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3 years ago

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Mr. Brady asked his students how long it took them to complete their homework from the previous night. He presented the results

in the line plot shown. How many minutes did the greatest number of students take to do their homework? How many combined hours did those particular students spend on homework explain.

1 answer:

zvonat [6]3 years ago

7 0

Answer:

The question is lacking the accompanying line plot, I found a matching line plot online:

Answers:

a) The number of minutes te greatest number of students took to do their homework = 40 minutes

b) The number of combined hours the students spent on their homework $= 5\frac{1}{3}\ hours$

Step-by-step explanation:

From the diagram, the greatest number of students ( 8 students) take 2/3 hours to do their homework. Converting this time to minutes:

1 hour = 60 minutes

∴ 2/3 hours = 60 × 2/3 = 120/3 = 40 minutes

∴ 40 minutes is the time the greatest number of students take to do their homework

b) combined hours those students did their homework is calculated as follows:

total time = (number of hours used by 1 student) × number of students

$Total\ time = \frac{2}{3} \times 8 = \frac{16}{3}\\= 5\frac{1}{3} \ hours$

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The answer is b because the ratio is blondes to brunettes.

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4 years ago

A bookbook can be classified as either non dash fictionnon-fiction or fictionfiction. Suppose that 9494% of booksbooks are clas

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Answer:

a) 0.8836

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c) 0.2342

Step-by-step explanation:

Books classified as fiction = 94% or 0.94 probability

Books classified as non-fiction = 1 - 0.94 = 0.06

a) for two books, we get: 0.94 * 0.94 = 0.8836 Probability

b) The probability that all five books are fiction is: 0.94^5

This equals 0.7339 Probability

c) Probability that one books is non-fiction, while the other 4 are fiction is determined in the following way:

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This does not account for the order at which the non-fiction book shows up. Such as the non-fiction book being the first book picked, or in another case - the non-fiction book being the last picked.

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4 years ago

*15 points please help easy question*

Igoryamba

Answer:

x = $\sqrt{13}$

Step-by-step explanation:

The segment from the vertex to the base is a perpendicular bisector.

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3 years ago

A subset $S \subseteq \mathbb{R}$ is called open if for every $x \in S$, there exists a real number $\epsilon > 0$ such that

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Answer:

Step-by-step explanation:

REcall that given sets S,T if we want to prove that $S\subseteqT$ , then we need to prove that for all x that is in S, it is in T.

a) Let (a,b) be a non empty interval and $x\in (a,b)$ . Then a<x <b. Let $\varepsilon = \min{\min\{b-x, x-a\}}{2}$ Consider $y \in (x-\varepsilon,x+\varepsilon)$ , then

$y$ and

$y>x-\varepsilon>x-(x-a) = a$ .

Then $y\in (a,b)$ . Hence, (a,b) is open.

Consider the complement of [a,b] (i.e $(a,b)^c$ ).

Then, it is beyond the scope of this answer that

$(a,b)^c = (-\infty,a) \cup (b,\infty)$ .

Suppose that $x\in (a,b)^c$ and without loss of generality, suppose that x < a (The same technique applies when x>b). Take $\varepsilon = \frac{a-x}{2}$ and consider $y \in (x-\varepsilon,x+\varepsilon)$ . Then

$y$

Then y \in (-\infty,a). Applying the same argument when $x \in (b,\infty)$ we find that [a,b] is closed.

c) Let I be an arbitrary set of indexes and consider the family of open sets $\{A_i\}_{i\in I}. Let [tex]B = \bigcup_{i\in I}A_i$ . Let $x \in B$ . Then, by detinition there exists an index $i_0$ such that $x\in A_{i_0}$ . Since $A_{i_0}$ is open, there exists a positive epsilon such that $(x-\varepsilon,x+\varepsilon)\subseteq A_{i_0} \subseteq B$ . Hence, B is open.

d). Consider the following family of open intervals $A_n = (a-\frac{1}{n},b+\frac{1}{n})$ . Let $B = \bigcap_{n=1}^{\infty}A_n$ . It can be easily proven that

$B =[a,b]$ . Then, the intersection of open intervals doesn't need to be an open interval.

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Note that the definition of an open set is the following:

if for every $x \in S$ , there exists a real number $\epsilon > 0$ such that $(x-\epsilon,x \epsilon) \subseteq S$ . This means that if a set is not open, there exists an element x in the set S such that for a especific value of epsilon, the subset (x-epsilon, x + epsilon) is not a proper subset of S. Suppose that S is the empty set, and suppose that S is not open. This would imply, by the definition, that there exists an element in S that contradicts the definition of an open set. But, since S is the empty set, it is a contradiction that it has an element. Hence, it must be true that S (i.e the empty set) is open. Hence $\mathbb{R}$ is also closed, by definition. If you want to prove that this are the only sets that satisfy this property, you must prove that $\mathbb{R}$ is a connected set (this is a topic in topology)

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bixtya [17]

Answer:

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Step-by-step explanation:

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