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8 days ago

A sample proportion of 0.44 is found. To determine the margin of error for this statistic, a simulation of 100 trials is run, ea

ch with a sample size of 100 and a point estimate of 0.44. The minimum sample proportion from the simulation is 0.32, and the maximum sample proportion from the simulation is 0.50. What is the margin of error of the population proportion using an estimate of the standard deviation?

Mathematics

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.580 80 repeating as fraction

Svet_ta [12734]

To start, we will shift the non-repeating segment of the decimal to the left side by dividing by a power of 10.

Then we will assign a variable to represent the value and also shift the repeating segment to the left.

Essentially, the concept here is that we can denote the repeating portion with a variable, let's say "x", and move forward with the calculation;

$\bf 0.580\overline{80}\implies \boxed{\cfrac{5.80\overline{80}}{10}}\qquad \textit{now, let's say }x= 5.80\overline{80}\\\\
-------------------------------$

$\bf thus\qquad \begin{array}{llll}
100\cdot x&=&580.80\overline{80}\\
&&575+5.80\overline{80}\\
&&575+x
\end{array}\qquad \implies 100x=575+x
\\\\\\
99x=575\implies x=\cfrac{575}{99}\qquad therefore\qquad \boxed{\cfrac{5.80\overline{80}}{10}}\implies \cfrac{\quad \frac{575}{99}\quad }{10}
\\\\\\
\cfrac{\quad \frac{575}{99}\quad }{\frac{10}{1}}\implies \cfrac{575}{99}\cdot \cfrac{1}{10}\implies \cfrac{575}{990}\implies \stackrel{simplified}{\cfrac{115}{198}}$

you can verify that using your calculator.

4 0

2 months ago

Read 2 more answers

Makayla is taking a math course and is working with the perimeter of rectangles. She knows the perimeter and length of her recta

lawyer [12517]

Answer:

$W=\dfrac{P}{2}-L$

Step-by-step explanation:

$P=2L+2W \qquad\text{given}\\\\P-2L=2W \qquad\text{subtract $2L$}\\\\\dfrac{P-2L}{2}=W \qquad\text{divide by 2}\\\\W=\dfrac{P}{2}-L \qquad\text{carry out the division}$

The given formula doesn't appear to match any of the answer choices.

W = (P/2) - L

5 0

2 months ago

Shane and Abha earned a team badge that required their team to collect no less than 20002000 cans for recycling. Abha collected

tester [12383]

Shane and Abha received a team badge for gathering at least 2000 cans for recycling.

This indicates that their collection must total a minimum of 2000 cans.

Abha managed to collect 178 more cans than Shane.

Let’s denote the number of cans Shane collected as S

So, Abha collected = S + 178

The inequality representing the number of cans collected by Shane can be expressed as:

$S+S+178\geq 2000$

= $2S+178\geq 2000$

$2S\geq 2000-178$

$2S\geq 1822$

$S\geq 911$

3 0

2 months ago

The name of the lesson is "Using Trigonometric Laws" The two non-parallel sides of an isosceles trapezoid are each 7 feet long.

AnnZ [12381]

The diagonal measures 20.68 ft; the shorter base is 17.21 ft. To understand this, we recognize that with base angles summing to 140°, each angle is 70°, given the isosceles trapezoid's properties. We can apply the Law of Cosines to find the diagonal's length, denoted as d. The length of the diagonal determines to be d = 20.68 ft. Determining the shorter base is somewhat more complex. By drawing an altitude from the upper vertices to the base, which measures 22 ft, we create two similar smaller right triangles requiring us to find the height and base measures related to each of the 70-degree angles and the hypotenuse of 7. By working through the calculations for height and base from one triangle, we subsequently find that 22 minus twice the base measure gets us to the shorter base's measure, arriving at x = 17.21 ft.

4 0

2 months ago

In a test of a printed circuit board using a random test pattern, an array of 16 bits is equally likely to be 0 or 1. Assume the

AnnZ [12381]

Answer:

A), B), and C) are clarified below.

Step-by-step explanation:

The inquiry involves using binary digits, employing probabilities that are equal for both conditions, by applying a random test pattern, where the formula is derived from p = q.

Simplifying gives us

P[k] = nCk / 2^n

A. Probability of all bits being 1s

16c16/2^16 = 1/65536

B. Probability of all bits being 0s

16c0/2^16 = 1/65536

C. The probability of having exactly 8 bits as 1s and the other 8 as 0s

16c8/2^16 = 12870/65536 => 0.1963 ≈ 19.63%

8 0

3 months ago