A 12-m3 oxygen tank is at 17°C and 850 kPa absolute. The valve is opened, and some oxygen is released until the pressure in the tank drops to 650 kPa. Calculate the mass of oxygen that has been released from the tank if the temperature in the tank does not change during the process.

Answer:

Step-by-step explanation:

Here is the recommended solution method:

  3.4 + 2(9.7 – 4.8x) = 61.2 . . . . . given

  3.4 + 19.4 - 9.6x = 61.2 . . . . . . . distribute 2 to 9.7 and -4.8x

  22.8 - 9.6x = 61.2 . . . . . . . . . . . . combine 3.4 and 19.4

  -9.6x = 38.4 . . . . . . . . . . . . . . . . . subtract 22.8 from both sides

  x = -4 . . . . . . . . . . . . . . . . . . . . . . divide both sides by -9.6

_____

Alternate solution method using different steps

You can also "undo" what is done to the variable, in reverse order. The variable has these operations performed on it:

So, another possible solution method is this:

  3.4 + 2(9.7 – 4.8x) = 61.2 . . . . . given

  2(9.7 -4.8x) = 57.8 . . . . . . . . . . . add the opposite of 3.4 (undo add 3.4)

  9.7 -4.8x = 28.9 . . . . . . . . . . . . . divide by 2 (undo multiply by 2)

  -4.8x = 19.2 . . . . . . . . . . . . . . . . . add the opposite of 9.7 (undo add 9.7)

  x = -4 . . . . . . . . . . . . . . . . . . . . . . divide by -4.8 (undo multiply by -4.8)

Linear equation solutions are indeed the points where the lines or planes describing various linear equations intersect or connect. The candidate solution of a set of linear equations is indeed the collection of all feasible solution' values again for variables, and further calculation can be defined as follows:

Given:

[tex]\to \bold{3.4 + 2(9.7 -4.8x) = 61.2}\\\\[/tex]

To find:

Solve the linear equation=?

Solution:

[tex]\to \bold{3.4 + 2(9.7 -4.8x) = 61.2}\\\\\to \bold{3.4 + 19.4 -9.6x = 61.2}\\\\\to \bold{22.8 -9.6x = 61.2}\\\\\to \bold{ -9.6x = 61.2- 22.8}\\\\\to \bold{ -9.6x = 38.4}\\\\\to \bold{ x =- \frac{38.4}{9.6}}\\\\\to \bold{ x =- 4}\\\\[/tex]

Therefore, the steps are:

Distribute 2 to [tex]\bold{9.7\ and\ -4.8x}[/tex].    

Combine [tex]\bold{3.4\ and \ 19.4}[/tex].              

Subtract [tex]22.8[/tex] from both sides.      

Divide both sides by[tex]-9.6[/tex].            

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Answer:

The maximum number of fruits bag Harper can buy are 3.

Step-by-step explanation:

Let there be x bags of fruits.

Let there be y boxes of chocolates

Cost of 1 bag of fruit = $4.75

So, cost of x bags = $4.75x

Cost of one box of crackers costs = $3.50

As per the given situation, the inequality forms:

[tex]4.75x+3.50y\leq 15[/tex]

So, the maximum number of bags of fruit Harper can buy, is when she buys no box of cracker.

So, putting y = 0 in above inequality , we have,

[tex]4.75x+3.50(0)\leq 15[/tex]

=> [tex]4.75x+0\leq 15[/tex]

=> [tex]4.75x \leq 15[/tex]

[tex]x\leq 3.15[/tex] rounding to 3.

Hence, the maximum number of fruits bag Harper can buy are 3.

Answer:

the [tex]95\%[/tex] confidence interval for the population proportion is:

[tex]\left [0.3469, \hspace{0.1cm} 0.4387\right][/tex]

Step-by-step explanation:

To solve this problem, a confidence interval of [tex](1-\alpha) \times 100\%[/tex] for the population proportion will be calculated.

[tex]$$Sample proportion: $\bar P=0.3928$\\Sample size $n=751$\\Confidence level $(1-\alpha)\times100\%=99\%$\\$\alpha: \alpha=0.01$\\Z values (for a 99\% confidence) $Z_{\alpha/2}=Z_{0.005}=2.5758$\\\\Then, the (1-\alpha) \times 100\%$ confidence interval for the population proportion is given by:\\\\\left [\bar P - Z_{\frac{\alpha}{2}}\sqrt{\frac{\bar P(1- \bar P)}{n}}, \hspace{0.3cm}\bar P + Z_{\frac{\alpha}{2}}\sqrt\frac{\bar P(1- \bar P)}{n} \right ][/tex]

Thus, the [tex]95\%[/tex] confidence interval for the population proportion is:

[tex]\left [0.3928 - 2.5758\sqrt{\frac{0.3928(1-0.3928)}{751}}, \hspace{0.1cm}0.3928 + 2.5758\sqrt{\frac{0.3928(1-0.3928)}{751}} \right ]=\left [0.3469, \hspace{0.1cm} 0.4387\right][/tex]

Answer:

3x + y = 20

Step-by-step explanation:

The equation of line passing through two points is determined by formula:

[tex]y-y_{1}=\frac{y_{2}-y_{1}}{x_{2} - x_{1}}(x - x_{1})[/tex]

Here, (x₁ , y₁) = (4, 8)

and (x₂, y₂) = (6, 2)

Putting these value in above formula. We get,

[tex]y-8=\frac{2-8}{6 - 4}(x - 4)[/tex]

⇒ [tex]y-8=\frac{-6}{2}(x - 4)[/tex]

⇒ ( y - 8) = -3 (x - 4)

⇒ y - 8 = -3x + 12

⇒ 3x + y = 20

which is required equation.

Answer:

the estimated answer is A: 9.75.

(the actual answer is 9.66, so rounding up makes it 9.75)

Answer:

The fist four terms are 0, -1, -4, -6.

Step-by-step explanation:

Since [tex] k\geq 0 [/tex], the fist four terms of the sequence correspond to those in wich [tex] k [/tex] takes the values 0, 1, 2 and 3, that is

Answer:

Step-by-step explanation:

[tex]\dfrac{x}{3}=\dfrac{4}{9} \qquad\text{has x-coefficient $\frac{1}{3}$}[/tex]

Multiply by the reciprocal of the x-coefficient. Then you have ...

[tex]x=\dfrac{4}{3}[/tex]

The required simplified expression for 11 x 10³ is 11,000. Option B is correct.

Given that,
A simplified expression form of the expression 11 x 10³, is to be determined.

The process in mathematics to operate and interpret the function or expression to make the function or expression simple or more understandable is called simplifying and the process is called simplification.

Here,
= 11 x 10³
Since 10³ = 10 x 10 x 10 = 1000
= 11 x 1000
= 11,000

Thus, the required simplified expression for 11 x 10³ is 11,000. Option B is correct.

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Answer:

4.00512 cents

Step-by-step explanation:

Given:

Power output = 48 Watts

Time for which owner paddles = 8 hours

Selling price of the electricity = 10.43 cents/kWh

Now,

Power = Energy × Time

or

Power generated = 48 × 8 = 384 Wh = 0.384 kWh

now,

Money earned will be = Power generated × selling price per kWh

or

Money earned = 0.384 kWh ×  10.43 cents/ kWh = 4.00512 cents

Answer:  15

Step-by-step explanation:

Given : Elena has 60 colored pencils, and Lucy has 26 colored pencils.

Total colored pencils = [tex]60+26=86[/tex]

Let 'x' denotes the number of pencils Lucy has when she get pencils from Elena , and Elena left with 'y' pencils.

Then, by considering the given information we have the following system :-

[tex]\text{Total pencils : }x+y=86-----(1)\\\\\text{Colored pencils left with Elena : }y=4+x-----(2)[/tex]

We can rewritten the equation (2) as

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

Now, add equation (1) and (2), we get

[tex]2y=90\\\\\Rightarrow\ y=\dfrac{90}{2}=45[/tex]

Put the value of y in (2), we get

[tex]45=4+x\\\\\Rightarrow\ x=45-4=41[/tex]

It means, the number of pencils Lucy has now =41

The number of pencils Elena give to Lucy= [tex]41-26=15[/tex]

Hence, Elena must give 15 pencils to lucy so that elena will have 4 more colored pencils than Lucy.

Approximately 0.02 mg of the drug miglitol (Glyset) would be detected in human breast milk.

Given that, approximately 0.02% of a 100-mg dose appears in human breast milk.

To calculate the quantity of drug detected in milligrams following a single dose, we can use the given information that approximately 0.02% of a 100-mg dose appears in human breast milk.

Step-by-step calculation:

1. Convert 0.02% to a decimal by dividing it by 100: 0.02/100 = 0.0002.

2. Multiply the decimal by the dose of the drug: 0.0002 * 100 mg = 0.02 mg.

Therefore, following a single dose, approximately 0.02 mg of the drug miglitol (Glyset) would be detected in human breast milk.

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Answer:

There is a 62.7% probability that a transaction can be completed in less than 100 milliseconds.

Step-by-step explanation:

This a probability problem.

We have the following probabilities:

-70% probability that a transaction is an addition of an item.

-30% probability that a transaction is a change to an item

-81% probability that an addition can be completed in less than 100 milliseconds

-20% probability that change can be completed in less than 100 milliseconds

The probability P that a transaction can be completed in less than 100 milliseconds is:

[tex]P = P_{1} + P_{2}[/tex]

In which [tex]P_{1}[/tex] is the probability that the transaction is an addition and it takes less than 100 milliseconds. So

[tex]P_{1} = 0.7*0.81 = 0.567[/tex]

[tex]P_{2}[/tex] is the probability that the transaction is a change and it takes less than 100 milliseconds. So

[tex]P_{2} = 0.2*0.3 = 0.06[/tex]

So

[tex]P = P_{1} + P_{2} = 0.567 + 0.06 = 0.627[/tex]

There is a 62.7% probability that a transaction can be completed in less than 100 milliseconds.

Answer:  The proof is given below.

Step-by-step explanation:  We are given to show that the following equality is true :

[tex]^{n+1}C_r=^nC_{r-1}+^nC_r.[/tex]

We know that

the number of combinations of n different things taken r at a time is given by

[tex]^nC_r=\dfrac{n!}{r!(n-r)!}.[/tex]

Therefore, we have

[tex]R.H.S.\\\\=^nC_{r-1}+^nC_r\\\\\\=\dfrac{n!}{(r-1)!(n-(r-1))!}+\dfrac{n!}{(r)!(n-r)!}\\\\\\=\dfrac{n!}{(r-1)!(n-r+1)!}+\dfrac{n!}{(r)!(n-r)!}\\\\\\=\dfrac{n!}{(r-1)!(n-r+1)(n-r)!}+\dfrac{n!}{r(r-1)!(n-r)!}\\\\\\=\dfrac{n!}{(r-1)!(n-r)!}\left(\dfrac{1}{n-r+1}+\dfrac{1}{r}\right)\\\\\\=\dfrac{n!}{(r-1)!(n-r)!}\left(\dfrac{r+n-r+1}{(n-r+1)r}\right)\\\\\\=\dfrac{n!}{(r-1)!(n-r)!}\times\dfrac{n+1}{(n-r+1)r}\\\\\\=\dfrac{(n+1)!}{r!(n-r+1)!}\\\\\\=\dfrac{(n+1)!}{r!((n+1)-r)!}\\\\\\=^{n+1}C_r\\\\=L.H.S.[/tex]

Thus, [tex]^{n+1}C_r=^nC_{r-1}+^nC_r.[/tex]

Hence proved.

Answer:

Part a) value of [tex]\lambda [/tex] such that all the solutions tend to zero equals 1.

Part b)

For a particular solution to tend to 0 will depend on the boundary conditions.

Step-by-step explanation:

The given differential equation is

[tex]y'+\lambda y=e^{-t}[/tex]

This is a linear differential equation of first order of form [tex]\frac{dy}{dt}+P(t)\cdot y=Q(t)[/tex] whose solution is given by

[tex]y\cdot e^{\int P(t)dt}=\int e^{\int P(t)dt}\cdot Q(t)dt[/tex]

Applying values we get

[tex]y\cdot e^{\int \lambda dt}=\int e^{\int \lambda dt}\cdot e^{-t}dt\\\\y\cdot e^{\lambda t}=\int (e^{(\lambda -1)t})dt\\\\y\cdot e^{\lambda t}=\frac{e^{(\lambda -1)t}}{(\lambda -1)}+c\\\\\therefore y(t)=\frac{c_{1}}{\lambda -1}(e^{-t}+c_{2}e^{-\lambda t})[/tex]

here [tex]c_{1},c_{2}[/tex] are arbitrary constants

part 1)

For all the function to approach 0 as t approaches infinity we have

[tex]y(t)=\lim_{t\to \infty }[\frac{c_{1}}{\lambda -1}(e^{-t}+c_{2}e^{-\lambda t})]\\\\y(\infty )=\frac{c_{1}}{\lambda -1}=0\\\\\therefore \lambda =1[/tex]

Part b)

For a particular solution to tend to 0 will depend on the boundary conditions as [tex]c_{1},c_{2}[/tex] are arbitrary constants

Answer:

a)The linear function for [tex]P(t)[/tex] is:

[tex]P(t) = 2000 - 100*t[/tex]

b)The exponential function for [tex]P(t)[/tex] is:

[tex]P(t) = 2000e^{-0.055t}[/tex]

Step-by-step explanation:

(a) Suppose that P(t) is a linear function. Find a formula for P(t):

[tex]P(t)[/tex] can be modeled by a linear function in the following format.

[tex]P(t) = P_{0} - r*t[/tex], in which [tex]P_{0}[/tex] is the initial number of bacteria cells in the dish, t is the time and r is the rate that the number decreases.

Since the dish initially contains 2000 bacteria cells, [tex]P_{0} = 2000[/tex]

We have

[tex]P(t) = 2000 - r*t[/tex]

An antibiotic is introduced and after 4 hour, there are now 1600 bacteria cells present. So [tex]P(4) = 1600[/tex]. With this information, we can find the value of r.

[tex]P(t) = 2000 - r*t[/tex]

[tex]1600 = 2000 - r*(4)[/tex]

[tex]4r = 400[/tex]

[tex]r = \frac{400}{4}[/tex]

[tex]r = 100[/tex]

So, the linear function for [tex]P(t)[/tex] is:

[tex]P(t) = 2000 - 100*t[/tex]

b) Suppose that P(t) is an exponential function. Find a formula for P(t)

[tex]P(t)[/tex] can also be modeled by an exponential function in the following format:

[tex]P(t) = P_{0}e^{rt}[/tex]

The values mean the same as in a). We use the fact that [tex]P(4) = 1600[/tex] to find r.

[tex]P(t) = 2000e^{rt}[/tex]

[tex]1600 = 2000e^{4r}[/tex]

[tex]e^{4r} = \frac{1600}{2000}[/tex]

[tex]e^{4r} = 0.8[/tex]

[tex]ln e^{4r} = ln 0.8[/tex]

[tex]4r = -0.22[/tex]

[tex]r = \frac{-0.22}{4}[/tex]

[tex]r = -0.055[/tex]

So, the exponential function for [tex]P(t)[/tex] is:

[tex]P(t) = 2000e^{-0.055t}[/tex]

Answer:

Claim :The proportion, p, of full-term babies born in their hospital that weigh more than 7 pounds is 36%.

n =  170

x = 56

We will use one sample proportion test  

[tex]\widehat{p}=\frac{x}{n}[/tex]

[tex]\widehat{p}=\frac{56}{170}[/tex]

[tex]\widehat{p}=0.3294[/tex]

The proportion, p, of full-term babies born in their hospital that weigh more than 7 pounds is 36%.

[tex]H_0:p \neq 0.36 \\H_a:p= 0.36[/tex]

Formula of test statistic =[tex]\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]

                                       [tex]=\frac{0.3294-0.36}{\sqrt{\frac{0.36(1-0.36)}{170}}}[/tex]

                                        =−0.8311

Now refer the p value from the z table

P-Value is .202987 (Calculated by online calculator)

Level of significance α = 0.05

Since p value < α

So we reject the null hypothesis .

Hence the claim is true

Final answer:

To find (g°f)(0), evaluate f(0) to get 6, then apply g to this result to also get 6. Hence, (g°f)(0) equals 6.

Explanation:

The question is asking to evaluate the composition of two functions, (g°f)(x), at the value x = 0. Composition of functions involves applying one function to the result of another. In this case, since f(x) = 2x + 6 and g(x) = x, we first find f(0), which is the result of plugging 0 into f(x).

Let's calculate:
f(0) = 2(0) + 6 = 6.
Next, we apply g to this result:
g(f(0)) = g(6) = 6.
Therefore, (g°f)(0) = 6.

Answer and Explanation:

Given : A lab technician is tested for her consistency by taking multiple measurements of cholesterol levels from the same blood sample.

The target accuracy is a variance in measurements of 1.2 or less. If the lab technician takes 16 measurements and the variance of the measurements in the sample is 2.2.

To find :

1) Does this provide enough evidence to reject the claim that the lab technician’s accuracy is within the target accuracy?

2)Compute the value of the appropriate test statistic ?

Solution :

1) n=16 number of sample

The target accuracy is a variance in measurements of 1.2 or less i.e. [tex]\sigma_1^2 =1.2[/tex]

The variance of the measurements in the sample is 2.2 i.e. [tex]\sigma_2^2=2.2[/tex]

According to question,  

We state the null and alternative hypotheses,

Null hypothesis [tex]H_o : \text{var}^2 \geq 1.2[/tex]

Alternative hypothesis [tex]H_a : \text{var}^2<1.2[/tex]

We claim the alternative hypothesis.

2) Compute the value of the appropriate test statistic.

Using Chi-square,

[tex]\chi =\frac{(n-1)\sigma_2^2}{\sigma_1^2}[/tex]

[tex]\chi =\frac{(16-1)(2.2)}{(1.2)}[/tex]

[tex]\chi =\frac{(15)(2.2)}{1.2}[/tex]

[tex]\chi =\frac{33}{1.2}[/tex]

[tex]\chi =27.5[/tex]

Therefore, The value of the appropriate test statistic is 27.5.

To determine if the lab technician's accuracy is within the target accuracy, a one-sample variance test can be performed using the chi-square statistic.

To determine if the lab technician's accuracy is within the target accuracy, we can perform a hypothesis test. We will use a one-sample variance test to compare the sample variance to the target variance.

The appropriate test statistic for a one-sample variance test is the chi-square statistic. The chi-square statistic is calculated by taking the sample variance and dividing it by the target variance, then multiplying by the degrees of freedom.

In this case, we have 16 measurements and a target variance of 1.2. The sample variance is 2.2. We can calculate the test statistic using the formula chi-square = (n-1) * (sample variance / target variance). Plugging in the values, we get chi-square = (16-1) * (2.2 / 1.2) = 29.67.

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The average speed for this additional distance is 6.57 mi/h

The given parameters include;

To find:

The average speed formula is given as;

[tex]average\ speed = \frac{total \ distance }{total \ time} \\\\average\ speed = \frac{16.5 \ miles \ +\ 5.5 \ miles}{\frac{16.5 \ mi}{21.8 \ mi/h}\ \ +\ \ \frac{5.5 \ mi}{v} } \\\\13.8 \ mi/h = \frac{(16.5 \ miles \ +\ 5.5 \ miles)}{(\frac{16.5 \ mi}{21.8 \ mi/h}\ \ +\ \ \frac{5.5 \ mi}{v} ) }\\\\13.8 = \frac{(16.5 \ +\ 5.5 )}{(\frac{16.5 }{21.8 }\ \ +\ \ \frac{5.5 }{v} ) }\\\\13.8 = \frac{22 }{0.757\ \ +\ \ \frac{5.5 }{v} }\\\\13.8 (0.757\ \ +\ \ \frac{5.5 }{v} ) = 22\\\\[/tex]

[tex]10.446 + \frac{75.9}{v} = 22\\\\\frac{75.9}{v} = 22-10.446\\\\\frac{75.9}{v} = 11.554\\\\v = \frac{75.9}{11.554} \\\\v = 6.57 \ mi/h[/tex]

Thus, the average speed for this additional distance is 6.57 mi/h

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The runner must run at an average speed of 6.57 mi/h to achieve the overall average speed of 13.8 mi/h for the entire biathlon race.

To solve for the average running speed needed to achieve an overall average speed of 13.8 mi/h in the biathlon, we will use the concept of weighted averages. The total time for both cycling and running can be expressed in terms of the distances traveled and the speeds for each segment.

First, we calculate the time taken to complete the cycling portion, which is 16.5 miles at an average speed of 21.8 mi/h. The formula for time is distance divided by speed:

Next, we need to find the running speed. Let's call the average running speed 'R'. Using the same formula for time:

The overall average speed for the entire race is the total distance divided by the total time. The total distance is 16.5 miles + 5.5 miles = 22 miles. Let 'T' represent the total time for the race.

Average speed for the entire race = Total distance / Total time = 22 miles / T

Given that the overall average speed must be 13.8 mi/h, we set up the equation:

13.8 mi/h = 22 miles / T

Now we can express 'T' as the sum of the cycling time and running time:

T = Time (cycling) + Time (running) = 0.7565 hours + (5.5 miles / R)

Substitute this into our average speed equation:

13.8 mi/h = 22 miles / (0.7565 hours + (5.5 miles / R))

Now, we solve for 'R', which represents the average running speed. We cross multiply and isolate 'R' to obtain:

(13.8 mi/h) (0.7565 hours + (5.5 miles / R)) = 22 miles

This simplifies to:

10.4469 hours + 75.9 miles / R = 22 miles

Subtract 10.4469 hours from both sides:

75.9 miles / R = 11.5531 miles

Finally, divide 75.9 by 11.5531 to find the average running speed 'R':

R = 6.57 mi/h

The runner must run at an average speed of 6.57 mi/h to have an overall average speed of 13.8 mi/h for the biathlon.

Answer:

a) It is a proposition .

b) It is not a proposition.

c) It is a proposition.

d) It is a proposition.

e) It is a proposition.

f) It is not a proposition.

Step-by-step explanation:

a) The integer 36 is even: It is a proposition, since this statement can be assigned a true value. If 36 is an even number, the statement is true, but if 36 is an odd number, the statement is false.

b) Is the integer 315 - 8 even ?: It is not a proposition, since this question cannot be assigned a true value.

c) The product of 3 and 4 is 11: It is a proposition, since this statement can be assigned a true value. If 3x4 = 11, the statement is true, but if 3x4 is not 11, the statement is false.

d) The sum of x and y is 12: It is a proposition, since, this statement can be assigned a true value. If x + y = 12, the statement is true, but if x + y is not 12, the statement is false.

e) If x> 2, then x 2> 3: It is a proposition, since, this statement can be assigned a truth value.

f) 52 - 5 + 3: It is not a proposition, since this statement cannot be assigned a true value.

Answer:

  z = (3/8)y - 11

Step-by-step explanation:

Undo what has been done to z, in reverse order. Here, 11 is added and the sum is multiplied by 8/3.

To undo the multiplication, multiply the equation by the inverse, 3/8:

  (3/8)(8/3)(z+11) = (3/8)y

  z +11 = (3/8)y . . . . . . . . . simplify

To undo the addition of 11, add the opposite of 11 to the equation.

  z +11 -11 = (3/8)y -11

  z = (3/8)y -11 . . . . . . . . . simplify

Answer:

a) [tex](3x+y)^5=243x^5+405x^4y+270x^3y^2+90x^2y^3+15xy^4+y^5[/tex].

b) The middle term in the expansion is [tex]\frac{6}{x}[/tex].

c) The coefficient of [tex]x^{11}[/tex] is 120.

Step-by-step explanation:

Remember that the binomial theorem say that [tex](x+y)^n=\sum_{k=0}^{n} \binom{n}{k}x^{n-k}y^{k}[/tex]

a) [tex](3x+y)^5=\sum_{k=0}^5\binom{5}{k}3^{n-k}x^{n-k}y^k[/tex]

Expanding we have that

[tex]\binom{5}{0}3^5x^5+\binom{5}{1}3^4x^4y+\binom{5}{2}3^3x^3y^2+\binom{5}{3}3^2x^2y^3+\binom{5}{4}3xy^4+\binom{5}{5}y^5[/tex]

symplifying,

[tex](3x+y)^5=243x^5+405x^4y+270x^3y^2+90x^2y^3+15xy^4+y^5[/tex].

b) The middle term in the expansion of [tex](\frac{1}{x} +\sqrt{x})^4=\sum_{k=0}^{4}\binom{4}{k}\frac{1}{x^{4-k}}x^{\frac{k}{2}}[/tex] correspond to k=2. Then [tex]\binom{4}{2}\frac{1}{x^2}x^{\frac{2}{2}}=\frac{6}{x}[/tex].

c) [tex](x^2+\frac{1}{x})^{10}=\sum_{k=0}^{10}\binom{10}{k}x^{2(10-k)}\frac{1}{x^k}=\sum_{k=0}^{10}\binom{10}{k}x^{20-2k}\frac{1}{x^k}=\sum_{k=0}^{10}\binom{10}{k}x^{20-3k}[/tex]

Since we need that 11=20-3k, then k=3.

Then the coefficient of [tex]x^{11}[/tex] is [tex]\binom{10}{3}=120[/tex]

Answer:

[tex]I=\$47.25[/tex]

Step-by-step explanation:

we know that

The simple interest formula is equal to

[tex]I=P(rt)[/tex]

where

I is the Final Interest Value

P is the amount of money borrowed

r is the rate of interest  

t is Number of Time Periods

in this problem we have

[tex]t=9/12\ years\\ P=\$900\\r=0.07[/tex]

substitute in the formula above

[tex]I=900(0.07(9/12))[/tex]

[tex]I=\$47.25[/tex]

Final answer:

Margo will pay $47.25 in interest on a $900 loan with a 7% annual interest rate after 9 months, after converting the rate to a decimal and the time to years for the calculation.

Explanation:

The question asks how much interest Margo will pay on a $900 loan with an annual interest rate of 7% after 9 months. To calculate this, we need to know that Interest = Principal × Rate × Time, where the principal is the amount borrowed, the rate is the annual interest rate (as a decimal), and the time is the period of the loan in years

First, convert the annual interest rate from a percentage to a decimal by dividing by 100: 7% / 100 = 0.07. Next, convert the loan period from months to years since the interest rate is annual. There are 12 months in a year, so 9 months is equal to 9/12 or 0.75 years.

Now, calculate the interest: $900 (Principal) × 0.07 (Rate) × 0.75 (Time) = $47.25. Therefore, Margo will pay $47.25 in interest.

Answer:

(a) in the step-by-step explanation

(b) The optimal solution is 8 chairs and 4 tables.

(c) Graph attached

Step-by-step explanation:

(a)

C: number of small chairs

T: number of large tables

Maximize Income = 9T + 3C

Restrictions:

Wood: 4T+C<=24

Glazing: 2T+C<=16

In the graph its painted in green the "feasible region" where lies every solutions that fit the restrictions.

One of the three points marked in the graph is the optimal solution.

Point 1 (C= 16, T= 0)

Income = 9*0+3*16=$ 48

Point 2 (C=8, T=4)

Income = 9*4+3*8 = $ 60

Point 3 (C=0, T=6)

Income = 9*6+3*0 = $ 54

The optimal solution is 8 chairs and 4 tables.

Answer:

a. If I Catch the 8:05 bus, then I will arrive on time for work.

b. If an integer is divisible by 9, then that integer is divisible by 3.

C. If the Cubs win tomorrow's game, then they win the pennant and if the Cubs have won the pennant it is because they will have won the game tomorrow.

Step-by-step explanation:

Final answer:

To rewrite the statements in 'If...Then' form: 1. If you catch the 8:05 bus, then you will be on time for work. 2. If a number is divisible by 9, then it is divisible by 3. 3. If the Cubs win the pennant, then they have won tomorrow's game.

Explanation:

Rewriting Statements in 'If...Then' Form

To rephrase the given statements in 'If...Then' form while identifying the sufficient and necessary conditions, we'll look at each statement individually:

Catching the 8:05 bus is a sufficient condition for being on time for work. This can be rewritten as: If you catch the 8:05 bus, then you will be on time for work.

Being divisible by 3 is a necessary condition for a number to be divisible by 9. The 'If...Then' form is: If a number is divisible by 9, then it is divisible by 3.

The Cubs will win the pennant only if they win tomorrow's game. In the 'If...Then' form: If the Cubs win the pennant, then they have won tomorrow's game.

Each of these rephrased statements now clearly shows the reliance of the consequent (the outcome) on the antecedent (the condition).

Answer:

The energy of each photon is [tex]6.468 \times 10^{-26}[/tex] Joule.

Step-by-step explanation:

Consider the provided information.

According to the plank equation:

[tex]E=h\nu[/tex]

Where E is the energy of photon, h is the plank constant and [tex]\nu[/tex] is the frequency.

It is given that [tex]h= 6.6 \times10^{-34}[/tex] and [tex]\nu=98MHz[/tex]

98Mhz = [tex]98\times 10^6Hz[/tex]

Substitute the respective value in plank equation.

[tex]E=6.6\times 10^{-34}\times 98\times 10^6[/tex]

[tex]E=6.6\times 98\times 10^{-34+6}[/tex]

[tex]E=646.8 \times10^{-28}[/tex]

[tex]E=6.468 \times 10^{-26}[/tex]

Hence, the energy of each photon is [tex]6.468 \times 10^{-26}[/tex] Joule.

Answer:

The amount of principal Jamie borrowed was $3600.

Step-by-step explanation:

Let the principal of loan is = Po

[tex]P=Po\times0.12\times0.5[/tex]

Here r = 12% or 0.12

t = 6 months or 0.5 years

Now as given : Po+Interest=3816

[tex]Po+0.06Po=3816[/tex]

[tex]=>1.06Po=3816[/tex]

So, Po=3600

Hence, the amount of principal Jamie borrowed was $3600.

Answer: 92378

Step-by-step explanation:

Given : At the grocery store, Hosea has narrowed down his selections to 7 vegetables, 6 fruits, 5 cheeses, and 6 whole grain breads.

Total items : 7+6+5+6=24

If he wants all cheeses  , then the remaining items needed to be select = 15-5=10

Total items left from which he will select = 24-5=19

No. of combinations of r things out of n : [tex]C(n;r)=\dfrac{n!}{r!(n-r)!}[/tex]

The combination of 10 things selecting from 19 things given by :-

[tex]C(5;5)\timesC(19;10)=\dfrac{19!}{10!(19-10)!}\\\\=(1)\times\dfrac{19\times18\times17\times16\times15\times14\times13\times12\times11\times10!}{10!9!}\ \because ^nC_n=1\\\\=92378[/tex]

Hence, the number of ways he can choose 15 items to buy if he wants all 5 cheeses =92378

Final answer:

If Hosea wants to buy all 5 cheeses, there are 19 items he can choose from the remaining categories. Using the combination formula, the number of ways to choose 10 items from a set of 19 is calculated to be 92378.

Explanation:

If Hosea wants to buy all 5 cheeses, he needs to choose 10 more items from the remaining categories (vegetables, fruits, and whole grain breads). There are 7 vegetables, 6 fruits, and 6 whole grain breads, which means he can choose from a total of 7+6+6 = 19 items in those categories. Since Hosea can only choose 10 more items, he needs to calculate the number of ways to choose 10 items from a set of 19. This can be calculated using the combination formula.

The number of ways to choose 'r' items from a set of 'n' items is given by the combination formula: C(n, r) = n! / (r! * (n-r)!). In this case, n = 19 (the number of items to choose from) and r = 10 (the number of items to choose). Evaluating the formula, we get: C(19, 10) = 19! / (10! * (19-10)!).

By simplifying the factorial expressions and performing the calculations, we find that there are 92378 ways for Hosea to choose his 15 items.

Answer:

16.

Step-by-step explanation:

The ratio is 4:1 so 4 / (4 + 1) = 4/5 of the total is green peppers.

So it is 20 * 4/5 = 16 .

Answer:

e) none

Step-by-step explanation:

The first thing that we need to realise is that this question is set to a practical example of a ball being tossed into the air for a period of time. Thus, we know that:

a) time cannot be negative

b) a ball cannot travel a negative distance in height from where it was tossed it into the air (given that this starting position is also the ground)

In the context of what the problem asks us to achieve, which is to find the domain, we now know that:

a) t ≥ 0

b) the function h(t) is only practical when h(t) ≥ 0

In other words, we know that the domain starts at t = 0, however we need to find when the ball drops back onto the ground. We can do this by solving h(t) = 0:

h(t) = -16t^(2) + 96t

0 = -16t^(2) + 96t

0 = -16t(t - 6) (Factorise -16t^(2) + 96t)

So, now we get:

-16t = 0, therefor t = 0

or

t - 6 = 0, therefor t = 6

Thus, the ball is on the ground at t = 0 seconds and t = 6 seconds; it is between these two values of t that the domain exists and that the problem is practical.

Now, looking at the multiple choice options, it seems as though none of them are correct, therefor the answer would be e) none.

The other way to work through this question (particularly if it is just multiple choice) is that, after realising that the ball is tossed into the air at t = 0 seconds and then drops at some point later, we could already discount a), b) and c) as answers since they include infinity in the domain, which is not practical for this problem.

Then, we could substitute t = 5 into the equation to get:

h(t) = -16(5)^2 + 96*5

h(t) = -16*5*5 + 96*5

h(t) = -80*5 + 96*5

Since we need h(t) to be 0 at the end value of the domain, this is not the correct answer. Thus, the answer is e) none.