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Murljashka [212]
2 years ago
5

The total number of credit hours that remain for a student to earn a degree after s semesters have been completed successfully i

s given by 130 - 9s.
Which statement is correct?
A.
The remaining credit hours decrease by 130 for each semester completed.

B.
The remaining credit hours increase by 9 for each semester completed.

C.
The remaining credit hours decrease by 9 for each semester completed.

D.
The remaining credit hours increase by 130 for each semester completed.
Mathematics
1 answer:
Scilla [17]2 years ago
4 0

Answer:

C.

The remaining credit hours decrease by 9 for each semester completed.

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Taylor used 2 cans of blue paint and 5 cans of yellow paint to decorate a wall. Each can contains 200 millilitres of paint.
rewona [7]

Answer:

C.

Step-by-step explanation:

200 X 5 = 1000

200 X 2 = 200

1000-200= 800.

Taylor uses 800 ml more of yellow paint compared the blue paint.

7 0
3 years ago
Enter the simplified form of the complex fraction in the box. Assume no denominator equals zero.
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Answer:

\frac{(9-x)}{3x}

Step-by-step explanation:

we have

\frac{\frac{2}{x-3}-\frac{3}{x}}{\frac{3}{x-3}}

step 1

Solve the numerator of the quotient

\frac{2}{x-3}-\frac{3}{x}=\frac{2x-3(x-3)}{(x-3)x}=\frac{2x-3x+9}{(x-3)x}=\frac{9-x}{(x-3)x}

step 2

substitute in the original expression

\frac{\frac{9-x}{(x-3)x}}{\frac{3}{x-3}}

\frac{(x-3)(9-x)}{3x(x-3)}

step 3

simplify

\frac{(9-x)}{3x}

5 0
2 years ago
A random sample of 64 observations produced a mean value of 84 and standard deviation of 5.5. The 90% confidence interval for th
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Answer:  The 90% confidence interval for the population mean μ is between 82.85 and 85.15,

Step-by-step explanation:

When population standard deviation is not given ,The confidence interval population proportion is given by (\mu ):-

\overline{x}\pm t^*\dfrac{s}{\sqrt{n}}

, where n= Sample size.

s= Sample standard deviation

\overline{x} = sample mean

t* = Critical t-value (Two-tailed)

As per given , we have

\overline{x}=84

n= 64

Degree of freedom : df = n-1=63  

s= 5.5

Significance level : \alpha=1-0.90=0.1

Two-tailed T-value for df = 63 and  \alpha=1-0.90=0.1 would be

t_{\alpha/2,df}=t_{0.05,63}=1.669  (By t-distribution table)

i.e. t*= 1.669

The 90% confidence interval for the population mean μ would be

84\pm (1.669)\dfrac{5.5}{\sqrt{64}}

=84\pm (1.669)\dfrac{5.5}{8}

\approx84\pm 1.15

=(84-1.15,\ 84+1.15)=(82.85,\ 85.15)

∴ The 90% confidence interval for the population mean μ is between 82.85 and 85.15,

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