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AlexFokin [52]
3 years ago
14

The yield of a certain chemical reaction is believed to be related to temperature. A study collected the yield from 15 such reac

tions selected at random to test the belief and producedthe following results.
Variable Estimate
Intercept 1 15.5 2.96
Temp 1 0.05 0.017
R-sq = 0.73
Which of the following is the appropriate test statistic for the test?
a. t=0.05/0.73
b. t=15.5/0.73
c. t=0.05/15.5
d. t=15.5/0.05
Mathematics
1 answer:
Nookie1986 [14]3 years ago
5 0

Answer:

t = 0.05 / 0.017

Step-by-step explanation:

Given the table :

Variable ____ Estimate

Intercept 1 ____ 15.5 ____2.96

Temp 1 _______0.05 ___ 0.017

R-sq = 0.73

The test statistic for a regression test :

t = b1 / E

Where ;

b1 = slope ; E = Standard error of the slope

From the result table ;

b1 = 0.05 ;

E = 0.017

The test statistic, t = 0.05 / 0.017

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Question 2 of 15. Step 1 of 1
azamat

9514 1404 393

Answer:

  A = -68.75p +2906.25

Step-by-step explanation:

We can solve for m and b using the given values of A and p.

  1600 = m·19 +b

  2150 = m·11 +b

Subtracting the second equation from the first, we have ...

  -550 = 8m

  m = -550/8 = -68.75

Substituting this into the first equation gives ...

  1600 = 19·(-68.75) +b

  b = 1600 +1306.25 = 2906.25

The desired linear equation is ...

  A = -68.75p +2906.25

__

<em>Additional comment</em>

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3 0
3 years ago
16.4 roundest to the nearest tenth
faltersainse [42]

Answer:

Step-by-step explanation:

16.0 you’re welcome anything 5 or over gets rounded up 4 is less

Than 5 so it gets rounded down

7 0
3 years ago
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Help me pizzazz help
Gelneren [198K]
9. 0.705
10. 0.005
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7 0
2 years ago
Solve and graph the absolute value inequality: |4x + 1| ≤ 5.
Andrews [41]
You would have to do it two times, one being negative and one being positive.
4x + 1 < 5
     -1     -1
4x < 4
x < 1

4x - 1 < 5
    +1    +1
4x < 6
x < 1.5

So the answer would be x<1 and x<1.5

I hope this helps!
7 0
3 years ago
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Solve using identities​
jeka57 [31]

Answer:

Solution given

Cos\displaystyle \theta_{1}=\frac{13}{15}

consider Pythagorean theorem

\bold{Sin²\theta+Cos²\theta=1}

Subtracting Cos²\thetaboth side

\displaystyle Sin²\theta=1-Cos²\theta

doing square root on both side we get

Sin\theta=\sqrt{1-Cos²\theta}

Similarly

Sin\theta_{1}=\sqrt{1-Cos²\theta_{1}}

Substituting value of Cos\theta_{1}

we get

Sin\theta_{1}=\sqrt{1-(\frac{-13}{15})²}

Solving numerical

Sin\theta_{1}=\sqrt{1-(\frac{169}{225})}

Sin\theta_{1}=\sqrt{\frac{56}{225}}

Sin\theta_{1}=\frac{\sqrt{56}}{\sqrt{225}}

Sin\theta_{1}=\frac{\sqrt{2*2*14}}{\sqrt{15*15}}

Sin\theta_{1}=\frac{2\sqrt{14}}{15}

Since

In III quadrant sin angle is negative

<h3>\bold{Sin\theta_{1}=-\frac{2\sqrt{14}}{15}}</h3>
3 0
2 years ago
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