column table with 4 rows. Column 1 is labeled x with entries 3, 4, 5, 6.

Column 2 is labeled y with entries 3, 4, 6, 7. A 2-column table with 4 rows. Column 1 is labeled x with entries 3, 4, 5, 6. Column 2 is labeled y with entries 6, 5, 4, 3. A 2-column table with 4 rows. Column 1 is labeled x with entries 3, 4, 5, 6. Column 2 is labeled y with entries 7, 6, 5, 3. A 2-column table with 4 rows. Column 1 is labeled x with entries 3, 4, 5, 6. Column 2 is labeled y with entries 2, 4, 5, 6.
Mathematics
2 answers:
yulyashka [42]4 years ago
7 0

Answer:

B

Step-by-step explanation:

    The other tables don't increase or decrease by only one number.  I'm pretty sure it's B. If it isn't, my apologies.

Sedbober [7]4 years ago
5 0

Answer:

the table :

(-3,8.5) .....x1 = -3 and y1 = 8.5

(-1,5.5) ....x2 = -1 and y2 = 5.5

slope = (y2 - y1) / (x2 - x1) = (5.5 - 8.5) / (-1 - (-3) = - 3 / (-1 + 3) = -3/2

coordinate plane :

(-3,0), (2,0)

notice how u have the same y values of 0....this means u have a horizontal line where no matter what x is, y will always be 0. A horizontal line has a slope of 0. And ur y intercept is (0,0)

the linear function that is in the table contains a negative slope and also has a steeper slope then the one on the coordinate plane <===

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The number of accidents per week at a hazardous intersection varies with mean 2.2 and standard deviation 1.4. This distribution
ankoles [38]

Answer:

a) Normal Distribution

b) 0.146

c) 0.070

Step-by-step explanation:

We are given the following information in the question:

The number of accidents per week at a hazardous intersection varies with mean 2.2 and standard deviation 1.4

a) \bar{x}: mean number of accidents per week at the intersection during a year (52 weeks)

According to central limit theorem, as the sample size becomes larger, the distribution of mean approaches a normal distribution.

Since we have a large sample, the approximate distribution of \bar{x} is a normal distribution with

\text{Mean} = 2.2\\\text{Standard Deviation} = \displaystyle\frac{\sigma}{\sqrt{n}} = \frac{1.4}{\sqrt{52}} = 0.19

b) P(mean is less than 2)

Formula:

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

P(x > 610)

P( \bar{x} < 2) = P( z < \displaystyle\frac{2 - 2.2}{0.19}) = P(z < -1.052)

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

P(\bar{x} < 2) = 0.146

c) P(fewer than 100 accidents at the intersection in a year)

P(x < 100)

P( \bar{x} < \frac{100}{52}) =P(\bar{x} < 1.92) =P(z < \displaystyle\frac{1.92 - 2.2}{0.19}) = P(z < -1.473)

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

P(x

7 0
3 years ago
A ball is thrown into the air. The path it takes is modeled by the equation: -3t+24t = h, where t is the time in seconds and h i
astra-53 [7]

Given:

The given equation is:

-3t^2+24t=h

Where, t is the time in seconds and h is the height of the ball above the ground, measured in feet.

To find:

The inequality to model when the height of the ball is at least 36 feet above the ground. Then find time taken by ball to reach at or above 36 feet.

Solution:

We have,

-3t^2+24t=h

The height of the ball is at least 36 feet above the ground. It means h\geq 36.

-3t^2+24t\geq 36

-3t^2+24t-36\geq 0

-3(t^2-8t+12)\geq 0

Splitting the middle term, we get

-3(t^2-6t-2t+12)\geq 0

-3(t(t-6)-2(t-6))\geq 0

-3(t-2)(t-6)\geq 0

The critical points are:

-3(t-2)(t-6)=0

t=2,6

These two points divide the number line in 3 intervals (-\infty,2),(2,6),(6,\infty).

Intervals      Check point           -3(t-2)(t-6)\geq 0           Result

(-\infty,2)               0                       (-)(-)(-)=(-)         False

(2,6)                    4                       (-)(+)(-)=+>0            True

(6,\infty)                  8                        (-)(+)(+)=(-)        False

The inequality is true for (2,6) and the sign of inequality is \geq. So, the ball is above 36 feet between 2 to 6 seconds.

6-2=4

Therefore, the required inequality is -3t^2+24t\geq 36 and the ball is 36 feet above for 4 seconds.

3 0
3 years ago
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