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kkurt [141]
3 years ago
8

1. The table shows a function.

Mathematics
1 answer:
Nataliya [291]3 years ago
7 0

Answer: a) Linear function

 b) adding 8 each time x increases by 1

c) (0,13)


Step-by-step explanation:

For the given table, the rate of change is constant throughout the table.

The rate of change = \frac{change\ in\ y}{change\ in\ x}=\frac{21-13}{1-0}=8

Similarly we can check for every interval, the rate of change remains constant .

Thus, it is a linear function and the pattern we observe here is "adding 8 each time x increases by 1".

We know that the ordered pair of y intercept = (0,y)

In the table at x=0, y=13

hence, the y intercept of the given function is (0,13)

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Find GCF of 40 and 120​
vagabundo [1.1K]

Answer:

Hi there!

<h2>Find the GCF of 40 and 120</h2>

  • The GCF of 40 and 120 is 40

40 ÷ 40 = 1

120 ÷ 40 = 3

<h3>Hope its help!</h3>
5 0
3 years ago
Can someone help me solve this please + ( − )2
Pavlova-9 [17]
-2

We know that a positive and a negative will end in a negative result. (+) (-) = (-)
4 0
3 years ago
A local company is concerned about the number of days missed by its employees due to ill ness. A Random Sample of 10 employees i
sukhopar [10]

Answer:

The incentive program does not cuts down on the number of days missed by employees.

Step-by-step explanation:

The dependent t-test (also known as the paired t-test or paired samples t-test) compares the two means associated groups to conclude if there is a statistically significant difference amid these two means.

A paired <em>t</em>-test would be used to determine whether the incentive program cuts down on the number of days missed by employees.

The hypothesis for the test can be defined as follows:

<em>H₀</em>: The incentive program does not cuts down on the number of days missed by employees, i.e. <em>d</em> ≥ 0.

<em>Hₐ</em>: The incentive program cuts down on the number of days missed by employees, i.e. <em>d</em> < 0.

From the information provided the data computed is as follows:

 n=10\\\bar d=-1.1\\SD_{d}=2.99

Compute the test statistic value as follows:

 t=\frac{\bar d}{SD_{d}/\sqrt{n}}

   =\frac{-1.1}{2.99/\sqrt{10}}\\\\=-1.16

The test statistic value is -1.16.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.

Compute the p-value of the test as follows:

p-value=P(t_{n-1}

*Use a t-table.

The p-value of the test is 0.1379.

The p-value of the test is very large for all the commonly used significance level. The null hypothesis will not be rejected.

Thus, there is not enough evidence to support the claim.

Conclusion:

The incentive program does not cuts down on the number of days missed by employees

4 0
3 years ago
Suppose that an object moves along the y-axis so that its location is y=x2+3x at time x. (Here y is in meters and x is in second
vfiekz [6]

Answer:

a) 13 m/s

b) (15  + h) m/s

c) 15 m/s

Step-by-step explanation:

if the location is

y=x²+3*x

then the average velocity from 3 to 7 is

Δy/Δx=[y(7)-y(3)]/(7-3)=[7²+3*7- (3²+3*3)]/4= 13 m/s

then the average velocity from x=6 to to x=6+h

Δy/Δx=[y(6+h)-y(6)]/(6+h-6)=[(6+h)²+3*(6+h)- (6²+3*6)]/h= (2*6*h+3*h+h²)/h=2*6+3= (15 + h) m/s

the instantaneous velocity can be found taking the limit of Δy/Δx when h→0. Then

when h→0 , limit Δy/Δx= (15 + h) m/s = 15 m/s

then v= 15 m/s

also can be found taking the derivative of y in x=6

v=dy/dx=2*x+3

for x=6

v=dy/dx=2*6+3 = 12+3=15 m/s

7 0
3 years ago
Please help with this question!!
xxMikexx [17]

\text{Let}\ k:y=m_1x+b_1\ \text{and}\ l:y=m_2x+b_2\\\\l\perp k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\\text{We have the points J(-24, -4) and K(-4, 6)}.\\\\\text{The formula of a slope:}\\\\m=\dfrac{y_2-y_1}{x_2-x_1}\\\\\text{substitute:}\\\\m_1=\dfrac{6-(-4)}{-4-(-24)}=\dfrac{10}{20}=\dfrac{1}{2}\\\\\text{therefore}\ m_2=-\dfrac{1}{\frac{1}{2}}=-2\\\\\text{The formula of a midpoint:}\\\\\left(\dfrac{x_1+x_2}{2};\ \dfrac{y_1+y_2}{2}\right)\\\\\text{substitute the coordinates of the points J and K:}

x=\dfrac{-24+(-4)}{2}=\dfrac{-28}{2}=-14\\\\y=\dfrac{-4+6}{2}=\dfrac{2}{2}=1\\\\\text{midpoint}\ (-14,\ 1)\\\\\text{The point-slope form:}\\\\y-y_1=m(x-x_1)\\\\\text{substitute}\ m=-2,\ x_1=-14\ \text{and}\ y_1=1:\\\\y-1=-2(x-(-14))\\\\\boxed{y-1=-2(x+14)}

8 0
3 years ago
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