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Nina [5.8K]
3 years ago
15

What is 11.11 as a mixed number ?

Mathematics
2 answers:
zvonat [6]3 years ago
7 0
11.11=11+0.11=11+\frac{11}{100}=\boxed{11 \frac{11}{100}}
maks197457 [2]3 years ago
6 0
11.11 = 1111/100 and simplifies to: 11 11/100
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In a study of government financial aid for college​ students, it becomes necessary to estimate the percentage of​ full-time coll
ICE Princess25 [194]

Answer:

a) n = 1037.

b) n = 1026.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

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The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

99% confidence level

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​(a) Assume that nothing is known about the percentage to be estimated.

We need to find n when M = 0.04.

We dont know the percentage to be estimated, so we use \pi = 0.5, which is when we are going to need the largest sample size.

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.04 = 2.575\sqrt{\frac{0.5*0.5}{n}}

0.04\sqrt{n} = 2.575*0.5

(\sqrt{n}) = \frac{2.575*0.5}{0.04}

(\sqrt{n})^{2} = (\frac{2.575*0.5}{0.04})^{2}

n = 1036.03

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n = 1037.

(b) Assume prior studies have shown that about 55​% of​ full-time students earn​ bachelor's degrees in four years or less.

\pi = 0.55

So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.04 = 2.575\sqrt{\frac{0.55*0.45}{n}}

0.04\sqrt{n} = 2.575*\sqrt{0.55*0.45}

(\sqrt{n}) = \frac{2.575*\sqrt{0.55*0.45}}{0.04}

(\sqrt{n})^{2} = (\frac{2.575*\sqrt{0.55*0.45}}{0.04})^{2}

n = 1025.7

Rounding up

n = 1026.

6 0
3 years ago
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nikklg [1K]
The general vertex form is this:
v(x) = a (x-h)2 + k
where (h,k) is the coordinates of the of vertex.
and a indicates the widening or shrinking of the function compared to another parabolic function. If a become bigger, the graph becomes narrower. If a becomes negative, the graph is reflected over the x-axis.

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The graph is made narrower.
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From the choices we only have:
<span>The graph of f(x) = x2 is made narrower</span>
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PLEASE HELP ASAP !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! PLEASEEEEEEEEEE AND DO NOT LOOK IT UP ON ANY SOURCE THANK YOU AND ILY
AleksandrR [38]
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