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andreev551 [17]
3 years ago
14

What is the 12th term in the addition pattern with formula 6n – 7?   A. 59   B. 65   C. 71   D. 72

Mathematics
2 answers:
Alina [70]3 years ago
8 0
To find this, plug the term you are trying to find into the variable, n.
Work:
6(12)-7
The answer is 65, or B
tatyana61 [14]3 years ago
5 0
Since we know what term nth will be, 12, we can solve this.
(6 x 12) - 7
72-7=65.
Your final answer is B.) 65.
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estimate a 15% tip on a dinner bill of $49.75 by rounding first bill amount to the nearest ten dollars
AlladinOne [14]

Answer:

60.00

Step-by-step explanation:

49.75 + 15%(7.46) = 57.21

7 0
3 years ago
What is (-10/-25)5+24x85=
Marina86 [1]

Answer:

2042

Step-by-step explanation:

Do you need a step by step?

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Olive oil has a density of 0.9 g/mL. What is the mass of 200 mL of olive oil?
lesantik [10]
Known, density= mass/ volume
The volume is given as 200ml
Set up an equation and cross-multiply
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3 0
3 years ago
. A square picture frame occupies an area of 112 ft2 . What is the length of each side of the picture in simplified radical form
saw5 [17]
Recall that a square is just a rectangle with all sides equal.

and the area of a rectangle is just length * width.

\bf \textit{area of a rectangle}\\\\
A=length\cdot width\quad 
\begin{cases}
length=x\\
width=x\\
A=112
\end{cases}\implies 112=x\cdot x\implies 112=x^2
\\\\\\
\sqrt{112}=x\implies \sqrt{16\cdot 7}=x\implies \sqrt{4^2\cdot 7}=x\implies 4\sqrt{7}=x
3 0
3 years ago
A student wanted to construct a 95% confidence interval for the mean age of students in her statistics class. She randomly selec
VMariaS [17]

Answer:

19.1-3.355\frac{1.5}{\sqrt{9}}=17.42    

19.1+3.355\frac{1.5}{\sqrt{9}}=20.78    

And the best option would be:

C. [17.42,20.78]

Step-by-step explanation:

Previous concepts

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".

\bar X=19.1 represent the sample mean

\mu population mean (variable of interest)

s=1.5 represent the sample standard deviation

n=9 represent the sample size  

Solution to the problem

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

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

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=9-1=8

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,8)".And we see that t_{\alpha/2}=

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

19.1-3.355\frac{1.5}{\sqrt{9}}=17.42    

19.1+3.355\frac{1.5}{\sqrt{9}}=20.78    

And the best option would be:

C. [17.42,20.78]

5 0
3 years ago
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