Researchers have created every possible "knockout" line in yeast. Each line has exactly one gene deleted and all the other genes present (Steinmetz et al. 2002). The growth rate - how fast the number of cells increases per hour - of each of these yeast lines has also been measured, expressed as a multiple of the growth rate of the wild type that has all the genes present. In other words, a growth rate greater than 1 means that a given knockout line grows faster than the wild type, whereas a growth rate less than 1 means it grows more slowly. Below is the growth rate of a random sample of knockout lines:

0.8, 0.98, 0.72, 1, 0.82, 0.63, 0.63, 0.75, 1.02, 0.97, 0.86

What is the standard deviation of growth rate this sample of yeast lines (answer to 3 decimals)?

Answer:

"The Hulk is not green AND the Iron Man is not red"

Step-by-step explanation:

DeMorgan's laws state that the negation of an statement whose structure is "p OR q" is "not p AND not q", and similarly, that the negation of an statement whose structure is "p AND q" is "not p OR not q". The statement we want to negate in our case is "The Hulk is green OR the Iron Man is red". This is an statement whose structure is of the type "p OR q", where p would be "The Hulk is green", and q would be "the Iron Man is red". So according to DeMorgan's laws, its negation should be the statement "not p AND not q". To put them in common english, not p would be "The Hulk is NOT green", and not q would be "The Iron Man is NOT red". So the statement "not p AND not q" is simply "The Hulk is not green AND the Iron Man is not red".

To negate the statement 'the Hulk is green or Iron Man is red' using DeMorgan's Laws, you rephrase it as 'the Hulk is not green and Iron Man is not red'. The formal representation switches the disjunction to a conjunction and applies negation to each individual proposition.

To write the negation of the statement 'the Hulk is green or Iron Man is red' using DeMorgan's Laws, we first need to understand what these laws state. DeMorgan's Laws tell us how to move a negation across a conjunction (and) or a disjunction (or). According to DeMorgan's Laws, the negation of a conjunction is the disjunction of the negations, and the negation of a disjunction is the conjunction of the negations.

In formal logic, the original statement can be represented as (G \/ R), where G represents 'the Hulk is green' and R represents 'Iron Man is red'. Applying DeMorgan's Law to negate this statement would involve negating the entire proposition and then switching the 'or' to 'and'. Thus, the negation of the statement would be \u00AC(G \/ R) which translates to \u00ACG \u2227 \u00ACR using DeMorgan's Laws. This means 'the Hulk is not green and Iron Man is not red'.

Using DeMorgan's Laws is a way to express logical equivalence between propositions. By utilizing these transformation rules, one can simplify complex logical expressions or restate them in a different form without changing their meaning, making them powerful tools in formal logic and mathematics.

Answer:   0.2

Step-by-step explanation:

Let F denotes the event that graduate senior will get offer from the first company and S denotes the event that graduate senior will get offer from the second company .

Then, we have : [tex]P(F)=0.6[/tex]       [tex]P(S)=0.5[/tex]

The probability that at least one will reject him ( neither first nor second ) = [tex]P(F'\cap S')=0.8[/tex]

Now, [tex]P(F\cup S)=1-P(F'\cap S')=1-0.8=0.2[/tex]

Hence, the probability that he gets at least one offer ( either first or second)= 0.2

Answer:

11.70%

Step-by-step explanation:

Given;

Interest paid = $44

Principle amount = $1,153

Time = 119 days = [tex]\frac{\textup{119}}{\textup{365}}\textup{days}[/tex]  = 0.326 years

Now,

the interest is calculated as:

interest = Principle × Rate of interest × Time

thus,

$44 = $1,153 × Rate of interest × 0.326

or

Rate of interest = 0.1170

or

in percentage = Rate × 100 = 0.1170 × 100 = 11.70%

Answer:

a. [tex]\bar X[/tex] is distributed [tex]N(84;4)[/tex]

b. [tex]P(\bar X \geq 87.6) = 0.03593[/tex]

c. [tex]P(\bar X \leq 79.2) = 0.00820[/tex]

d. [tex]P(\79.2 \leq \bar X \leq 87.6) = 0.95587[/tex]

Step-by-step explanation:

a.

The central limit theorem states that, for large n, the sampling distribution of the sample mean is approximately normal with mean [tex]\µ[/tex] and variance [tex]\frac{\sigma^2}{n}[/tex], then, the sample mean is distributed as a normal random variable with means [tex]\mu_{\bar X}=\mu=84[/tex] and variance [tex]\sigma^2_{\bar X}=\frac{\sigma^2}{n}=\frac{16^2}{64}=4[/tex].

b.

[tex]P(\bar X \geq 87.6) = 0.03593[/tex]

c.

[tex]P(\bar X \leq 79.2) = 0.00820[/tex]

d.

[tex]P(\79.2 \leq \bar X \leq 87.6) = 0.95587[/tex]

Answer:

  80

Step-by-step explanation:

The three highest exam scores have an average of 80. To maintain an average of 80, the final exam score must be at least 80.

  (83 +67 +90)/3 = 240/3 = 80

_____

Essentially, the final exam score counts as two tests, and the lowest test score is thrown out.

The lowest score the student can obtain is 80

The student would take 5 tests. The total marks obtainable is 500( 100 x 5). The student wants to score at least 80 in the course. So she needs to have at least a total score of 400(80 x 5).

In this calculation 52, which is the lowest score would be excluded. This is because the professor would replace this score with her exam score. The student has the scores for 3 tests already.

Total scores of the student for the 3 tests = 83 + 67 + 90 = 240

Sum of scores on the remaining two tests = 400 - 240 = 160

Average score on the remaining two tests = 160/ 2 =  80

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Answer:

b) 3

c) 5

d) 4

e) 3

f) 3 days to 5 days

Step-by-step explanation:

Total number of conferences  = 84

Part a) We can make a pie chart to represent the distribution of number of days. The area of each sector represents the percentage of duration compared to all events. For example, the conferences that lasted 2 days occurred for 4.7619% or approximately 5% of the time.

Part b) The first quartile

The number of days for which the conferences last as per the given data in ascending order are listed below in the second image. In order to find First Quartile, we first need to find the Median. Since, total number of quantities is 84, the median will be the average of middle two values (42nd and 43rd). These values are made bold in the second image below. So median of the data is 4. Median divides the data into halves, both of which are colored separately in the image. First Quartile is the middle value of the first half. Since, values in first are 42, the middle value will be the average of central two values (21st and 22nd). These are made bold and colored red. Thus, First Quartile of the data is 3.

Part c) The third quartile

Similar to last step, the 3rd Quartile is the middle value of second half of the data. The second half of the data is colored purple. Number of values in this half are again 42, so middle value will be the average of digits at 21st and 22nd place, which comes out to be 5. Therefore, 3rd quartile is 5.

Part d) The 65th percentile.

65th percentile means, 65% of the data values are below this point. 65% of the 84 is 54.6. This means 54.6 values should be below the 65th percentile. Thus, 65th percentile occurs at 55th position. Counting from the start, the digit at 55th position comes out to be 4. So, the 65th percentile is 4.

Part e) The 40th percentile

40th percentile means, 40% of the data values are below this point. 40% of the 84 is 33.6. This means 33.6 values should be below the 40th percentile. Thus, 40th percentile occurs at 34th position. Counting from the start, the digit at 34th position comes out to be 3. So, the 40th percentile is 3.

Part f) Middle 50% of the conferences

Since, First Quartile is 25th percentile and Third Quartile is 75th percentile, in between these two Quartiles 50% of the data is present. The difference of first quartile and third quartile is known as IQR, Inter Quartile range and is a common measure of spread in stats.

Therefore, for the given data:

The middle 50% of the conferences last from 3 days to 5 days

a. The organized chart is attached below.

b. The First Quartile of the data is 3.

c. The 3rd quartile is 5.

d. The 65th percentile is 4.

e.  The 40th percentile is 3

f. The middle 50% of the conferences last from 3 days to 5 days

a. We can create a pie chart to show how many days each event lasted. Each slice of the pie represents the percentage of time that event took compared to all the events.

For example, if conferences that lasted 2 days happened for about 5% of the time, their slice in the pie chart would be around that size.

b. According to the information provided, the conferences last for a certain number of days. To find the first quartile, we first need to find the median. Since there are a total of 84 quantities, the median would be the average of the middle two values (the 42nd and 43rd values). These values are highlighted in bold in the image. So, the median of the data is 4. The median divides the data into two halves, which are shown in different colours in the image.

The First Quartile is the middle value of the first half. Since the values in the first are 42, the middle value will be the average of the central two values (21st and 22nd). These are made bold and coloured red.

Thus, the First Quartile of the data is 3.

c. Similar to the last step, the 3rd Quartile is the middle value of the second half of the data. The second half of the data is coloured purple. The number of values in this half is again 42, so the middle value will be the average of digits at 21st and 22nd place, which comes out to be 5.

Therefore, the 3rd quartile is 5.

d. 65th percentile means, 65% of the data values are below this point. 65% of the 84 is 54.6.

This means 54.6 values should be below the 65th percentile. Thus, the 65th percentile occurs at the 55th position.

Counting from the start, the digit at the 55th position comes out to be 4. So, the 65th percentile is 4.

e. 40th percentile means, 40% of the data values are below this point. 40% of the 84 is 33.6. This means 33.6 values should be below the 40th percentile.

Thus, the 40th percentile occurs at the 34th position.

Counting from the start, the digit at the 34th position comes out to be 3.

So, the 40th percentile is 3.

f. Based on the given data, we can say that the middle 50% of the conferences last between 3 days and 5 days.

As the first Quartile is the 25th percentile and the Third Quartile is the 75th percentile, in between these two Quartiles 50% of the data is present. The difference of the first quartile and the third quartile is known as IQR, Inter Quartile range and is a common measure of spread in stats.

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Answer:

6.75g

Step-by-step explanation:

The first step is converting everything to grams, by rules of three. Then we add.

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: 0.0025kg to g

Each kg has 1000g, so:

1kg - 1000g

0.0025kg - xg

x = 1000*0.0025 = 2.5g

0.0025kg = 2.5g

Second step:  1750 mg

Each g has 1000mg, so:

1g - 1000mg

xg - 1750mg

1000x = 1750

[tex]x = \frac{1750}{1000}[/tex]

x = 1.75g

1750 mg = 1.75g

2.25g - OK

Third step: 825,000 μg to g

Each g has 1,000,000 ug, so:

1g - 1,000,000 ug

xg - 825,000 ug

1,000,000x = 825,000

[tex]x = \frac{825,000}{1,000,000}[/tex]

x = 0.25g

825,000 μg = 0.25g

Final step: adding everyting

0.25g + 2.25g + 1.75g + 2.50g = 6.75g

Taking into account the change of units,  the sum results in 7.325 g.

The rule of three is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them.

That is, what is intended with it is to find the fourth term of a proportion knowing the other three.  

If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other) , the direct rule of three must be applied.

To solve a direct rule of three, the following formula must be followed, being a, b and c known data and x the variable to be calculated:

a ⇒ b

c ⇒ x

So: [tex]x=\frac{cxb}{a}[/tex]

The direct rule of three is the rule applied in this case where there is a change of units.

To perform in this case the conversion of units, you must first know that 1 kg= 1000 g, 1 mg= 0.001 g and 1 μg=1×10⁻⁶ g. So:

1 kg ⇒ 1000 g

0.0025 kg ⇒ x

So:  [tex]x=\frac{0.0025 kgx1000 g}{1 kg}[/tex]

Solving:

x= 2.5 g

So, 0.0025 kg is equal to 2.5 g.

1 mg ⇒ 0.001 g

1750 mg ⇒ x

So:  [tex]x=\frac{1750 mgx0.001 g}{1 mg}[/tex]

Solving:

x= 1.75 g

So, 1750 mg is equal to 1.75 g.

1 μg ⇒ 1×10⁻⁶ g

825000 μg ⇒ x

So:  [tex]x=\frac{825000 ugx1x10^{-6} g}{1 ug}[/tex]

Solving:

x= 0.825 g

So, 825000 μg is equal to 0.825 g.

Now you add all the values ​​in the same unit of measure:

2.5 g + 1.75 g + 2.25 g + 0.825 g= 7.325 g

Finally, the sum results in 7.325 g.

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Answer and Step-by-step explanation:

A = {l,m,n,o,p}

B = {o,p,q,r}

C = {r,s,t,u}

(A ∪ B) ∩ C  

(A ∪ B) = {l,m,n,o,p,q,r}

C = {r,s,t,u}

(A ∪ B) ∩ C  = {r}

A ∩ (C ∪ B)

(C ∪ B)  = {o,p,q,r,s,t,u}

A = {l,m,n,o,p}

A ∩ (C ∪ B)  = {o,p}

(A ∩ B) ∪ C

(A ∩ B) = {o,p}

C = {r,s,t,u}

(A ∩ B) ∪ C = {o,p,r,s,t,u}

At a newsstand, out of 46 customers, 27 bought the Daily News, 18 bought the Tribune, and 6 bought both papers. Use a Venn diagram to answer the following questions:

only daily news: 21 (27-6)

only tribune: 12 (18-6)

Total newspaper: 39 (21+12+6)

Other than newspapers: 7 (46 - 39)

How many customers bought only one paper? 21+12 = 33

How many customers bought something other than either of the two papers? 7

equal, equivalent, or neither.

{d,o,g}: {c,a,t}  equivalent

{run} : {{r,u,n}  equal

{t,o,p} :{p,o,t}  equal

Answer:

The patient would receive 1.05mg of the drug weekly.

Step-by-step explanation:

First step: How many mcg of the drug would the patient receive daily?

The problem states that he takes three doses of 50-mcg a day. So

1 dose - 50mcg

3 doses - x mcg

x = 50*3

x = 150 mcg.

He takes 150mcg of the drug a day.

Second step: How many mcg of the drug would the patient receive weekly?

A week has 7 days. He takes 150mcg of the drug a day. So:

1 day - 150mcg

7 days - x mcg

x = 150*7

x = 1050mcg

He takes 1050mcg of the drug a week.

Final step: Conversion of 1050 mcg to mg

Each mg has 1000 mcg. How many mg are there in 1050 mcg? So

1mg - 1000 mcg

xmg - 1050mcg

1000x = 1050

[tex]x = \frac{1050}{1000}[/tex]

x = 1.05mg

The patient would receive 1.05mg of the drug weekly.

The patient receives 0.15 milligrams of pergolide mesylate daily and thus 1.05 milligrams weekly.

To calculate the weekly dosage of pergolide mesylate in milligrams, we first need to understand the daily dosage. The patient takes three 50-mcg tablets daily, which is 150 mcg daily. Knowing that 1 mg (milligram) is equal to 1000 mcg (micrograms), we can convert the daily dose to milligrams by dividing by 1000 which is 0.15 mg daily. To find the weekly dosage, we multiply this daily total by 7 (since there are 7 days in a week), which gives us 1.05 milligrams weekly.

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Step-by-step explanation:

When you divide an integer number by n, you get a remainder of either 0, 1, 2, ..., n-1 (for example 5 divided by 2 leaves a remainder of 1, or 13 divided by 5 leaves a remainder of 3, or 16 divided by 2 leaves a remainder of 0, and so on).

So there are n different remainders we could get when dividing an integer number by n. If we are given n+1 numbers, they each leave a certain remainder when divided by n. Since there are only n possible remainders, and we have n+1 numbers, by the pigeonhole principle we know there must be at least 2 numbers that leave the same remainder when divided by n. Call them numbers a and b, and let's call r the remainder they leave when divided by n. So both a and b are of the form:

[tex] a=kn+r[/tex] (for some integer k)

[tex] b=ln+r[/tex] (for some integer l)

(this is exactly what it means to leave a remainder of r when divided by n)

And so their difference is

[tex] a-b=kn+r-(ln+r)=kn-ln=(k-l)n[/tex]

Which is divisible by n by definition of being divisible (or think of it as a-b being a multiple of n, so it's divisible by n).

Answer:

The required factors are: x, (x + 6) and (x - 3).

Step-by-step explanation:

As per the question,

The given polynomial is:

[tex]x^{3}+3x^{2}-18x[/tex]

Now,

BY factorization, we get

[tex]x^{3}+3x^{2}-18x[/tex]

[tex]=x(x^{2}+3x-18)[/tex]

By splitting the mid-term, that is split 3x like:

3x = 6x - 3x

Therefore,

[tex]x(x^{2}+6x-3x-18)[/tex]

Now on further solving by taking common factor out, we get

[tex]=x[x(x+6)-3(x+6)][/tex]

[tex]=x(x+6)(x-3)[/tex]

Therefore, the given second polynomial (x - 4), is not a factor of given polynomial [tex]x^{3}+3x^{2}-18x[/tex].

Hence, the given polynomial has three factor x, (x + 6) and (x - 3).

Answer:

Let V a vector space. And B a subset of elements in V.

B is a basis for V if satisfies the following conditions:

1. V= span(B). It means that every element of V can be written as a finite linear combination of elements of B.

2. B is a linear independent subset.

Answer:

$89.60 will be saved by purchasing $224 of cloths be at the register

Step-by-step explanation:

This problem can be solved by a rule of three.

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.

In this problem, we have the following measures:

-The total money

-The percentage of money

As the percentage of money increases, so does the total money. It means that the relationship between the measures is direct.

The problem states that Shirtbarn is having a sale where everything in the store is 40% off. It means that in every purchase, you save 40% of the money.

How much will be saved by purchasing $224 of cloths be at the register?

40% of 224 will be saved. So

$224 - 1

$x - 0.4

x = 224*0.4

x = $89.60

$89.60 will be saved by purchasing $224 of cloths be at the register

Answer:

The proof will be as follows

      Statement                                  Reason

1. -2(x+1) = 8                                      Given

2.  -2(x+1)/ -2 = 8/-2                          Division Property of equality

3. x+1 = -4                                          Simplifying

4. x+1-1 = -4-1                                      Subtraction property of Equality

5. x = -5                                             Simplifying

The option of Distributive property of equality will not be used ..

Answer:

It took 14 hours to get out of the well.

Step-by-step explanation:

Consider the provided information.

The well was 70-feet deep, the frog climbs 7 feet per hour, and it slips back 2 feet while resting.

In first hr frog climbs 7 feet and slips back 2 feet, that means it climbs 5 feet in an hour.

In 2 hour it climbs 2 × 5 = 10 feet, in 3 hours it climbs 3 × 5 = 15 feet and so on.

Now we need to find the time taken by the frog to get out of the well.

By observing the above pattern we can say that after 13 hours it can climb 65 feet.

13 × 5 = 65 feet

In next hour frog will climb 65 feet + 7 feet = 72 feet.

That means after 14 hours frog can get out of the well.

Hence it took 14 hours to get out of the well.

Final answer:

There are 108 students in the class, calculated using the data from the class survey with the principle of inclusion-exclusion.

Explanation:

To find out how many students are in the class, we should analyze the survey data provided. We know that:

29 students watched TV on Monday.

24 students watched TV on Tuesday.

25 students watched TV on Wednesday.

13 students watched TV on only Monday.

9 students watched TV on only Tuesday.

10 students watched TV on only Wednesday.

12 students watched TV on all three days.

14 students watched TV on both Monday and Tuesday.

Since 14 watched on both Monday and Tuesday, and 12 of those also watched on Wednesday, there are 14 - 12 = 2 who watched on Monday and Tuesday but not Wednesday.

We can now calculate the total number of students using the principle of inclusion-exclusion:

Start with the total number of students who watched TV each day: 29 + 24 + 25 = 78.

Subtract the students who were counted twice because they watched on two different days: 78 - 2 (from Monday and Tuesday) = 76.

Subtract the students who watched on all three days, as they have been counted three times: 76 - 12 = 64.

Add back the number of students who watched TV on all three days to account for their previous subtraction: 64 + 12 = 76.

Add the students who watched TV on only one specific day to avoid double-counting those who were included in the two and three days' viewership: 76 + 13 (Monday only) + 9 (Tuesday only) + 10 (Wednesday only) = 108.

Therefore, there are 108 students in the class.

Answer:

12 units

Step-by-step explanation:

The given function is

[tex]S(q) = 5q^2 + 1000q + 100[/tex]

where, S(q) is the price in dollars at which q units are supplied.

We need to find the quantity supplied in a month when the company sets the price of its olive oil press at $12,820.

Substitute S(q)=12820 in the given function.

[tex]12820 = 5q^2 + 1000q + 100[/tex]

Subtract both sides by 12820.

[tex]0= 5q^2 + 1000q - 12720[/tex]

Taking out GCF.

[tex]0= 5(q^2 + 200q - 2544)[/tex]

Now solve the equation for q by splitting the middle term.

[tex]0= 5(q^2 + 212q-12q - 2544)[/tex]

[tex]0= 5(q(q + 212)-12(q + 212))[/tex]

[tex]0= 5(q + 212)(q-12)[/tex]

Using zero product property we get

[tex]q + 212=0\Rightarrow q=-212[/tex]

[tex]q-12=0\Rightarrow q=12[/tex]

q is number of units. So the value of q can not be negative.

Therefore the quantity supplied in a month when the company sets the price of its olive oil press at $12,820 is 12 units.

Answer: There are 133 chickens and 20 pigs.

Step-by-step explanation:

Let x be the number of chickens and y be the number of pigs.

Given : Number of heads of pigs and chicken = 153

Number of feet = 346

Since one chicken has 2 legs and one pig has 4 legs.

By considering the given information, we have the following system of equations:-

[tex]x+y=153------(1)\\\\ 2x+4y=346---------(2)[/tex]

Multiply 2 on both sides of (1), we get

[tex]2x+2y=306-----(3)[/tex]

Subtract (3) from (2), we get

[tex]2y=40\\\\\Rightarrow\ y=\dfrac{40}{2}=20[/tex]

Put value of y in (1), we get

[tex]x+20=153\\\\\Rightarrow\ x=153-20=133[/tex]

Hence, there are 133 chickens and 20 pigs.

Final answer:

The problem is a system of linear equations in Mathematics, where we find that there are 133 chickens and 20 pigs in the barnyard after setting up and solving the equations based on the given number of heads and feet.

Explanation:

Let's denote the number of chickens as C and the number of pigs as P. Therefore, we have two equations based on the given information:

C + P = 153 (since each animal has one head)

2C + 4P = 346 (since chickens have 2 feet and pigs have 4 feet)

By solving these equations, we can find the values for C and P. Multiplying the first equation by -2 and adding it to the second equation, we eliminate C and get:

-2C - 2P = -306
2C + 4P = 346
-----------------
2P = 40

Dividing both sides by 2, we find that P = 20. Substituting this value back into the first equation, we get C + 20 = 153, which means C = 133.

Therefore, there are 133 chickens and 20 pigs in the barnyard.

Answer:

The cost to rent a trailer for 2.9 ​hours is $27.2.

The cost to rent a trailer for 3 ​hours is $28.

The cost to rent a trailer for 8.5 ​hours is $72.

The cost to rent a trailer for 9 ​hours is $76.

Step-by-step explanation:

It is given that the charge to rent a trailer is $20 for up to 2 hours plus $8 per additional hour or portion of an hour.

Let x be the number of hours.

The cost to rent a trailer for x ​hours is defined as

[tex]C(x)=\begin{cases}20 & \text{ if } x\leq 2 \\ 20+8(x-2) & \text{ if } x>2 \end{cases}[/tex]

For x>2,

[tex]C(x)=20+8(x-2)[/tex]

Substitute x=2.9 in the cost function.

[tex]C(x)=20+8(2.9-2)=27.2[/tex]

The cost to rent a trailer for 2.9 ​hours is $27.2.

Substitute x=3 in the cost function.

[tex]C(x)=20+8(3-2)=28[/tex]

The cost to rent a trailer for 3 ​hours is $28.

Substitute x=8.5 in the cost function.

[tex]C(x)=20+8(8.5-2)=72[/tex]

The cost to rent a trailer for 8.5 ​hours is $72.

Substitute x=9 in the cost function.

[tex]C(x)=20+8(9-2)=76[/tex]

The cost to rent a trailer for 9 ​hours is $76.

All ordered​ pairs, in the form of (hours,​ cost) are (2.9, 27.2), (3,28), (8.5, 72) and (9,76).

The graph of all ordered​ pairs is shown below.

Final answer:

The costs to rent a trailer for 2.9 and 3 hours are both $28, for 8.5 and 9 hours are both $76. To plot this function, one must mark these costs against the rental hours, showing a flat rate for the first two hours and additional charges thereafter.

Explanation:

The question asks for the cost of renting a trailer for varying numbers of hours and then requires plotting a graph with these costs against the hours. The rental system has a flat rate of $20 for the first 2 hours, and an additional charge of $8 for each hour or part of an hour thereafter.

To graph all ordered pairs, plot points for each of the time periods mentioned with their corresponding costs. Note how the graph demonstrates incremental jumps after the first 2 hours, reflecting the additional $8 charge per hour or part thereof.

Answer: Hi, first, the cannon only gives the cannonball the initial velocity, and when the cannonball is in the air, the only force acting on the ball is the gravitational force.

First, let's compute the initial velocity, if the ball is fired with angle A (measured from the ground, or +x in this case) and velocity V0. then the vector of the velocity is (cos(A)*V0, Sin(A)*V0)

now start describing all the equations.

Acceleration, we know that an object in the air will fall with acceleration g = 9.8 m/s.

then a(t) = (0, -g)

Velocity: integrating the acceleration over the time, we obtain v(t) = (0,-g*t) +C

where C is a integration constant, equal to the initial velocity. Then v(t) = (cos(A)*V0, Sin(A)*V0 - g*t)

Position; For the position we need to integrate again over time, then:

p(t)= (cos(A)*V0*t, Sin(A)*V0*t - [tex]g*\frac{t^{2} }{2}[/tex]) + K.

where again K is an integration constant, in this case the initial position, that  write it as (X0,Y0).

The height of the cannonball after 2 seconds is the y component valued in t=2

height = Y0 + Sin(A)*V0*2 - [tex]g*\frac{2^{2} }{2}[/tex].

where you can put the angle A and the initial velocity V0 to obtain the height.

Answer:

Step-by-step explanation:

Michael Perez deposited a total of $2,000 with two savings institutions.

One pays interest at a rate of 5% per year whereas the other pays interest at a rate of 8% per year.

Let x denote the amount of money invested at 5%

y = 2000 - x ---(1)

Let y denote the amount of money invested at 8%

so 5x/100 + 8(2000-x)/100 = $130 ----(2)

Michael earned $130 in interest during a single year.

Answer:

The maximum profit is $1600 at x=30 and y=10.

Step-by-step explanation:

Let x be the number of units of product X.

y be the number of units of product Y.

The profit for X is $35 and the profit for Y is $55.

Maximize [tex]Z = 35x + 55y[/tex]               ..... (1)

It requires 4 pounds of material and 2 hours of labor to produce one unit of X. It requires 4 pounds of material and 6 units of labor to produce one unit of Y.

Total material = 4x+4y

Total labor = 2x+6y

There are 160 pounds of material and 120 hours of labor available.

[tex]4x+4y\leq 160[/tex]             .... (2)

[tex]2x+6y\leq 120[/tex]            ..... (3)

[tex]x\geq 0,y\geq 0[/tex]

The related line of inequality (2) and (3) are solid line because the sign of equality "≤" contains all the point on line in the solution set.

Check the inequalities by (0,0).

[tex]4(0)+4(0)\leq 160[/tex]

[tex]0\leq 160[/tex]

This statement is true.

[tex]2x+6y\leq 120[/tex]  

[tex]2(0)+6(0)\leq 120[/tex]  

[tex]0\leq 120[/tex]  

It means shaded region of both inequalities contain (0,0).

The extreme points of common shaded region are (0,0), (0,20), (40,0) and (30,10).

At (0,0),

[tex]Z = 35(0) + 55(0)=0[/tex]

At (0,20),

[tex]Z = 35(0) + 55(20)=110[/tex]

At (40,0),

[tex]Z = 35(40) + 55(0)=140[/tex]

At (30,10),

[tex]Z = 35(30) + 55(10)=1600[/tex]

Therefore the maximum profit is $1600 at x=30 and y=10.

Answer:

(a) [tex]p \wedge -q[/tex]

(b) [tex]\neg p \Rightarrow \neg q[/tex]

(c) [tex]q \wedge -p[/tex]

Step-by-step explanation:

(a) You drive over 65 miles per hour, but you do not get a speeding ticket, it can be represented by: [tex]p \wedge -q[/tex]

(b) If you do not drive over 65 miles per hour, then you will not get a speeding ticket, it can be represented by: [tex]\neg p \Rightarrow \neg q[/tex]

(c) You get a speeding ticket, but you did not drive over 65 miles per hour, can be represented by: [tex]q \wedge -p[/tex]

Answer:

[tex]x\ =\ \dfrac{209}{30}[/tex]

[tex]y\ =\ \dfrac{29}{18}[/tex]

[tex]z\ =\ \dfrac{64}{15}[/tex]

Step-by-step explanation:

Given equations are

15x + 15y + 10z = 106

5x + 15y + 25z = 135

15x + 10y - 5z = 42

The augmented matrix by using above equations can be written as

[tex]\left[\begin{array}{ccc}15&15&10\ \ |106\\5&15&25\ \ |135\\15&10&-5|42\end{array}\right][/tex]

[tex]R_1\ \rightarrow\ \dfrac{R_1}{15}[/tex]

[tex]=\ \left[\begin{array}{ccc}1&1&\dfrac{10}{15}|\dfrac{106}{15}\\5&15&25|135\\15&15&-5|42\end{array}\right][/tex]

[tex]R_1\rightarrowR_2-5R1\ and\ R_3\rightarrow\ R_3-15R_1[/tex]

[tex]=\ \left[\begin{array}{ccc}1&1&\dfrac{10}{15}|\dfrac{106}{15}\\\\0&10&\dfrac{65}{3}|\dfrac{299}{3}\\\\0&0&-15|-64\end{array}\right][/tex]

[tex]R_2\rightarrow\ \dfrac{R_2}{10}[/tex]

[tex]=\ \left[\begin{array}{ccc}1&1&\dfrac{10}{15}|\dfrac{106}{15}\\\\0&1&\dfrac{65}{30}|\dfrac{299}{30}\\\\0&0&-15|-64\end{array}\right][/tex]

[tex]R_3\rightarrow\ \dfrac{R_3}{-15}[/tex]

[tex]=\ \left[\begin{array}{ccc}1&1&\dfrac{10}{15}|\dfrac{106}{15}\\\\0&1&\dfrac{65}{30}|\dfrac{299}{30}\\\\0&0&1|\dfrac{64}{15}\end{array}\right][/tex]

[tex]R_1\rightarrow\ R_1-R_2[/tex]

[tex]=\ \left[\begin{array}{ccc}1&0&\dfrac{-3}{2}|\dfrac{17}{30}\\\\0&1&\dfrac{65}{30}|\dfrac{299}{30}\\\\0&0&1|\dfrac{64}{15}\end{array}\right][/tex]

[tex]R_1\rightarrow\ R_1+\dfrac{3}{2}R_3[/tex]

[tex]=\ \left[\begin{array}{ccc}1&0&0|\dfrac{209}{30}\\\\0&1&\dfrac{65}{30}|\dfrac{299}{30}\\\\0&0&1|\dfrac{64}{15}\end{array}\right][/tex]

[tex]R_2\rightarrow\ R_2-\dfrac{65}{30}R_3[/tex]

[tex]=\ \left[\begin{array}{ccc}1&0&\0|\dfrac{209}{30}\\\\0&1&0|\dfrac{29}{18}\\\\0&0&1|\dfrac{64}{15}\end{array}\right][/tex]

Hence, we can write from augmented matrix,

[tex]x\ =\ \dfrac{209}{30}[/tex]

[tex]y\ =\ \dfrac{29}{18}[/tex]

[tex]z\ =\ \dfrac{64}{15}[/tex]

Answer:

[tex]L\{f(t)\}=\frac{8(\sqrt3s+1)}{s^2+1}[/tex]

Step-by-step explanation:

Given : [tex]f(t)=16\cos (t-\frac{\pi}{6})[/tex]

To find : ℒ{f(t)} by first using a trigonometric identity ?

Solution :

First we solve the function,

[tex]f(t)=16\cos (t-\frac{\pi}{6})[/tex]

Applying trigonometric identity, [tex]\cos (A-B)=\cos A\cos B+\sin A\sin B[/tex]

[tex]f(t)=16(\cos t\cos (\frac{\pi}{6})+\sin t\sin(\frac{\pi}{6})[/tex]

[tex]f(t)=16(\frac{\sqrt3}{2}\cos t+\frac{1}{2}\sin t)[/tex]

[tex]f(t)=\frac{16}{2}(\sqrt3\cos t+\sin t)[/tex]

[tex]f(t)=8(\sqrt3\cos t+\sin t)[/tex]

We know, [tex]L(\cos at)=\frac{s}{s^2+a^2}[/tex] and [tex]L(\sin at)=\frac{a}{s^2+a^2}[/tex]

Applying Laplace in function,

[tex]L\{f(t)\}=8\sqrt3L(\cos t)+8L(\sin t)[/tex]

[tex]L\{f(t)\}=8\sqrt3(\frac{s}{s^2+1})+8(\frac{1}{s^2+1})[/tex]

[tex]L\{f(t)\}=\frac{8\sqrt3s+8}{s^2+1}[/tex]

[tex]L\{f(t)\}=\frac{8(\sqrt3s+1)}{s^2+1}[/tex]

Therefore, The Laplace transformation is [tex]L\{f(t)\}=\frac{8(\sqrt3s+1)}{s^2+1}[/tex]

Final answer:

In this college-level mathematics question, the task is to find ℒ{f(t)} by employing a trigonometric identity. By rewriting the function f(t) = 16cos(t−π/6) in terms of sine and utilizing a trigonometric identity, we find ℒ{f(t)} = 16s / (s^2 + 1).

Explanation:

To find ℒ{f(t)} by utilizing a trigonometric identity, first rewrite the function f(t) = 16cos(t−π/6) in terms of sine. Use the trigonometric identity cos(a) = sin(a + π/2) to rewrite cos(t−π/6) as sin(t−π/6 + π/2). This simplifies to sin(t−π/3). Thus, ℒ{f(t)} = 16 * ℒ{cos(t−π/6)}

= 16 * ℒ{sin(t−π/3)} = 16 * (s / (s^2 + 1)).

Therefore, the answer is 16s / (s^2 + 1).

Answer:

a) 996.6 mg

b) 498.96 mg

c) 4

d) 2

Step-by-step explanation:

Given:

Dose to be given = 10 mg/kg/day

Number of dose to be divided = 2

weight of the patient = 220 pounds

now,

1 pound = 0.453 kg

thus,

weight of the patient = 220 × 0.4536 = 99.792 kg

a) Amount of Drug patient should receive per day = dose × weight of patient

or

Amount of Drug patient should receive per day = 10 × 99.792

or

Amount of Drug patient should receive per day = 997.92 mg

b) Now, the dose is divided in to 2 per day

thus,

The amount of drug received per dose = [tex]\frac{\textup{Drug received per day}}{\textup{Number of dose per day}}[/tex]

or

The amount of drug received per dose = [tex]\frac{\textup{997.92 mg}}{\textup{2}}[/tex]

or

The amount of drug received per dose = 498.96 mg

c) weight of capsule = 250 mg

Thus,

capsules received by patient per day = [tex]\frac{\textup{Dose per day in mg}}{\textup{weight of capsule in mg}}[/tex]  

or

capsules received by patient per day = [tex]\frac{\textup{997.92}}{\textup{250}}[/tex]  

or

capsules received by patient per day = 3.99168 ≈ 4

d) Capsules to be received per dose = [tex]\frac{\textup{Amount of drug per dose in mg}}{\textup{weight of capsule in mg}}[/tex]  

or

capsules received by patient per dose = [tex]\frac{\textup{498.86}}{\textup{250}}[/tex]  

or

capsules received by patient per dose = 1.99544 ≈ 2

Answer:

a) The value of the Annual Payment is A=$17,258.80

b) Is the picture in the attachment file

c) As you can see it in the picture with each payment, balance comes down, due it is the interest base, Interest portion comes down too.

Step-by-step explanation:

Hi

a) First of all, we are going to list the Knowns: [tex]VP=47000[/tex], [tex]i=5[/tex]% and [tex]n=3[/tex], Then we can use [tex]A=\frac{VP}{\frac{1-(1+i)^{-n} }{i} } =\frac{47000}{\frac{1-(1+0.05)^{-3} }{0.03} }=17258.80[/tex]. So this is the value of the Annual Payment

Joan Messines's annual payment on her $47,000 loan at 5% interest over 3 years is $17,158.11. The interest portion of each payment declines over time due to the decreasing loan balance, leading to a smaller interest calculation base in each subsequent year.

Joan Messines borrowed $47,000 at a 5% annual rate of interest to be repaid over 3 years. The loan is amortized into three equal, annual, end-of-year payments.

Calculation of the Annual Loan Payment

To calculate the annual payment, we use the formula for an annuity:

PV = PMT [(1 - (1 + r)^-n) / r]

Where:

PV is the present value of the loan (initial loan amount).

PMT is the annual payment.

r is the annual interest rate (expressed as a decimal).

n is the number of years.

Rearranging the formula to solve for PMT yields:

PMT = PV / [(1 - (1 + r)^-n) / r]

Substitute PV = $47,000, r = 0.05 (5%), and n = 3:

PMT = $47,000 / [(1 - (1 + 0.05)^-3) / 0.05]

PMT = $17,158.11 (rounded to the nearest cent).

Loan Amortization Schedule

Year 1: Interest = $47,000 * 5% = $2,350; Principal = $17,158.11 - $2,350 = $14,808.11; Remaining Balance = $47,000 - $14,808.11 = $32,191.89

Year 2: Interest = $32,191.89 * 5% = $1,609.59; Principal = $17,158.11 - $1,609.59 = $15,548.52; Remaining Balance = $32,191.89 - $15,548.52 = $16,643.37

Year 3: Interest = $16,643.37 * 5% = $832.17; Principal = $17,158.11 - $832.17 = $16,325.94; Remaining Balance = $16,643.37 - $16,325.94 = $317.43

Why the Interest Portion Declines Over Time

The interest portion of each payment declines with the passage of time because as the loan principal is paid down, there is a smaller balance on which interest is calculated. This results in a decreasing interest payment and an increasing principal payment with each subsequent payment until the loan is paid off.

Final answer:

The qualitative statements are A. The flower is red. and C. The candy was sour. The quantitative statements are B. The bug is 5cm long. and D. You have three sisters.

Explanation:

The qualitative statement refers to descriptions that do not involve numerical values or measurements. On the other hand, quantitative statements involve numerical values or measurements. Based on these definitions, the qualitative statements in the given options are:

A. The flower is red.

C. The candy was sour.

The quantitative statements in the given options are:

B. The bug is 5cm long.

D. You have three sisters.

Answer:

83%

Step-by-step explanation:

100% - 17% = 83%

Cutting the number of pages by 17% leaves 83% of the original number of pages.

After reducing the number of pages by 17%, the final version of the book contains 83% of the original number of pages.

If the author of a book was told to cut the number of pages by 17%, we want to determine what percent of the original pages remained in the final version. To do this, we subtract the percentage of the pages cut from 100% (which represents the original number of pages).

100% - 17% = 83%.

Therefore, after a 17% cut, 83% percent of the pages are left in the final version of the book.

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Answer:

In the files are the truth tables.

c) and d) are the same.

The student's question is about constructing truth tables for four different logical expressions, which involves the step-by-step process of calculating the truth values based on the logical operators for conjunction, disjunction, conditional, and negation.

Constructing Truth Tables

To construct truth tables for the given logical expressions, we enumerate all the possible truth values for the propositions p, q, and r, and calculate the truth values of the complex expressions based on logical operators like conjunction (∧), disjunction (∨), conditional (⇒), and negation (¬).

a) (p ∧ q) ∨ r

This expression involves the conjunction of p and q, followed by the disjunction with r. To construct the truth table, we first list all possible binaries (true/false) for p, q, and r, then determine the result of p ∧ q, and finally the disjunction with r.

b) (p ∨ q) ⇒ (p ∧ r)

The expression starts with the disjunction of p and q, which implies (p ∧ r). Calculate the truth values step by step.

c) (p ⇒ q) ∨ (¬p ⇒ q)

Here, we have two conditional expressions connected by disjunction. The truth value of each conditional is ascertained separately, and then combined using the disjunction ∨.

d) (p ⇒ q) ∧ (¬p ⇒ q)

Similarly, for the conjunction, the truth of both conditionals must hold. Construct each column gradually, concluding with the combined result.

By following logical operators' rules and systematically filling in every row, we can complete these truth tables to see which combinations of propositions render the expressions true or false.