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Kazeer [188]
3 years ago
11

What is 12 x 11 i need this now

Mathematics
2 answers:
dimaraw [331]3 years ago
5 0

Answer:

that is 132

Step-by-step explanation:

12 over 11 then multiply

Aleksandr [31]3 years ago
4 0

Answer:

bru its 132

Step-by-step explanation:

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Gregg is a wedding photographer. He charges $500 for his basic package and $150 for each hour he is at the wedding. Which equati
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The equation that best fits the situation is w = 150 * 5 (or the other way around). Since the numbers that would be used are already stated, the equation would be as simple as that. We know that the operation would be multiplication because of the term used "150 for EACH hour" which usually indicates the operation is multiplication.

Hope I helped and good luck!
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Write the ratio as a ratio of a whole numbers using fractional notation 6 1/2 to 15 1/6
motikmotik
\bf 6\frac{1}{2}:15\frac{1}{6}\qquad 
\begin{cases}
6\frac{1}{2}\implies \cfrac{6\cdot 2+1}{2}\\\\
15\frac{1}{6}\implies \cfrac{15\cdot 6+1}{6}
\end{cases}\implies \cfrac{ \frac{6\cdot 2+1}{2}}{\frac{15\cdot 6+1}{6}}\\\\
-----------------------------\\\\
\cfrac{\frac{a}{b}}{\frac{c}{{{ d}}}}\implies \cfrac{a}{b}\cdot \cfrac{{{ d}}}{c}\qquad thus
\\\\\\
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5 0
4 years ago
(Hypothetical.) Suppose five measurements of the volume of a liquid yield the following data (in milliliters).
Digiron [165]

Answer:

The 95% confidence interval for the measurements is [48.106, 53.494].

Step-by-step explanation:

The average M of this sample is

M=\dfrac{1}{5}\sum_{i=1}^5x_i=\dfrac{1}{5}(52+48+49+52+53)\\\\\\M=\dfrac{254}{5}=50.800

The standard deviation s of this sample is

s=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^5(x_i-M)^2}

s=\sqrt{\dfrac{1}{4}\cdot [(52-50.8)^2+(48-50.8)^2+(49-50.8)^2+(52-50.8)^2+(53-50.8)^2]}

s=\sqrt{\dfrac{1}{4}\cdot [(1.44)+(7.84)+(3.24)+(1.44)+(4.84)]}=\sqrt{\dfrac{18.8}{4}}=\sqrt{4.7}\\\\\\s=2.168

The degrees of freedom are

df=n-1=5-1=4

Then, the critical value of t for a 95% CI and 4 degrees of freedom is t=2.776.

The margin of error of the CI is:

E=t\cdot s/\sqrt{n}=2.776\cdot 2.17/\sqrt{5}=6.024/2.236=2.694

Then, the lower and upper bounds of the CI are:

LL=50.800-2.694=48.106\\\\UL=50.800+2.694=53.494

The 95% confidence interval for the measurements is [48.106, 53.494].

4 0
4 years ago
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