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Norma-Jean [14]
2 years ago
5

Make t the subject q= 1+3t/2-t

Mathematics
2 answers:
jok3333 [9.3K]2 years ago
5 0

Answer:

t = dfrac-1 + 2 q3 + q. Solve the equations: t = dfrac-1 + 2 q3 + q Answer: t = dfrac-1 + 2 q3 + q.

Step-by-step explanation:

Alinara [238K]2 years ago
3 0

Answer:

t = dfrac-1 + 2 q3 + q.

Step-by-step explanation:

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Solve K/2 + 5 - 2k = 6k
pantera1 [17]

Answer:

k = 2/3 is the answer!

Step-by-step explanation:

can you mark me branliest please

4 0
3 years ago
Suppose you have a set of measurements. What additional information would you need to describe the accuracy of the measurements?
Andru [333]

Answer:

D

Step-by-step explanation:

6 0
3 years ago
How do i solve this rational expression r-4/5r=1/5r+1
NNADVOKAT [17]

r - 4/5r  =  1/5r + 1

Subtract  1/5r from each side:    r - 4/5r - 1/5r  =  1

Combine all the 'r' terms on the left side: 

                                 r - 5/5r  =  1

                                 r  -  r  =  1

                                       0  =  1

There is no value of 'r' that can make ' 0 = 1 ' a true statement.
So the equation has <em>no solution</em>.


4 0
3 years ago
How do u do this equation
Sladkaya [172]

Answer:

8 kittens


Step-by-step explanation:

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7 0
3 years ago
Assume that the national credit card interest rate is 12.83 percent. A study of 62 college students finds that their average int
meriva

Answer:

b)  95 percent confidence interval for this single-sample t test

[11.64, 16.36]

Step-by-step explanation:

Explanation:-

Given data  a study of 62 college students finds that their average interest rate is 14 percent with a standard deviation of 9.3 percent.

Sample size 'n' =62

sample mean x⁻ = 14  

sample standard deviation 'S' = 9.3

<u>95 percent confidence interval for this single-sample t test</u>

The values are (x^{-} - t_{0.05} \frac{S}{\sqrt{n} } ,x^{-}+t_{0.05}\frac{S}{\sqrt{n} })  the <u>95 percent confidence interval for the population mean  'μ'</u>

Degrees of freedom γ=n-1=62-1=61

t₀.₀₅ = 1.9996 at 61 degrees of freedom

(14 - 1.9996\frac{9.3}{\sqrt{62} } ,14+1.9996\frac{9.3}{\sqrt{62} })

(14-2.361 , 14 + 2.361)

[(11.64 , 16.36]

<u>Conclusion:-</u>

95 percent confidence interval for this single-sample t test

[11.64, 16.36]

<u></u>

6 0
3 years ago
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