Consider randomly selecting a student at a certain university, and let A denote the event that the selected individual
has a Visa credit card and B be the analogous event for a MasterCard. Suppose that , P(A)= 0.6 and P(B)=0.4.
a. Could it be the case that P( A ∩ B )=0.5, why or why not?

b. From now on, suppose that P( A ∩ B )=0.3 What is the probability that student has one of these two types of cards?

c. What is the probability that the selected student has neither type of card?

d. Describe in terms of A and B the event that the select student has a visa card, but not a mastercard? and then calulate the probability of this event.

e. Calcuate th probability that the selected student has exactly one of the two types of cards?

Answer:

Step-by-step explanation:

[tex]$$a. Sam had pizza last night and Chris finished her homework.\\p\wedge q\\\\$b. Chris did not finish her homework and Pat watched the news this morning.$\\\neg q \wedge r\\\\$c. Sam did not have pizza last night or Chris did not finish her homework.$\\\neg p \vee \neg q\\\\[/tex]

[tex]$$d. Either Chris finished her homework or Pat watched the news this morning, but not both.$\\q\vee r\\\\$e. If Sam had pizza last night then Chris finished her homework.$\\p \rightarrow q\\\\$f. Pat watched the news this morning only if Sam had pizza last night.$\\p\leftrightarrow r\\\\[/tex]

[tex]$$g. Chris finished her homework if Sam did not have pizza last night.$\\\neg p \rightarrow q\\\\$h. It is not the case that if Sam had pizza last night, then Pat watched the news this morning.$\\\neg (p\rightarrow r)\\\\[/tex]

[tex]$$i. Sam did not have pizza last night and Chris finished her homework implies that Pat watched the news this morning.$\\(\neg p \wedge q) \Rightarrow r\\\\ $j. q\Rightarrow r$\\Chris finished her homework implies that Pat watched the news this morning.\\\\[/tex]

[tex]$$k. p \Rightarrow (q \wedge r)$\\Sam has pizza last night implies that Chris finished her homework and Pat watched the news this morning.$\\\\[/tex]

[tex]$$l. \neg p \Rightarrow (q \vee r)$\\Sam did not have pizza last night implies that Chris finished her homework or Pat watched the news this morning.$\\\\[/tex]

[tex]$$m. r \Rightarrow (p \vee q)$\\Pat watched the news this morning implies that Sam had pizza last night or Chris finished her homework$[/tex]

Final answer:

Logical formulas for the given statements about Sam, Chris, and Pat are provided using logical operators such as 'and', 'or', 'not', and 'implies'. The logical symbols ∧, ∨, ¬, and ⇒ are utilized to form the statements. Some formulas are also expressed in words to match their logical predictions with linguistic intuitions.

Explanation:

The logical formulas for the statements given about Sam, Chris, and Pat using p, q, and r are as follows:

a) Sam had pizza last night and Chris finished her homework: p ∧ q

b) Chris did not finish her homework and Pat watched the news this morning: ¬q ∧ r

c) Sam did not have pizza last night or Chris did not finish her homework: ¬p ∨ ¬q

d) Either Chris finished her homework or Pat watched the news this morning, but not both: q ⊕ r

e) If Sam had pizza last night then Chris finished her homework: p ⇒ q

f) Pat watched the news this morning only if Sam had pizza last night: r ⇒ p

g) Chris finished her homework if Sam did not have pizza last night: ¬p ⇒ q

h) It is not the case that if Sam had pizza last night, then Pat watched the news this morning: ¬(p ⇒ r)

i) Sam did not have pizza last night and Chris finished her homework implies that Pat watched the news this morning: (¬p ∧ q) ⇒ r

Now let's express some formulas in words:

j) q ⇒ r: If Chris finished her homework, then Pat watched the news this morning.

k) p ⇒ (q ∧ r): If Sam had pizza last night, then Chris finished her homework and Pat watched the news this morning.

l) ¬p ⇒ (q ∨ r): If Sam did not have pizza last night, then Chris finished her homework or Pat watched the news this morning.

m) r ⇒ (p ∨ q): If Pat watched the news this morning, then Sam had pizza last night or Chris finished her homework.

Answer:

a) False. b) False. c) No, it's Rational. d) pi=355/113 e) 5i (for Complex Set of Numbers)

Step-by-step explanation:

a) Since there is the empty set. And an axiom assures us the existence of this Set. "There is a set such no element belongs to it"

∅ has no elements.

b) A perfect number is a positive integer equals to the sum of proper divisions

The smallest is 6. Since the proper divisors of 6={3,2,1}. 6=1+2+3

28 is a perfect number, but not the smallest. It is perfect since

28 proper divisors={14,7,4,2} 28=1+2+3+4+5+6+7

c) No, An Irrational number cannot be written as a fraction a/b where "a" and "b" are integers. 1.5 is a rational one, since 3/2 =1.5

d)

[tex]\pi[/tex]=22/7 -1/791= 355/113

e)  Not for Real Numbers, since it is not defined for Real numbers. But for the set of Complex 5i

Answer: Ok, we have two numbers, and one of them is an odd integer, and the other is even.

Lets call M to the odd integer and N the even.

We know that a even integer can be written as 2k, where k is a random integer, and a odd integer can be written as 2j + 1, where j is also a random integer.

then M = 2k, N= 2j+1

then the product of M and N is: M*N = 2*k*(2*j + 1) = 2*(k*2*j + k)

is obvious to see that (k*2*J + k) is a integer, because k and j are integers.

then if we call g = ( k*2*J + k), we can write M*N=2g, and we already know that this is an even number. So M*N is a even integer.

Answer:

a) [tex]636_{7}[/tex] and  [tex]641_{7}[/tex]

b) [tex]11111_{2}[/tex] and  [tex]100001_{2}[/tex]

c) [tex]554_{6}[/tex] and  [tex]1000_{6}[/tex]

d)[tex]44_{5}[/tex] and  [tex]101_{5}[/tex]

e)[tex]3333_{4}[/tex] and  [tex]10001_{4}[/tex]

f) [tex]404_{6}[/tex] and  [tex]410_{6}[/tex]

Step-by-step explanation:

The logics followed in order to find them:

Base numbers work exactly the same as base 10 numbers (the ones we use on a daily basis). Take for example, in a decimal system, we have 10 digits available:

0,1,2,3,4,5,6,7,8,9.

when counting, when we get to the 9, we start over again, but writting a 1 to the left.

0,1,2,3,4,5,6,7,8,9,10,11,12....20,21,22,23....,90,91,92,93,94,95,96,97,98,99,100

when reaching 99, we have no more digits to use, so we start the count again and go from 99 to 100.

The same works with any other base number, take for example the base 7 numbet. When counting in base 7, we only have 7 digits available: (0,1,2,3,4,5,6) So when we re0_{7},ach the digit 6, we go immediately to 10, like this:

[tex]0_{7},1_{7},2_{7},3_{7},4_{7},5_{7},6_{7},10_{7},11_{7},12_{7},13_{7},14_{7},15_{7},16_{7},20_{7}...[/tex]

notice how it went from 6 to 10 and from 16 to 20. This is because in base 7, there is no such thing as digits from 7 to 9, so we don't have any other option to go directly to 20.

So, on part A) if we were working with decimal numbers, the  previous value for 640 would be 639, but notice that in base 7, there is no such thing as the digit 9, so the greatest digit we can use there would be 6, therefore, the previous value for the [tex]640_{7}[/tex] number would be [tex]636_{7}[/tex]. The next number would be [tex]641_{7}[/tex] because the 1 does exist for a base 7 system.

The same logics is followed for the rest of the problems.

One less than the given number is known as preceding number whereas one more than the given number is the succeeding number.

The logics followed in order to find them:

Base numbers work exactly the same as base 10 numbers (the ones we use on a daily basis). Take for example, in a decimal system, we have 10 digits available:

0,1,2,3,4,5,6,7,8,9.

When counting, when we get to the 9, we start over again, but writing a 1 to the left.

0,1,2,3,4,5,6,7,8,9,10,11,12....20,21,22,23....,90,91,92,93,94,95,96,97,98,99,10

When reaching 99, we have no more digits to use, so we start the count again and go from 99 to 100.

The same works with any other base number, take for example the base 7 number. When counting in base 7, we only have 7 digits available: (0,1,2,3,4,5,6) So when 0_{7},each the digit 6, we go immediately to 10, like this:

It went from 6 to 10 and from 16 to 20. This is because in base 7, there is no digits from 7 to 9, so we have directly to 20.

So, if working with decimal numbers, the previous value for 640 would be 639, but notice that in base 7,

There is no such thing as the digit 9, so the greatest digit we can use there 6,

Therefore, the previous value[tex]636_7[/tex] for the number would be [tex]636_7[/tex] . The next number would be [tex]632_7[/tex] because the 1 does exist for a base 7 system.

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Answer:

42 hamburgers

Step-by-step explanation:

Your answer is 42.

6 x 7 = 42

Answer:

It will take 0.1175 years or the number of employees at this company  to increase by 10%.

Step-by-step explanation:

We are given that the number of employees at a certain company is described by the  function [tex]P(t)= 300 (1.5)^{2t}[/tex]

Initial no. of employees = 300

Increase% = 10%

So, New no. of employees = [tex]300+\frac{10}{100} \times 300[/tex]

                                            = [tex]330[/tex]

Now we are supposed find how long does it take for the number of employees at this company to increase by 10%.

So, [tex]330= 300 (1.5)^{2t}[/tex]

[tex]\frac{330}{300}= (1.5)^{2t}[/tex]

[tex]1.1= (1.5)^{2t}[/tex]

[tex]t=0.1175[/tex]

So, it will take 0.1175 years or the number of employees at this company  to increase by 10%.

Answer: There are 104 silvery minnow population in the Rio Grande River.

Step-by-step explanation:

Since we have given that

Number of tagged silvery minnows = 54

Number of captured silvery minnows = 62

It includes 12 tagged silvery minnows.

So, Number of silvery minnows without tags = 62 -12 =50

So, Good estimate of the silvery minnow population in the Rio Grande River would be

[tex]54+50\\\\=104[/tex]

Hence, there are 104 silvery minnow population in the Rio Grande River.

Answer:

C. 800

Step-by-step explanation:

[tex]\sqrt{100} =10; \ \sqrt{400}=20; \ \sqrt{144}=12; \ \sqrt{800}=28.284271247461900976033774484194.[/tex]

To construct an 85% confidence interval with a maximum error of 0.06 pounds (with a standard deviation of 1.3 pounds), a sample size of 275 males over age 25 would be required.

The subject of this question revolves around the mathematical concept of

confidence intervals

by utilizing the formula to find the appropriate sample size. As per the question about meat consumption among males over age 25, the company wish to construct an

85% confidence interval

with a maximum error of 0.06 pounds, with the standard deviation being 1.3 pounds. To find the minimum sample size necessary for the research, we'll have to use the following formula:

n=z²*σ²/E²

, where z is the z-score representative of the desired confidence level, σ is the known standard deviation, and E is the maximum acceptable error. Looking up the Z table or Z score table, 85% confidence corresponds to a z score of approximately 1.44. Substituting these values into the formula gives us n=1.44² * 1.3² / 0.06². This results to roughly 274.94, but since we can't have a fraction of a person, we round up to the nearest whole number, giving us a minimum sample size of 275.

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Answer:

a) injective but not surjective. b) neither injective nor surjective.

Step-by-step explanation:

A function is injective if there aren't repeated images. To check if a function is injective we are going to suppose that for two values in the domain the image is equal, then we need to find that the two values are equal.

a) f : N → N with f(n) = n+2.

suppose that for n and m natural numbers, f(n)=f(m). Then

n+2 = m+2

n+2-2 = m

n = m.

Then, f(n) is injective.

Now, a function is surjective if every term m in the codomain there exists a pre-image of that element, that is to say, there exists an n such that f(n) = m. That is, the range of the function is equal to the codomain.

In this case, f(n) is not surjective. For example, if we would have that

n+2 = 1

n = 1-2

n = -1

but -1 is not a natural number, then for m=1 we don't have a pre image in f.

b) f: P({1, 2, 3}) → N with f(A) = |A|  (amount of elements in the subset A).

Now, for {1,2,3} we can have subsets of 0, 1, 2 or 3 elements. Then, the range of the function f is {1, 2, 3} (0 is not included because 0 is not natural). Then, the range is not all the natural numbers and therefore the function is not surjective.

Now, let's check if f is injective. Let {1} and {2} subsets of {1, 2, 3}. Then

f({1}) = |{1}| = 1.

f ({2}) = |{2}| = 1.

We have two different subsets with the same image, then f is not injective.

Answer:

[tex]v_{1}.v_{2} = 0.804[/tex]

In terms of the class, the dot product represents the weighed class average.

Step-by-step explanation:

The two vectors are:

-[tex]v_{1}:[/tex] The weight of each of the semester's exams.

[tex]v_{1} = (10%, 15%, 25%, 50%)[/tex]

In decimal:

[tex]v_{1} = (0.10, 0.15, 0.25, 0.50)[/tex]

-[tex]v_{2}:[/tex] The class average on each of the exams

In decimal:

[tex]v_{2} = (0.75, 0.91, 0.63, 0.87)[/tex]

-----------------------

Dot product:

Suppose there are two vectors, u and v

u = (a,b,c)

v = (d,e,f)

There dot product between the vectors u and v is:

u.v = (a,b,c).(d,e,f) = ad + be + cf

------------------

So

[tex]v_{1}.v_{2} = (0.10, 0.15, 0.25, 0.50).(0.75, 0.91, 0.63, 0.87) = 0.10*0.75 + 0.15*0.91 + 0.25*0.63 + 0.50*0.87 = 0.804[/tex]

[tex]v_{1}.v_{2} = 0.804[/tex]

In terms of the class, the dot product represents the weighed class average.

Answer:

The area of triangle is 25 square units.

Step-by-step explanation:

Given information: Vertices of the triangle are (2,1), (10,-1), and (-1,8).

Formula for area of a triangle:

[tex]A=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|[/tex]

The given vertices are (2,1), (10,-1), and (-1,8).

Using the above formula the area of triangle is

[tex]A=\frac{1}{2}|2(-1-8)+10(8-1)+(-1)(1-(-1))|[/tex]

[tex]A=\frac{1}{2}|2(-9)+10(7)+(-1)(1+1)|[/tex]

[tex]A=\frac{1}{2}|-18+70-2|[/tex]

On further simplification we get

[tex]A=\frac{1}{2}|50|[/tex]

[tex]A=\frac{1}{2}(50)[/tex]

[tex]A=25[/tex]

Therefore the area of triangle is 25 square units.

Answer:

The probability that a Tech High student fails math is 0.1006

Step-by-step explanation:

21% students are taking class with Mary.

Every year 4% of Mary’s students fail,

15% students are taking class with Tom

Every year 6% of Tom’s students fail.

64% students are taking class with Alex

Every year 13% of Alex's students fail.

Now we are supposed to find the probability that a Tech High student fails math.

[tex]P(\text{Tech High student fails math}) = P(\text{taking class with Mary})\times P(\text{Mary students fail})+P(\text{taking class with Tom}) \times P(\text{Tom students fail})+P(\text{taking class with Alex}) \times P(\text{Alex student fail})[/tex]

[tex]P(\text{Tech High student fails math}) = 0.21 \times 0.04+0.15 \times 0.06+0.64 \times 0.13[/tex]

[tex]P(\text{Tech High student fails math}) =0.1006[/tex]

Hence the probability that a Tech High student fails math is 0.1006

Answer:

The vector equation of the line is

l(t)= "-3,3,-5" + t "-2,-4,-4", where t is a parameter.

Answer:

The dimensions are 68 meters by 42 meters.

Step-by-step explanation:

Let the length be x

Let the width be y

Perimeter = 220 meter

So, [tex]2x+2y=220[/tex]

or [tex]x+y=110[/tex] or [tex]x=110-y[/tex]    ...........(1)

Area = 2856 square meter

So, [tex]xy=2856[/tex]     ............(2)

Substituting the value of x from (1) in (2)

[tex](110-y)y=2856[/tex]

=> [tex]110y-y^{2}=2856[/tex]

=> [tex]y^{2}-110y+2856=0[/tex]

Solving this equation, we get roots as y = 68 and y = 42

So, if we put y = 68, we get x = 42

If we put y = 42, we get x = 68

As length is longer than width, we will take length = 68 meters and width = 42 meters.

Hence, the dimensions are 68 meters by 42 meters.

Answer:

Intersection of collection of any closed set is a closed set.    

Step-by-step explanation:

Let F be a collection of arbitrary closed sets and let [tex]B_i[/tex] be closed set belonging to F.

We define a closed set as the set that contains its limit point or in other words it can be described that the complement or not of a closed set is an open set.

Thus, we can write R as

[tex]R =\bigcap\limits_{B_i \in F }^{} B_i[/tex]

Now, applying De-Morgan's Theorem, we have

[tex]R^c = (\bigcap\limits_{B_i \in F }^{} B_i)^c[/tex]

[tex]R^c = \bigcup\limits_{B_i \in F }^{} B_i^c[/tex]

Since we knew[tex]B_i[/tex] are closed set, thus, [tex]B_i^c[/tex] is an open set.

We also know that union of all open set is an open set.

Thus, [tex]R^c[/tex] is an open set.

Thus, R is a closed set.

Hence, the theorem.  

The intersection of any collection of closed sets is indeed closed. This can be proven using the properties of complements in topology and by employing a contradiction approach where the assumption that the intersection is not closed leads to a contradiction, hence proving it must be closed.

The question addresses whether the intersection of any collection of closed sets is closed. In topology, a closed set is one where its complement (the elements not in the set) is open. One way to approach the proof is considering De Morgan's Laws, which in topology state that the intersection of closed sets is the complement of the union of their complements, which are open sets. Since the union of open sets is open, it follows that the complement (the intersection of our original sets) is closed.

For example, consider two closed intervals on the real line, A and B. The intersection, A B, would be the set of all points that are in both A and B. The fact that both A and B contain their boundary points ensures that their intersection also contains boundary points, maintaining closedness.

Proof strategy involves contradiction: assume the intersection is not closed. This would imply that its complement is not open, violating the definition that complements of closed sets are open, thus leading to a contradiction and proving the proposition is true.

Answer:   0.0019

Step-by-step explanation:

Let x be the random variable that represents the number of years of employment at this department store.

Given : The number of years of employment at this department store is normally distributed,

Population mean : [tex]\mu=5.7[/tex]

Standard deviation : [tex]\sigma=1.8[/tex]

Z-score : [tex]z=\dfrac{x-\mu}{\sigma}[/tex]

Now, the z-value corresponding to 10 :  [tex]z=\dfrac{10-5.7}{1.8}\approx2.39[/tex]

P-value = [tex]P(x>10)=P(Z>2.89)=1-P(z\leq2.89)[/tex]

[tex]=1-0.9980737=0.0019263\approx0.0019\text{ (Rounded to nearest ten thousandth)}[/tex]

Hence, the probability that a cashier selected at random has worked at the store for over 10 years = 0.0019

Answer:

[tex]18234=2\times 3\times 3\times 1013[/tex]

Step-by-step explanation:

We are given that a number 18234

We have to find the prime factorization of the number

Prime factorization : The number written  is in  the product of prime numbers is called prime factorization.

In order to find the prime factorization we will find the factors of given number

[tex]18234=2\times 3\times 3\times 1013[/tex]

Hence, the prime factorization of [tex]18234=2\times 3\times 3\times 1013[/tex]

Answer:

17. C1 = 1    and    C2 = 0

18. C1 = 4/3    and    C2 = -1/3

Step-by-step explanation:

See it in the picture

Answer:

0.25

Step-by-step explanation:

We have a total of ten student, and three students are randomly selected (without replacement) to participate in a survey. So, the total number of subsets of size 3 is given by 10C3=120.

On the other hand A=Exactly 1 of the three selected is a freshman. We have that three students are freshman in the classroom, we can form 3C1 different subsets of size 1 with the three freshman; besides B=Exactly 2 of the three selected are juniors, and five are juniors in the classroom. We can form 5C2 different subsets of size 2 with the five juniors. By the multiplication rule the number of different subsets of size 3 with exactly 1 freshman and 2 juniors is given by

(3C1)(5C2)=(3)(10)=30 and

Pr(A∩B)=30/120=0.25

Final answer:

To find the probability that exactly one freshman and exactly two juniors are selected, we calculate the combination of selecting one from three freshmen and two from five juniors, then divide by the total combinations of selecting three students from ten. The probability is 0.25.

Explanation:

We need to find the probability Pr(A∩B) where:

For both events A and B to occur simultaneously, we must select one freshman and two juniors in our three student picks. The number of ways to choose one freshman out of three is C(3,1), and the number of ways to choose two juniors out of five is C(5,2). The total number of ways to choose any three students out of ten is C(10,3). Hence, the probability is:

Pr(A∩B) = (C(3,1) × C(5,2)) / C(10,3)

Calculating this gives:

Pr(A∩B) = (3 × 10) / 120 = 30 / 120

Pr(A∩B) = 0.25

Answer:

405.26%

Step-by-step explanation:

We have been given that in 1970 the male incarceration rate in the U.S. was approximately 190 inmates per 100,000 population. In 2008 the rate was 960 inmates per 100,000 population.

[tex]\text{Percent increase}=\frac{\text{Final amount}-\text{Initial amount}}{\text{Initial amount}}\times 100[/tex]

[tex]\text{Percent increase}=\frac{960-190}{190}\times 100[/tex]

[tex]\text{Percent increase}=\frac{770}{190}\times 100[/tex]

[tex]\text{Percent increase}=4.052631578947\times 100[/tex]

[tex]\text{Percent increase}=405.2631578947\%[/tex]

[tex]\text{Percent increase}\approx 405.26\%[/tex]

Therefore, the percent increase in the male incarceration rate during the given period is 405.26%.

Answer:

[tex]\text{Customer that bought only one paper}=33[/tex]

[tex]\text{Customer that bought something other than either of the two papers}=7[/tex]

Step-by-step explanation:

We have been given that at a newsstand, out of 46 customers, 27 bought the Daily News, 18 bought the Tribune, and 6 bought both papers.

[tex]\text{Customer that bought only Daily news}=27-6[/tex]

[tex]\text{Customer that bought only Daily news}=21[/tex]

[tex]\text{Customer that bought only Tribune}=18-6[/tex]

[tex]\text{Customer that bought only Tribune}=12[/tex]

The customer that bought only one paper would be the sum of customers, who bought only Daily news or Tribune.

[tex]\text{Customer that bought only one paper}=21+12[/tex]

[tex]\text{Customer that bought only one paper}=33[/tex]

Therefore, 33 customers bought only one paper.

[tex]\text{Customer that bought something other than either of the two papers}=46-(27+18-6)[/tex]

[tex]\text{Customer that bought something other than either of the two papers}=46-(45-6)[/tex]

[tex]\text{Customer that bought something other than either of the two papers}=46-39[/tex]

[tex]\text{Customer that bought something other than either of the two papers}=7[/tex]

Therefore, 7 customers bought something other than either of the two papers.

Answer:

  1, 5, 8

Step-by-step explanation:

1 is a vertical angle with angle 4, so is congruent.

5 is an alternate interior angle with angle 4, so is congruent.

8 is a corresponding angle with angle 4, so is congruent.

To determine the angles that are congruent to angle 4, we need to find angles with the same measure.

In geometry, two angles are congruent if they have the same measure. To determine which angles are congruent to angle 4, we need to find other angles that have the same measure as angle 4.

Let's assume angle 4 measures x degrees. In this case, any angle that also measures x degrees would be congruent to angle 4.

if angle 4 measures 60 degrees, any other angle that measures 60 degrees.

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Answer:

4

Step-by-step explanation:

8 *4 is the same as 4*8; commutative property

Answer:

[tex]4[/tex]

Step-by-step explanation:

8 x 4 is exactly the same as 4 x 8, both equaling 32.

No matter how insert 4 and 8, it will always be the same

[tex]x \times y = y \times x [/tex]

^^^

To estimate the result of 542,817 - 27,398, we can round to the nearest ten thousand and perform the subtraction, yielding an estimation of 510,000.

The student wants to estimate the result of 542,817 - 27,398. One way of doing this is by rounding to the nearest ten thousand. In this case, 542,817 becomes 540,000 and 27,398 becomes 30,000.

Now, perform the calculation: 540,000 - 30,000 = 510,000. Hence, the estimated result of 542,817 - 27,398 is roughly 510,000.

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Answer:

B. It is greater than the median.

Step-by-step explanation:

If n is an odd, prime integer and 10<n<19, then the true statements about the mean of all possible values of n are:

B. It is greater than the median.

Here, n can be 11 , 12 , 13 , 14 , 15 , 16 , 17 , 18.

But as n is odd , it can be 11, 13, 15 and 17.

And also it is given that n is prime, so we will cancel out 15.

Now we are left with 11, 13 and 17.

Their mean is = [tex]\frac{11+13+17}{3}[/tex] = 13.66

Median is 13, that is smaller than 13.66.

So, we can see that the mean is greater than median.

Answer:

The concentration of the drug is 5ug/uL

Step-by-step explanation:

The first step of the problem is the conversion of the quantities of the drug in mg and mL to ug and uL.

In a rule of three problem, the first step is identifying the measures and how they are related, if their relationship is direct of inverse.

When the relationship between the measures is direct, as the value of one measure increases, the value of the other measure is going to increase too.

When the relationship between the measures is inverse, as the value of one measure increases, the value of the other measure will decrease.

Unit conversion problems, like this one, is an example of a direct relationship between measures.

First step: Conversion of 125mg to ug

Each mg has 1,000ug. So:

1mg - 1,000ug

125mg - xug

x = 1,000*125

x = 125,000 ug

Second step: Conversion of 25 mL to uL

Each mL has 1,000uL. So:

1mL - 1,000uL

25mL - x uL

x = 25*1,000

x = 25,000uL

Concentration:

[tex]C = \frac{125,000 ug}{25,000uL} = 5ug/uL[/tex]

The concentration of the drug is 5ug/uL

The concentration of the drug in terms of  μg/μL is 5 μg/μL

The given parameters are:

125 mg of drug in each 25 mL

The concentration (k) of the drug is then calculated as:

[tex]k = \frac{125 mg}{25mL}[/tex]

Divide

[tex]k = 5\frac{mg}{mL}[/tex]

The units of the injection and the drug is in mill-.

So, the concentration can be rewritten as:

[tex]k = 5\frac{\mu g}{\mu L}[/tex]

Hence, the concentration of the drug in terms of  μg/μL is 5 μg/μL

Read more about concentration at:

https://brainly.com/question/14485544

Divide:

62.98 / 185.95 = 0.338693

Multiply the decimal by 100:

0.338693 x 100 = 33.8693%

Round the answer as needed.

Answer:

length of the third side may be either 3.4 m or 2.8 m.

Step-by-step explanation:

Measure of two sides of a right triangle are 1.4 m and the other side is 3.1 m.

then we have to find the measure of third side.

If the third side of the triangle is its hypotenuse then

(Third side)² = (1.4)² + (3.1)²

Third side = [tex]\sqrt{(1.4)^{2}+(3.1)^{2}}[/tex]

                 = [tex]\sqrt{1.96+9.61}[/tex]

                 = [tex]\sqrt{11.57}[/tex]

                 = 3.4 m

If the third side is one of the perpendicular sides of the triangle and 3.1 m is hypotenuse, then

(Third side)² + (1.4)² = (3.1)²

(Third side)²= (3.1)² - (1.4)²

Third side = [tex]\sqrt{(3.1)^{2}-(1.4)^{2}}[/tex]

                 = [tex]\sqrt{9.61-1.96}[/tex]

                 = [tex]\sqrt{7.65}[/tex]

                 = 2.76 m

                 ≈ 2.8 m

Therefore, length of the third side may be either 3.4 m or 2.8 m.