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barxatty [35]
3 years ago
10

Help. (Unprofessional people will be reported and so will wrong answers/people trying to be funny)

Mathematics
1 answer:
valkas [14]3 years ago
4 0
I think the answer is 10+3x8+5^2
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What is the average rate of change of the function f(x) on the interval 6 < x < 8
Rainbow [258]

Answer:

Step-by-step explanation:

The

average rate of change

of f(x) over an interval between 2 points (a ,f(a)) and (b ,f(b)) is the slope of the

secant line

connecting the 2 points.

To calculate the average rate of change between the 2 points use.

∣

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

f

(

b

)

−

f

(

a

)

b

−

a

a

a

∣

∣

∣

−−−−−−−−−−−−−−−

f

(

9

)

=

9

2

−

6

(

9

)

+

8

=

35

and

f

(

4

)

=

4

2

−

6

(

4

)

+

8

=

0

The average rate of change between (4 ,0) and (9 ,35) is

35

−

0

9

−

4

=

35

5

=

7

This means that the average of all the slopes of lines tangent to the graph of f(x) between (4 ,0) and (9 ,35) is 7.

8 0
3 years ago
B = 3x + 53°<br> Solve for x and then find the measure of
Ira Lisetskai [31]

Answer: sorry

Step-by-step explanation:

There isn't enough information for anyone to answer, sorry.

4 0
4 years ago
Peiling and John saved the same amount of money. After Peiling lent $28 to John, John had 5 times as much money as Peiling. How
patriot [66]

Answer:

140$

Step-by-step explanation:

so if peilling lent 28$ and then john has 5x as much that means john had 4x as much before getting the 28$ which would mean that john had 112$ dollars so now with that knowledge just add the 28$ to get 140$ or just do 5 x 28 to get 140

6 0
3 years ago
Philips Semiconductors is a leading European manufacturer of integrated circuits. Integrated circuits are produced on silicon wa
shepuryov [24]

Answer:

Step-by-step explanation:

1) The null hypothesis is,

H_0: The mean thickness of teh wafers for the five positions are equal

i.e, H_0:\mu_1=\mu_2=\mu_3=\mu_4=\mu_5

2)

The alternative hypothesis is,

H_1: There is an evidence of a difference in the mean thickness of the wafers for the five positions

3)

Let us consider the level of significance \alpha=0.01

from the Minitab outout

One-way ANOVA:C1 versus C2

source         DF            SS                 MS               F            P

C2                  4      1417.73          354.43        51.00       0.00

Error           145     1007.77              6.95

Total           14      2425.50

S = 2.636       R - S = 58.45%     R - Sq(adj) = 57.31%

Individual 95% CIs For Mean Based on Pooled StDev

level         N         Mean            StDev    -,----------,----------,----------,----------

1              30    240.53                2.62    (--,--)

2             30     243.73                2.79             (--,--)

3             30     246.07                2.90                           (--,--)

4             30     249.10                 2.66                                        (--,--)

5             30     247.07                 2.15                                  (--,--)

                                                               -,----------,----------,----------,----------

                                                        240.0   243.0   246.0   249.0

Pooled StDev = 2.64

4)

The test statistic is, F = 51

5)

The P-value is approximately 0

6)

Here, the P - value is less than the level of significance

\therefore \,P-value

So, we do not accept our null hypothesis H_0

7)

Therefore, we conclude that there is an evidence of a difference in the mean thickness of the wafers for the five positions at level of significance \alpha=0.05

b)

chek attachment

we observe that,

The mean thickness of the wafer for position 1 is significant with position 2,

position 18, position 19 and position 28.

The mean thickness of the wafer for position 2 is significant with position 18,

position 19 and position 28.

The mean thickness of the wafer for position 18 is significant with the

position 19.

But the mean thickness of the wafer for position18 is not significant with

position 28.

The mean thickness of the wafer for position 19 is significant with position 28.

7 0
3 years ago
The width of this rectangle is measured as 19.4mm correct to 1 decimal place.
Afina-wow [57]

Answer:

reducing 9.35  to  19.44  is rounded to 19.4  

Step-by-step explanation:

9.35  to  19.44  is rounded to 19.4  

19.35  = 19.4

19.36  = 19.4

19.37 = 19.4

19.38 = 19.4

19.39 = 19.4

19.40 = 19.4

19.41  = 19.4

19.42 = 19.4

19.43 = 19.4

19.44 = 19.4

Hence Lower bound of   of the rectangle  = 19.35 mm

6 0
3 years ago
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