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Kay [80]
3 years ago
11

Suppose you add or subtract two quadratic trinomials that use the same variable. What are the possible classifications for the s

um or difference? Explain.
Please help me asap. T.T
Mathematics
1 answer:
Leno4ka [110]3 years ago
3 0

Answer:

quadratic monomial, quadratic trinomial, constant monomial, linear monomial, quadratic binomial, linear binomial.

Step-by-step explanation:

two quadratic trinomials each have a degree-2, degree-1, and degree-0 term. It is possible that the coefficients of all or some of the terms cancel each other while adding or subtracting the polynomials.

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Suppose the method of tree ring dating gave the following dates A.D. for an archaeological excavation site. Please show your wor
natita [175]

Answer:

a) \bar X=\frac{1245+1245+1321+1191+1295+1330+1239+1250+1228}{9}= 1260.444

b) s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And if we replace we got:

s = 45.580

c) For this case since the sampel size is n=9 <30 we can use the t distribution in order to find the confidence interval.

d) 1260.444-2.306 \frac{45.580}{\sqrt{9}}= 1225.408

1260.444+2.306 \frac{45.580}{\sqrt{9}}= 1295.480

Step-by-step explanation:

For this case we ave the following data:

1245 1245 1321 1191 1295 1330 1239 1250 1228

Part a

We can calculate the sample mean with the following formula:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

And if we replace we got:

\bar X= \frac{1245+1245+1321+1191+1295+1330+1239+1250+1228}{9}= 1260.444

Part b

For this case we can use the following formula in order to find the sample standard deviation:

s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And if we replace we got:

s = 45.580

Part c

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

For this case since the sampel size is n=9 <30 we can use the t distribution in order to find the confidence interval.

Part d

The confidence interval for the mean is given by the following formula:  

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)  

We need to find the degrees of freedom given by:

df = n-1 = 9-1=8

Since the Confidence is 0.95 or 95%, the value of \alpha=1-0.95=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,8)".And we see that t_{\alpha/2}=2.306

Now we have everything in order to replace into formula (1):

1260.444-2.306 \frac{45.580}{\sqrt{9}}= 1225.408

1260.444+2.306 \frac{45.580}{\sqrt{9}}= 1295.480

   

   

8 0
4 years ago
Draw the reflection of the following quadrilateral over the line m.
alexira [117]
These are the points and just join them

6 0
3 years ago
Given a function f(x)=2x^2+3, what is the average rate of change of f on the interval [2, 2+h]?
AveGali [126]

Answer:

4x+2h

Step-by-step explanation:

The average rate of change of a continuous function,

f

(

x

)

, on a closed interval  

[

a

,

b

]

is given by

f

(

b

)

−

f

(

a

)

b

−

a

So the average rate of change of the function  

f

(

x

)

=

2

x

2

+

1

on  

[

x

,

x

+

h

]

is:

A

r

o

c

=

f

(

x

+

h

)

−

f

(

x

)

(

x

+

h

)

−

(

x

)

     

=

f

(

x

+

h

)

−

f

(

x

)

h

 

 

 

 

 

...

.

.

[

1

]

     

=

2

(

x

+

h

)

2

+

1

−

(

2

x

2

+

1

)

h

     

=

2

(

x

2

+

2

x

h

+

h

2

)

+

1

−

2

x

2

−

1

h

     

=

2

x

2

+

4

x

h

+

2

h

2

−

2

x

2

h

     

=

4

x

h

+

2

h

2

h

     

=

4

x

+

2

h

Which is the required answer.

Additional Notes:

Note that this question is steered towards deriving the derivative  

f

'

(

x

)

from first principles, as the definition of the derivative is:

f

'

(

x

)

=

lim

h

→

0

 

f

(

x

+

h

)

−

f

(

x

)

h

This is the function we had in [1], so as we take the limit as  

h

→

0

we get the derivative  

f

'

(

x

)

for any  

x

, This:

f

'

(

x

)

=

lim

h

→

0

 

4

x

+

2

h

     

=

4

x

4 0
3 years ago
Analyzing Equations (Pls help)
scoundrel [369]

Answer:

I think 3 ,4 ,and 5 are to only correct ones

6 0
3 years ago
What are the values of x and y?
anygoal [31]
see the attached figure with letters

we have that 
cos(alfa) triangle ABD=12/y
and
cos(alfa) triangle ABC=y/28
then
12/y=y/28----------> y²=12*28=336----------> y=√336=4√21
y=4√21

<span>applying the Pythagorean theorem  triangle ABC
</span>(16+12)²=x²+y²----------? x²=28²-(4√21)²--------> x²=784-336
x=√448=8√7

the answer is 
x=8√7
y=4√21

4 0
3 years ago
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