A test tube contains 25 bacteria, 5 of which are can stay alive for atleast 30 days, 10 of which will die in their second day. 10 of which are already dead.
Given that a randomly chosen bacteria for experiment is alive. What is the probability it will still be alive after one week?
(a)1 ⁄ 3(b)2 ⁄3
(c) 1⁄5 (d)4 ⁄5

Answer:

D. $105

Step-by-step explanation:

Use definition:

[tex]\text{Average of }n\text{ numbers}=\dfrac{\text{The sum of }n\text{ numbers}}{n}[/tex]

The average wage for 15 consecutive days is $91, then by definition

[tex]\$91=\dfrac{\text{The sum of wages for 15 days}}{15}\Rightarrow \\ \\\text{The sum of wages for 15 days}=\$91\cdot 15=\$1,365[/tex]

The average wage for first 7 consecutive days is $87, then by definition

[tex]\$87=\dfrac{\text{The sum of wages for first 7 days}}{7}\Rightarrow \\ \\\text{The sum of wages for first 7 days}=\$87\cdot 7=\$609[/tex]

The average wage for last 7 consecutive days is $93, then by definition

[tex]\$93=\dfrac{\text{The sum of wages for last 7 days}}{7}\Rightarrow \\ \\\text{The sum of wages for last 7 days}=\$93\cdot 7=\$651[/tex]

Now,

[tex]\text{The sum of wages for 15 days}=\text{The sum of wages for first 7 days }+\\ \\+\text{ The wage for 8th day}+\text{The sum of wages for last 7 days}\\ \\\$1,365=\$609+\text{ The wage for 8th day}+\$651\\ \\\text{ The wage for 8th day}=\$1,365-\$609-\$651=\$105[/tex]

Answer:

$20.9

Step-by-step explanation:

We have been given a formula [tex]C(x)=19x+1900[/tex], which represents the cost of x items.

First of all, we will find cost of 1000 items by substituting [tex]x=1000[/tex] in our given formula as:

[tex]C(1000)=19(1000)+1900[/tex]

[tex]C(1000)=19,000+1900[/tex]

[tex]C(1000)=20,900[/tex]

To find average cost per item, we will divide total cost by number of items as:

[tex]\text{Average cost per item}=\frac{\$20,900}{1000}[/tex]

[tex]\text{Average cost per item}=\$20.9[/tex]

Therefore, the average cost per item would be $20.9.

Answer: 26 items

Step-by-step explanation:

Let x denotes the last year's output.

Given : A company has determined that it must increase production of a certain line of goods by 112 times last years production.

The new output will be 2885 items.

According to the above statement we have the following equation :-

[tex]112x=2885[/tex]

Divide both sides by 112, we get

[tex]x=25.7589285714\approx26[/tex]

Hence, the  last year's output =26 items

Answer:

150 units;

Maximum revenue: $62,500.

Step-by-step explanation:

We have been given that a company’s total revenue from manufacturing and selling x units of their product is given by [tex]y=-3x^2+900x-5,000[/tex]. We are asked to find the number of units sold that will maximize the revenue.

We can see that our given equation in a downward opening parabola as leading coefficient is negative.

We also know that maximum point of a downward opening parabola is ts vertex.

To find the number of units sold to maximize the revenue, we need to figure our x-coordinate of vertex.

We will use formula [tex]\frac{-b}{2a}[/tex] to find x-coordinate of vertex.

[tex]\frac{-900}{2(-3)}[/tex]

[tex]\frac{-900}{-6}[/tex]

[tex]150[/tex]

Therefore, 150 units should be sold in order to maximize revenue.

To find the maximum revenue, we will substitute [tex]x=150[/tex] in our given formula.

[tex]y=-3(150)^2+900(150)-5,000[/tex]

[tex]y=-3*22,500+135,000-5,000[/tex]

[tex]y=-67,500+135,000-5,000[/tex]

[tex]y=62,500[/tex]

Therefore, the maximum revenue would be $62,500.

Answer:

Option B. 5 hours.

Step-by-step explanation:

Steve reads 80 pages in 2 hours and 40 minutes.

1 hour = 60 minutes

2 hour 40 minutes = (2 × 60) + 40 = 160 minutes

Speed of reading of Steve = [tex]\frac{80}{160}[/tex]

                                             = 0.5 page per minute

Cassandra reads twice as fast as Steve.

Therefore, speed of reading of Cassandra = 0.5 × 2

                                                                       = 1 page per minute

Therefore, time to read 300 pages by Cassandra = 300 × 1

                                                                                   = 300 minutes

Now convert minutes to hours

[tex]\frac{300}{60}[/tex]

= 5 hours

Cassandra will take 5 hours to read 300 page book.

Answer:

D had 310 points.

The final ranking is: B,A,D,C

Step-by-step explanation:

The problems states that:

There are 115 votes.

Each ballot has a vote for first place, that counts 4 points, a vote for second place, that counts 3 points, a vote for third place, that counts 2 points and a vote for fourth place that counts 1 point. This means that each ballot has 4+3+2+1 = 10 points.

Since there are 115 voters, there are 115 ballots. This means that the total points of A,B,C and D combined must be equal to 115*10 = 1150. So:

[tex]A + B + C + D = 1150[/tex]

We already know that:

A had 320 points, B had 330 points, and C had 190 points.

So:

[tex]A + B + C + D = 1150[/tex]

[tex]320 + 330 + 190 + D = 1150[/tex]

[tex]D = 310[/tex]

B had the most points, followed by A, D and C. So the final ranking is: B,A,D,C

Answer: 72

Step-by-step explanation:

Given : An experiment has 3 stages: A, B, and C.

If stage A has 6 outcomes, stage B has 4 outcomes, and stage C has 3 outcomes.

Then, by using the fundamental principle of counting (Total outcomes= product of all outcomes of each stage ) , we have

The number of outcomes the entire experiment have:-

[tex]6\times4\times3=72[/tex]

Hence, the number of outcomes the entire experiment =72

Final answer:

To find the total number of outcomes for an experiment with 3 stages having 6, 4, and 3 outcomes respectively, you multiply the outcomes of each stage together, resulting in 6 × 4 × 3 = 72 possible outcomes.

Explanation:

To determine how many outcomes the entire experiment has when an experiment has stages A, B, and C with different numbers of outcomes, you multiply the number of outcomes for each stage. Stage A has 6 outcomes, stage B has 4 outcomes, and stage C has 3 outcomes. Therefore, the total number of outcomes for the entire experiment is calculated as follows:

Stage A outcomes: 6

Stage B outcomes: 4

Stage C outcomes: 3

Multiply the outcomes of each stage:

Total outcomes = 6 (Stage A) × 4 (Stage B) × 3 (Stage C) = 72 possible outcomes.

This is similar to how the sample space is determined in other contexts, such as flipping a coin and rolling a die, where you would also multiply the number of outcomes to get the size of the sample space.

Answer:

x1= 1

Step-by-step explanation:

The Cramer's rule say that x1=[tex]\frac{det(A1)}{det(A)}[/tex] where A1 is the matrix A change the column 1 by the vector b.

Then A1= [tex]\left[\begin{array}{cccc}31&6&-3&20\\2&2&-5&-7\\21&2&7&20\\11& 2&-11&-4\end{array}\right][/tex].

Using Octave we have that  det(A1)=-3840 and det(A)=-3840.

Then x1=[tex]\frac{-3840}{-3840}=1[/tex].

Answer:

[tex]x=\pm\sqrt{77n+53}[/tex]

Step-by-step explanation:

Given : [tex]x^2\equiv 53\mod 77[/tex]

To find : All the square roots ?

Solution :

The primitive roots modulo is defined as

[tex]a\equiv b\mod c[/tex]

Where, a is reminder

b is dividend

c is divisor  

Converting equivalent into equal,

[tex]a-b=nc[/tex]

Applying in [tex]x^2\equiv 53\mod 77[/tex],

[tex]x^2\equiv 53\mod 77[/tex]

[tex]x^2-53=77n[/tex]

[tex]x^2=77n+53[/tex]

[tex]x=\pm\sqrt{77n+53}[/tex]

We have to find the possible value in which the x appear to be integer.

The possible value of n is 4.

As [tex]x=\pm\sqrt{77(4)+53}[/tex]

[tex]x=\pm\sqrt{308+53}[/tex]

[tex]x=\pm\sqrt{361}[/tex]

[tex]x=\pm 9[/tex]

Answer:

C

Step-by-step explanation:

1. 5.99 = 6

2. 2.3 = 2.25

3. 6 - 2.25 = 3.75

Answer:

Step-by-step explanation:

Given that a basic cellular phone plan costs $30 per month for 50 calling minutes

i.e. C(x) = [tex]30+0.50(x-50)[/tex], where x = calling minutes

Here 30 is fixed upto 50 calls after that cost increases at 0.50 per minute talk time.

[tex]C(x) = 30+0.5x-25 = 0.5x+5\\[/tex]

When monthly cost is atleast 35 and atmost 40 we have

[tex]35\leq C(x)\leq 40\\35\leq 0.5x+5\leq 40\\30\leq 0.5x\leq 35\\60\leq x\leq 70[/tex]

i.e. talking time must be atleast 60 minutes and atmost 70 minutes

Multiply his max heart rate by both 75% and 90%.

168 x 0.75 = 126

168 x 0.90 = 151.2, round to 151

The minimum is 126

The maximum is 151

Answer:

Step-by-step explanation:

The average obtained at the end of the course will be:

[tex]\frac{85+78+84+x}{4} = av[/tex]

Where x is the grade obtained in the final examination and av is the final average. To obtain an A, av has to be at least 90, av≥90, and to obtain an B, av has to be at least 80, av≥80

So, if we get 100 on the final average:

x = 100,

av = (85+78+84+100)/4 = 86,75 and  86,75∠90.

Answer: No, the higher grade obtained would be 86,75.

To earn a B, av≥80:

[tex]av= \frac{85+78+84+x}{4 \ }\geq  80\\ \frac{247+x}{4}\geq80   \\247+x\geq 80*4\\ x\geq 320-247\\ x\geq 73[/tex]

Answer: I must have a 73 or MORE to earn a B in the course

Answer:

[tex]\frac{1333}{2000}[/tex]

Step-by-step explanation:

We want to compute the probabily of being flipping coin1 in the third day. Observe that in day zero (the day were the coin to be flipped is chosen randomly with equal probability to be coin1 or coin2), day 1 and day 2 we will flip coin1 or coin2. So, there are 8 possible scenarios to consider:

[tex]S1=(1,1,1,1)\\S2=(1,1,2,1)\\S3=(1,2,1,1)\\S4=(1,2,2,1)\\S5=(2,1,1,1)\\S6=(2,1,2,1)\\S7=(2,2,1,1)\\S8=(2,2,2,1)[/tex]

Where S1 is the scenario where we flip coin1 everyday. S2 is the scenario where we flip coin1 the day zero and first day, coin2 the second day, and again coin1 the third day. S3,...,S8 are defined the same.

Observe that the probability to flip coin1 the third day is equal to the sum of [tex]P(S1)+P(S2)+...+P(S8).[/tex] To compute this probabilities we will define:

[tex]P(1,1)=[/tex]Probability to flip coin1 one day given that coin1 was flipped the day before.

[tex]P(1,2)=[/tex]Probability to flip coin2 one day given that coin1 was flipped the day before.

[tex]P(2,1)=[/tex]Probability to flip coin1 one day given that coin2 was flipped the day before.

[tex]P(2,2)=[/tex]Probability to flip coin2 one day given that coin2 was flipped the day before.

Then, using the question information, we can conclude that

[tex]P(1,1)=0.7, P(1,2)=0.3, P(2,1)=0.6, P(2,2)=0.4[/tex]

With this we can compute P(S1),...,P(S8) as follows:

[tex]P(S1)=\frac{1}{2}P(1,1)*P(1,1)*P(1,1)=\frac{1}{2}*(\frac{7}{10})^3=\frac{343}{2000}\\\\P(S2)=\frac{1}{2}P(1,1)*P(1,2)*P(2,1)=\frac{1}{2}*\frac{7}{10}*\frac{3}{10}*\frac{6}{10}=\frac{126}{2000}\\\\P(S3)=\frac{1}{2}P(1,2)*P(2,1)*P(1,1)=\frac{1}{2}*\frac{3}{10}*\frac{6}{10}*\frac{7}{10}=\frac{126}{2000}\\\\P(S4)=\frac{1}{2}P(1,2)*P(2,2)*P(2,1)=\frac{1}{2}*\frac{3}{10}*\frac{4}{10}*\frac{6}{10}=\frac{72}{2000}\\\\[/tex]

[tex]P(S5)=\frac{1}{2}P(2,1)*P(1,1)*P(1,1)=\frac{1}{2}*\frac{6}{10}*\frac{7}{10}*\frac{7}{10}=\frac{294}{2000}\\\\P(S6)=\frac{1}{2}P(2,1)*P(1,2)*P(2,1)=\frac{1}{2}*\frac{6}{10}*\frac{3}{10}*\frac{6}{10}=\frac{108}{2000}\\\\P(S7)=\frac{1}{2}P(2,2)*P(2,1)*P(1,1)=\frac{1}{2}*\frac{4}{10}*\frac{6}{10}*\frac{7}{10}=\frac{168}{2000}\\\\P(S8)=\frac{1}{2}P(2,2)*P(2,2)*P(2,1)=\frac{1}{2}*\frac{4}{10}*\frac{4}{10}*\frac{6}{10}=\frac{96}{2000}\\\\[/tex]

Finally, the probability to flip coin1 the third day is

[tex]P(S1)+...+P(S8)=\frac{1333}{2000}[/tex]

The probability that the coin flipped on the 3rd day after the initial flip is coin 1 is 65%.

To determine the probability that the coin flipped on the 3rd day after the initial flip is coin 1, let's analyze the scenario:

There is a 0.5 chance that we start with either coin 1 or coin 2 on the first day.

If we start with coin 1 and flip heads (with probability 0.7), we will flip coin 1 again on the second day.

If we start with coin 1 and flip tails (with probability 0.3), we will flip coin 2 on the second day.

If we start with coin 2 and flip heads (with probability 0.6), we will flip coin 1 on the second day.

If we start with coin 2 and flip tails (with probability 0.4), we will flip coin 2 on the second day.

To get to coin 1 on the third day, we have the following cases:

Start with coin 1, flip heads (0.5*0.7).

Start with coin 2, flip heads (0.5*0.6).

The total probability of flipping coin 1 on the third day is the sum of the probabilities of these two independent events: (0.5*0.7) + (0.5*0.6) = 0.35 + 0.3 = 0.65 or 65%.

Answer:

Step-by-step explanation:

A has one red and blue marble and B has one red and blue marble.

Hence selecting one marble is equally likely with prob = 0.5

Since A and B are independent the joint event would be product of probabilities.

Let A be the amount A wins.

If each selects one, the sample space would be

             (R,R)  (R,B)  (B,R) (B,B)

Prob     0.25  0.25  0.25  0.25

A              3       -2     -2       1

E(A)      0.75   -0.5    -0.5   0.25   =    0

The game is a fair game with equal expected values for A and B.

It does not matter whether to be A or B

The game described is fair, as both A and B have an expected value of $1. This is calculated using the probability of each outcome times its respective payoff. There is no advantage in being either A or B.

The marble game presented in the question involves probability and expectation of winnings. First, we must figure out the possible outcomes. There are four possibilities: (1) both A and B present red marbles, (2) both present blue marbles, (3) A presents red and B presents blue, and (4) A presents blue and B presents red.

When calculating the expected payoff for each player, we assume that each possibility carries an equal weight, as the question doesn't offer any bias towards a certain color. Therefore, the expected payoff (E) for each player is calculated by adding the product of each possible payout and its probability. For A, [tex]E(A) = (1/4)*$3 + (1/4)*$1 + (1/4)*$0 + (1/4)*$0 = $1.\\ For B, E(B) = (1/4)*$0 + (1/4)*$0 + (1/4)*$2 + (1/4)*$2 = $1.[/tex]

Given these calculations, we can see that the expected value of the game is equal for A and B, hence it doesn't matter who is A and who is B.

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Answer:

Ans. Since the annual rate is not compunded (for example, compounded monthly) you will have to pay in interest $27.41, and the total payment is $1,061.41

Step-by-step explanation:

Hi, since the balance is 1 month overdue, it means that you owe 2 months of interest to this obligation, but before we start finding the interest of your credit card, first let´s find the effective monthly equivalent rate for that 17% annual interest rate.

The formula is as follows.

[tex]r(monthly)=(1+r(annual))^{\frac{1}{12} }-1[/tex]

Therefore

[tex]r(monthly)=(1+0.17)^{\frac{1}{12} }-1 =0.01317[/tex]

So your monthly interest rate is 1.317%. Now let´s find the amount of interests that you have to pay for 2 months. This is the formula.

[tex]Interest=Present Value(1+r(monthly))^{n} )-PresentValue[/tex]

Where "n" is the period of time in months that you owe to the financial institution. The result of that is:

[tex]Interest=1,034(1+0.01317)^{2} -1,034=24.41[/tex]

This way, interest are = $27.41 and the total amount that you will have to pay is:

[tex]Payment=Present Value+Interest=1,034+27.41=1,061.41[/tex]

Best of luck.

Answer:

$925.20

Step-by-step explanation:

Loan Amount, P = $19,500

Rate of interest, r = 3.9%

Time, t = 6 years

Payment mode, n = Quarterly (4)

payment to amortize, EMI = ?

Formula: [tex]EMI=\dfrac{P\cdot \frac{r}{n}}{1-(1+\frac{r}{n})^{-n\cdot t}}[/tex]

where,

n = 4 , Rate of interest , r = 0.039

Put the values into formula

[tex]EMI=\dfrac{19500\cdot \frac{0.039}{4}}{1-(1+\frac{0.039}{4})^{-4\cdot 6}}[/tex]

[tex]EMI=915.20[/tex]

Hence, The payment to amortize the debt is $915.20

Answer:

Part (A) True

Part (B) False

Part (C) True

Step-by-step explanation:

Consider the provided information.

If both the statements are either true or false then the biconditionals are true. Otherwise biconditionals are false.

Part (A) 1 + 1 = 3 if and only if monkeys can fly.

Consider the first statement: 1+1=3 (This is a False statement)

Consider the second statement: "monkeys can fly"  (This is also a False statement)

True: First statement is false and the second statement is also false, Thus, making the biconditional true.

Part (B) If birds can fly, then 1 + 1 = 3.

Consider the first statement: "birds can fly" (This is true statement)

Consider the second statement: 1 + 1 = 3 (This is a False statement)

False: First statement is true, but second statement is false, making everything false.

Part (C) If 1 + 1= 3, then pigs can fly.

Consider the first statement: 1+1=3 (This is a False statement)

Consider the second statement: "pigs can fly"  (This is also a False statement)

True: First statement is false and the second statement is also false, Thus, making the biconditional true.

The given sequence :  0,4,18,48,100,180,......

We can write the terms of the sequence as :

[tex]\text{Ist term }:a_1=1(1^2-1)=0\\\\\text{IInd term }: a_2=2^2(2-1)=4(2-1)=4\\\\\text{IIIrd term }:a_3=3^2(3-1)=9(2)=18\\\\\text{IVth term }:a_4=4^2(4-1)=(16)(3)=48\\\\\text{Vth term }:a_5=5^2(5-1)=25(4)=100\\\\\text{VIth term }:a_6=6^2(6-1)=36(5)=180[/tex]

From the above presentation of the terms, the explicit formula in terms of n for the nth term will be :-

[tex]a_n=n^2(n-1)[/tex]

Put n= 7 , we get

[tex]a_7=7^2(7-1)=49(6)=294[/tex]

Therefore, the seventh term of the given sequence = 294

Answer:

3gallons (gal) = 11.35 liters

Step-by-step explanation:

This can be solved as a rule of three problem.

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.

Each gallon has 3.78L. So

1 gallon - 3.78L

3 gallons - xL

[tex]x = 3*3.78[/tex]

[tex]x = 11.35[/tex] liters

3gallons (gal) = 11.35 liters

Answer:

a) must be met

Step-by-step explanation:

We have two conditions:

a) For every [tex]b\in B[/tex] and [tex]\epsilon>0[/tex], there exists [tex]a\in A[/tex], such that [tex]a<b+\epsilon[/tex].

b) There exists [tex]a\in A[/tex] and [tex]b\in B[/tex] such that  [tex]a<b[/tex].

We will prove that conditon a) is equivalent to [tex]inf(A)\leq inf(B)[/tex]

If a) is not satisfied, then it would exist [tex]b\in B[/tex] and [tex]\epsilon >0[/tex] such that, for every [tex]a\in A[/tex], [tex]a\geq b+\epsilon[/tex]. This implies that [tex]b+\epsilon[/tex] is a lower bound for A and in consequence

[tex]inf(A)\geq b+\epsilon > b\geq inf(B)[/tex]

Then, [tex]inf(A) \leq inf(B)[/tex] implies a).

If [tex]inf(A) \leq inf(B)[/tex] is not satisfied then, [tex]inf(A) > inf(B)[/tex] and in consequence exists [tex]b\inB[/tex] such that [tex]b-inf(A)=\epsilon >0[/tex]. Then [tex]b-\epsilon=inf(A)[/tex] and, for every [tex]a\in A[/tex],

[tex]b-\epsilon =inf(A)\leq a[/tex].

So, a) is not satisfied.

In conclusion, a) is equivalent to [tex]inf(A)\leq inf(B)[/tex]

Finally, observe that condition b) is not an appropiate condition to determine if [tex]inf(A)\leq inf(B)[/tex] or not. For example:

Answer:

The answer is: There are only 143034 books left after the students return to their classroom

Step-by-step explanation:

Lets call X= Number of books left in the library after the students return to their classroom.

B=Total of Books in the library

S=Total of books taken by the 3rd grade students

If each student  of the 3rd grage class take 2 books, the total of book taken by the 3rd grade students would be:

S=2 books per student * Total of students of the 3rdgrade

S=2*21=42

Then we must substract Total of Books in the library minus Total of books taken by the 3rd grade students. That would be:

X=B-S

X=143076-42

X=143034

I hope that this answer will help you

Answer:

The scale factor used to go from P to Q is 1/4

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

If two figures are similar, then the ratio of its areas i equal to the scale factor squared

Let

z ----> the scale factor

x -----> area of polygon Q

y -----> area of polygon P

[tex]z^{2}=\frac{x}{y}[/tex]

we have

[tex]y=72\ units^2[/tex]

Find the area of polygon Q

Divide the the area of polygon Q in two triangles and three squares

The area of the polygon Q is equal to the area of two triangles plus the area of three squares

see the attached figure N 2

Find the area of triangle 1

[tex]A=(1/2)(1)(2)=1\ units^2[/tex]

Find the area of three squares (A2,A3 and A4)

[tex]A=3(1)^2=3\ units^2[/tex]

Find the area of triangle 5

[tex]A=(1/2)(1)(1)=0.5\ units^2[/tex]

The area of polygon Q is

[tex]x=1+3+0.5=4.5\ units^2[/tex]

Find the scale factor

[tex]z^{2}=\frac{x}{y}[/tex]

we have

[tex]y=72\ units^2[/tex]

[tex]x=4.5\ units^2[/tex]

substitute and solve for z

[tex]z^{2}=\frac{4.5}{72}[/tex]

[tex]z^{2}=\frac{1}{16}[/tex]

square root both sides

[tex]z=\frac{1}{4}[/tex]

therefore

The scale factor used to go from P to Q is 1/4

The scaled factor did Diego used to go from P to Q is 1/4.

Given

The area of polygon P is 72 square units.

If two figures are similar, then the ratio of its areas is equal to the scale factor squared.

[tex]\rm z^2=\dfrac{x}{y}[/tex]

Where; z is the scale factor, x = area of polygon Q, y = area of polygon P.

Therefore,

The area of the first triangle is;

[tex]\rm Area = \dfrac{1}{2} \times base \times height\\\\Area = \dfrac{1}{2} \times 1 \times 2\\\\Area = 1 \ units[/tex]

The area of three squares (A2, A3, and A4)

[tex]\rm Area = 3(1)^2\\\\Area = 3\ square \ units[/tex]

The area of the 5th triangle is;

[tex]\rm Area = \dfrac{1}{2} \times base \times height\\\\Area = \dfrac{1}{2} \times 1 \times 1\\\\Area = 0.5 \ units[/tex]

Then,

The area of the polygon is;

x = 1 + 3 + 0.5 = 4.5 units

Therefore,

The scaled factor did Diego used to go from P to Q is;

[tex]\rm z^2=\dfrac{4.5}{72}\\\\z^2=\dfrac{1}{16}\\\\z=\dfrac{1}{4}\\\\[/tex]

Hence, the scaled factor did Diego used to go from P to Q is 1/4.

To know more about the Scale factor click the link given below.

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Answer:

Mass of glucose will be 36 gram

Step-by-step explanation:

We have given volume of solution V = 200 ml =0.2 L

Morality = 1 M

Molar mass of glucose = 180 g/mol

We know that morality is given by

[tex]molarity=\frac{number\ of\ moles}{volume\ of\ solution\ in\ liters}[/tex]

[tex]1=\frac{number\ of\ moles}{0.2}[/tex]

[tex]number\ of\ moles\ n =0.2[/tex]

We know that number of moles is given by

[tex]n=\frac{mass\ in\ gram}{molar\ mass}[/tex]

[tex]0.2=\frac{mass\ in\ gram}{180}[/tex]

Mass = 36 gram  

Answer:[tex]a = \frac{F}{m}[/tex]

[tex]v = \frac{m}{d}[/tex]

[tex]t = \frac{A - P}{Pr}[/tex]

[tex]h = \frac{2A}{a + b}[/tex]

[tex]W = \frac{P - 2L}{2}[/tex]

[tex]y = 2m - x[/tex]

[tex]y = -1.5x + 4[/tex]

[tex]t = \frac{v - u}{a}[/tex]

Step-by-step explanation:

[tex]F = ma[/tex]

[tex]\frac{F}{m} = \frac{ma}{m}[/tex]

[tex]\frac{F}{m} = a[/tex]

[tex]d = \frac{m}{v}[/tex]

[tex]vd = m[/tex]

[tex]\frac{vd}{d} = \frac{m}{d}[/tex]

[tex]v = \frac{m}{d}[/tex]

[tex]A = P + Prt[/tex]

[tex]A - P = Prt[/tex]

[tex]\frac{A - P}{Pr} = \frac{Prt}{Pr}[/tex]

[tex]\frac{A - P}{Pr} = t[/tex]

[tex]A = \frac{1}{2}h(a + b)[/tex]

[tex]2A = h(a + b)[/tex]

[tex]\frac{2A}{a + b} = \frac{h(a + b)}{a + b}[/tex]

[tex]\frac{2A}{a + b} = h[/tex]

[tex]P = 2(L + W)[/tex]

[tex]P = 2(L) + 2(W)[/tex]

[tex]P = 2L + 2W[/tex]

[tex]P - 2L = 2W[/tex]

[tex]\frac{P - 2L}{2} = \frac{2W}{2}[/tex]

[tex]\frac{P - 2L}{2} = W[/tex]

[tex]m = \frac{x + y}{2}\\2m = x + y\\2m - x = y[/tex]

[tex]3x + 2y = 8\\2y = -3x + 8\\\frac{2y}{2} = \frac{-3x + 8}{2}\\y = -1.5x + 4[/tex]

[tex]a = \frac{v - u}{t}\\at = v - u\\\frac{at}{a} = \frac{v - u}{a}\\t = \frac{v - u}{a}[/tex]

Answer:

The limit of this function does not exist.

Step-by-step explanation:

[tex]\lim_{x \to 3} f(x)[/tex]

[tex]f(x)=\left \{ {{9-3x} \quad if \>{x \>< \>3} \atop {x^{2}-x }\quad if \>{x\ \geq \>3 }} \right.[/tex]

To find the limit of this function you always need to evaluate the one-sided limits. In mathematical language the limit exists if

[tex]\lim_{x \to a^{-}} f(x) = \lim_{x \to a^{+}} f(x) =L[/tex]

and the limit does not exist if

[tex]\lim_{x \to a^{-}} f(x) \neq \lim_{x \to a^{+}} f(x)[/tex]

Evaluate the one-sided limits.

The left-hand limit

[tex]\lim_{x \to 3^{-} } 9-3x= \lim_{x \to 3^{-} } 9-3*3=0[/tex]

The right-hand limit

[tex]\lim_{x \to 3^{+} } x^{2} -x= \lim_{x \to 3^{+} } 3^{2}-3 =6[/tex]

Because the limits are not the same the limit does not exist.

Recall that we can write

[tex]f(x)^{g(x)}=e^{\ln f(x)^{g(x)}}=e^{g(x)\ln f(x)}[/tex]

so that when we take the derivative, we get by the chain rule

[tex]\left(f(x)^{g(x)}\right)'=(g(x)\ln f(x))' e^{g(x)\ln f(x)}=(g(x)\ln f(x))'f(x)^{g(x)}[/tex]

Here, we have [tex]f(x)=3-\sin2x[/tex] and [tex]g(x)=\sqrt[3]{x}=x^{1/3}[/tex]. By the product rule,

[tex]\left(x^{1/3}\ln(3-\sin2x)\right)'=\left(x^{1/3}\right)\ln(3-\sin2x)+x^{1/3}(\ln(3-\sin2x))'[/tex]

[tex]=\dfrac13x^{-2/3}\ln(3-\sin2x)+x^{1/3}\dfrac{(3-\sin2x)'}{3-\sin2x}[/tex]

[tex]=\dfrac13x^{-2/3}\ln(3-\sin2x)+x^{1/3}\dfrac{2\cos2x}{\sin2x-3}[/tex]

[tex]=\dfrac1{3x^{2/3}}\left(\ln(3-\sin2x)+3x\dfrac{2\cos2x}{\sin2x-3}\right)[/tex]

[tex]=\dfrac1{3\sqrt[3]{x^2}}\left(\ln(3-\sin2x)+\dfrac{6x\cos2x}{\sin2x-3}\right)[/tex]

So we have

[tex]y'=\dfrac{(3-\sin2x)^{\sqrt[3]{x}}}{3\sqrt[3]{x^2}}\left(\ln(3-\sin2x)+\dfrac{6x\cos2x}{\sin2x-3}\right)[/tex]

By means of logarithmic differentiation, the derivative of function [tex]y = (3 - \sin 2x)^{\sqrt[3]{x}}[/tex] is equal to [tex]y' = \left[\frac{\ln (3 - \sin 2x)}{\sqrt[3]{x^2} } + \frac{2\cdot \sqrt[3]{x}\cdot \cos 2x}{3 - \sin 2x} \right]\cdot (3 - \sin 2x)^{\sqrt[3]{x} }[/tex].

How to use logarithmic differentiation in a function containing trascendent functions

In this problem we find the case of a function that cannot be differentiated easily by direct differentiation, due to presence of trascendent function, we shall use an alternative approach known as logarithmic differentiation. First, write the entire expression described in the statement:

[tex]y = (3 - \sin 2x)^{\sqrt[3]{x}}[/tex]

Second, use logarithms and its properties to modify the expression:

[tex]\ln y = \ln (3 - \sin 2x)^{\sqrt[3]{x}}[/tex]

[tex]\ln y = \sqrt[3]{x}\cdot \ln (3 - \sin 2x)[/tex]

Third, use derivative rules and clear variable y':

[tex]\frac{y'}{y} = \frac{\ln (3 - \sin 2x)}{\sqrt[3]{x^2} } + \frac{2 \cdot \sqrt[3]{x} \cdot \cos 2x}{3 - \sin 2x}[/tex]

[tex]y' = \left[\frac{\ln (3 - \sin 2x)}{\sqrt[3]{x^2} } + \frac{2\cdot \sqrt[3]{x}\cdot \cos 2x}{3 - \sin 2x} \right]\cdot y[/tex]

Fourth, substitute on y and write the resulting expression:

[tex]y' = \left[\frac{\ln (3 - \sin 2x)}{\sqrt[3]{x^2} } + \frac{2\cdot \sqrt[3]{x}\cdot \cos 2x}{3 - \sin 2x} \right]\cdot (3 - \sin 2x)^{\sqrt[3]{x} }[/tex]

Thus, the derivative of function [tex]y = (3 - \sin 2x)^{\sqrt[3]{x}}[/tex] is equal to [tex]y' = \left[\frac{\ln (3 - \sin 2x)}{\sqrt[3]{x^2} } + \frac{2\cdot \sqrt[3]{x}\cdot \cos 2x}{3 - \sin 2x} \right]\cdot (3 - \sin 2x)^{\sqrt[3]{x} }[/tex].

Answer:  5

Step-by-step explanation:

Given : Jerome has two bags of different candies, one with 15 pieces of candy and one with 20 pieces.

Every prize will have one kind of candy, the same number of pieces.

Then to find the greatest number of pieces of candies possible in each price , we need to find the greatest common factor of 15 and 20.

Prime factorization of 15 and 20 :-

[tex]15=3\times5\\\\20=2\times2\times5[/tex]

∴ Greatest common factor of 15 and 20 = 5

Hence , the greatest number of pieces of candies possible in each prize = 5

Answer:

The number of all possible subsets of D is 8.

Step-by-step explanation:

Consider the provided set D={c,a,t}

The subset of D contains no elements: {  }

The subset of D contains one element each: {c} {a} {t}

The subset of D contains two elements each: {c, a} {a, t} {c, t}

The subset of D contains three elements: {c, a, t)

Hence, all possible subsets of D are { }, {c}, {a}, {t}, {c, a}, {a, t}, {c, t}, {c, a, t}

Therefore, number of all possible subsets of D is 8.

Or we can use the formula:

The number of subsets of the set is [tex]2^n[/tex] If the set contains ‘n’ elements.

There are 3 elements in the provided set, thus use the above formula as shown:

2³=8

Hence, the number of all possible subsets of D is 8.

Answer:

29 is answer.

Step-by-step explanation:

Given that the function​ s(t) represents the position of an object at time t moving along a line. Suppose s(2)=150 and s(5)=237.

To find average velocity of the object over the interval of time [1,3]

We know that derivative of s is velocity and antiderivative of velocity is position vector .

Since moving along a line equation of s is

use two point formula

[tex]\frac{s-150}{237-150} =\frac{t-2}{5-2} \\s=29t-58+150\\s=29t+92[/tex] gives the position at time t.

Average velocity in interval (1,3)

=[tex]\frac{1}{3-1} (s(3)-s(1))\\=\frac{1}{2} [87+58-29-58]\\=29[/tex]

The average velocity of the object over the interval [1, 3] is 43.5 units of distance per unit of time.

The average velocity of an object over an interval of time is found by dividing the change in position by the change in time. In this case, we want to find the average velocity of the object over the interval [1, 3].

To do this, we first need to find the change in position during this interval. Given that s(2) = 150 and s(5) = 237, we can calculate the change in position as follows:

s(3) - s(1) = (s(5) - s(3)) + (s(3) - s(1))

Therefore, the average velocity is (s(3) - s(1)) / (3 - 1). By substituting the given values, we find that the average velocity is (237 - 150) / (3 - 1) = 87 / 2 = 43.5 units of distance per unit of time.