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2 years ago

Consider the statements below. Which are propositions? Mark all that apply.

Mathematics

1 answer:

allochka39001 [22]2 years ago

5 0

The following statements are considered to be propositions:

2 + 2 = 7

There are more men than women at BYU-Idaho.

2 + 2 = 4

Ron hates spiders.

<h3>What is deductive reasoning?</h3>

Deductive reasoning can be defined as a type of logical reasoning that typically involves drawing conclusions based on a given set of rules and conditions or from one or more premises (factual statements) that are assumed to be generally (universally) true.

<h3>What is a proposition?</h3>

A proposition can be defined as a type of statement (assertion) that is typically used to express an opinion or a judgement, with either a true or false answer.

This ultimately implies that, a proposition refers to a type of statement (assertion) that is either a true or false.

In this context, we can infer and logically deduce that the following statements are considered to be propositions:

2 + 2 = 7

There are more men than women at BYU-Idaho.

2 + 2 = 4

Ron hates spiders.

Read more on propositions here: brainly.com/question/24158168

#SPJ1

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if the sum of a number and nine is tripled the results is seven less than twice the number. find the number

aliina [53]

Here is the equation you need:

3(x + 9) = 2x - 7

You finish.

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3 years ago

Brian invests ?1900 into a savings account. The bank gives 3.5% compound interest for the first 2 years and 4.9% thereafter. How

Scorpion4ik [409]

let's check how much is it after 2 years firstly.

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &1900\\ r=rate\to 3.5\%\to \frac{3.5}{100}\dotfill &0.035\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{yearly, thus once} \end{array}\dotfill &1\\ t=years\dotfill &2 \end{cases} \\\\\\ A=1900\left(1+\frac{0.035}{1}\right)^{1\cdot 2}\implies A=1900(1.035)^2\implies A=2035.3275$

Brian invested the money for 6 years, so now let's check how much is that for the remaining 4 years.

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &2035.3275\\ r=rate\to 4.9\%\to \frac{4.9}{100}\dotfill &0.049\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{yearly, thus once} \end{array}\dotfill &1\\ t=years\dotfill &4 \end{cases}$

$\bf A=2035.3275\left(1+\frac{0.049}{1}\right)^{1\cdot 4}\implies A=2035.3275(1.049)^4 \\\\\\ A\approx 2464.54\implies \boxed{\stackrel{\textit{rounded up }}{A=2465}}$

4 0

3 years ago

The scatter plot shows the number of absences in a week for classes of different sizes. Trevor concluded that there is a positiv

QveST [7]

This is right the answer is A.

8 0

3 years ago

For every integer k from 1 to 10, inclusive the "k"th term of a certain sequence is given by (−1)(k+1)∗(12k). If T is the sum of

Katena32 [7]

Answer:

Option D. is the correct option.

Step-by-step explanation:

In this question expression that represents the kth term of a certain sequence is not written properly.

The expression is $(-1)^{k+1}(\frac{1}{2^{k}})$ .

We have to find the sum of first 10 terms of the infinite sequence represented by the expression given as $(-1)^{k+1}(\frac{1}{2^{k}})$ .

where k is from 1 to 10.

By the given expression sequence will be $\frac{1}{2},\frac{(-1)}{4},\frac{1}{8}.......$

In this sequence first term "a" = $\frac{1}{2}$

and common ratio in each successive term to the previous term is 'r' = $\frac{\frac{(-1)}{4}}{\frac{1}{2} }$

r = $-\frac{1}{2}$

Since the sequence is infinite and the formula to calculate the sum is represented by

$S=\frac{a}{1-r}$ [Here r is less than 1]

$S=\frac{\frac{1}{2} }{1+\frac{1}{2}}$

$S=\frac{\frac{1}{2}}{\frac{3}{2} }$

S = $\frac{1}{3}$

Now we are sure that the sum of infinite terms is $\frac{1}{3}$ .

Therefore, sum of 10 terms will not exceed $\frac{1}{3}$

Now sum of first two terms = $\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$

Now we are sure that sum of first 10 terms lie between $\frac{1}{4}$ and $\frac{1}{3}$

Since $\frac{1}{2}>\frac{1}{3}$

Therefore, Sum of first 10 terms will lie between $\frac{1}{4}$ and $\frac{1}{2}$ .

Option D will be the answer.

3 0

3 years ago

Please help me answer one of these!

babunello [35]

This seems to be referring to a particular construction of the perpendicular bisector of a segment which is not shown. Typically we set our compass needle on one endpoint of the segment and compass pencil on the other and draw the circle, and then swap endpoints and draw the other circle, then the line through the intersections of the circles is the perpendicular bisector.

There aren't any parallel lines involved in the above described construction, so I'll skip the first one.

2. Why do the circles have to be congruent ...

The perpendicular bisector is the set of points equidistant from the two endpoints of the segment. Constructing two circles of the same radius, centered on each endpoint, guarantees that the places they meet will be the same distance from both endpoints. If the radii were different the meets wouldn't be equidistant from the endpoints so wouldn't be on the perpendicular bisector.

3. ... circles of different sizes ...

[We just answered that. Let's do it again.]

Let's say we have a circle centered on each endpoint with different radii. Any point where the two circles meet will then be a different distance from one endpoint of the segment than from the other. Since the perpendicular bisector is the points that are the same distance from each endpoint, the intersection of circles with different radii isn't on it.

4. ... construct the perpendicular bisector ... a different way?

Maybe what I first described is different; there are no parallel lines.

8 0

3 years ago