PLEASE HELP!!
Demonstrate your understanding of how to solve exponential equations by rewriting the base. Solve the problem below fully and explain all the steps...

25^3k = 625

Answer:

The solution of the given equation is [tex]\sqrt[3]{4}-\sqrt[3]{2}[/tex].

Step-by-step explanation:

According to the cardano's method, the solution of the equation is x=u-v. If the equation is

[tex]x^3+px=q[/tex]

Where [tex]u^3-v^3=q[/tex]

[tex]3uv=p[/tex]

The given equation is

[tex]x^3+6x=2[/tex]

Here p=6 and q=2.

[tex]u^3-v^3=2[/tex]                 .... (1)

[tex]3uv=6[/tex]

[tex]uv=2[/tex]

Taking cube both the sides.

[tex]u^3v^3=8[/tex]

Multiply both sides by 4.

[tex]4u^3v^3=32[/tex]             .... (2)

Taking square both the sides of equation (1).

[tex](u^3-v^3)^2=2^2[/tex]

[tex](u^3)^2-2u^3v^3+(v^3)^2=4[/tex]       .... (3)

Add equation (2) and (3).

[tex](u^3)^2-2u^3v^3+(v^3)^2+4u^3v^3=4+32[/tex]

[tex](u^3+v^3)^2=36[/tex]

Taking square root both the sides.

[tex]u^3+v^3=6[/tex]             .... (4)

On adding equation (1) and (4), we get

[tex]2u^3=8[/tex]

[tex]u^3=4[/tex]

[tex]u=\sqrt[3]{4}[/tex]

On subtracting equation (1) and (4), we get

[tex]-2v^3=-4[/tex]

[tex]v^3=2[/tex]

[tex]v=\sqrt[3]{2}[/tex]

The solution of the equation is

[tex]x=u-v=\sqrt[3]{4}-\sqrt[3]{2}[/tex]

Therefore the solution of the given equation is [tex]\sqrt[3]{4}-\sqrt[3]{2}[/tex].

Final answer:

Solving x³+6x=2 using Cardano's method involves rewriting the equation to match the standard form of a depressed cubic equation, calculating the required constants, and finally, applying these constants to find the roots.

First, let's rewrite the equation x³+6x-2 = 0 as x³+6x = 2 to match the standard form of a depressed cubic equation which is x³ +px = q. Here, p = 6 and q = 2.

Next, we calculate the value t = sqrt[(q/2)² + (p/3)³]. So, t = sqrt[(1)² + (2)³] = sqrt[1 + 8] = 3.

Using these values, we can now calculate the roots. We know the roots are given by the formulaes u-v where u = cubicroot(q/2 + t) and v = cubicroot(q/2 - t). So, u = cubicroot(1 + 3) = 2, and v = cubicroot(1 - 3) = - root(2).

Therefore, the roots of the given polynomial equation are x = u - v = 2 - (- root(2)) = 2 + root(2).

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Step-by-step explanation:

hi I have answered ur question

Answer:

Step-by-step explanation:

Sugar and Salt even thought they both dissolve in water they both dissolve in different ways. When salt dissolves in water, its individual types of ions are torn apart from each other, while Sugar molecules stay together when dissolved in water, and therefore the molecules remain the same when dissolved in water.

This being said in science using your senses can be just as valuable as using calculations. In this case both Sugar and Salt taste differently. Sugar is sweet while Salt is salty. Therefore tasting the substance can be the easiest and most accurate way of determining the substance.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

Final answer:

To determine if an unknown sample contains sugar, salt, or both, examine their solubility properties and chemical reactivity. Comparing the density of the sample against reference values of pure substances can provide preliminary identification, while testing for chloride ions with silver nitrate confirms the presence of salt.

Explanation:

Identifying Sugar and Salt in an Unknown Sample

To determine whether an unknown sample contains sugar, salt, or both, we must identify the physical and chemical properties that distinguish these substances. Salt (sodium chloride) and sugar (sucrose) have distinct solubility properties and chemical reactivity, which we can use to identify them when dissolved in water.

Solubility and Density Test

Both sugar and salt are highly soluble in water, but we can compare their densities to make a preliminary identification. A known volume of each substance is weighed and their densities calculated. Salt generally has a greater density than sugar. If the unknown sample has a certain mass, comparing it with the reference densities may provide an initial indication.

Chemical Reactivity Test

To confirm the identity of the substances, a chemical reactant such as silver nitrate can be introduced to the water solution of the unknown sample. If a white precipitate forms, it indicates the presence of chloride ions, which suggests the presence of salt. Since sugar does not produce a precipitate with silver nitrate, its absence would indirectly indicate the presence of sugar.

Performing these tests will allow us to determine if the unknown sample is sugar, salt, or a mixture of both. The greater the discrepancy between the calculated density and the known densities of pure sugar or pure salt, the more likely it is that the sample is a mixture.

Answer:

Step-by-step explanation:

Assuming A as x axis and B as y axis the equations are

[tex]2x+0.5y\geq 90\quad \left(1\right)\\0.75x+5y\geq 200\quad \left(2\right)\\0.75x+1.5y\leq 150\quad \left(3\right)\\x,y\geq 0\quad \left(4\right)[/tex]

Solving equations (2) and (3) we get

x=171.429 y=14.286

Solving equations (1) and (3) we get

x=22.857 y=88.571

Solving equations (1) and (2) we get

x=36.364 y=34.545

The area enclosing the above three points is the feasible region.

Answer:

Step-by-step explanation:

See if the attachment below helps you with this.

Check the picture below.

so the hyperbola looks more or less like so, with a = 6, and its center at the origin.

[tex]\bf \textit{hyperbolas, vertical traverse axis } \\\\ \cfrac{(y- k)^2}{ a^2}-\cfrac{(x- h)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h, k\pm a)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2}\\ asymptotes\quad y= k\pm \cfrac{a}{b}(x- h) \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]

[tex]\bf \begin{cases} a=6\\ h=0\\ k=0\\ \stackrel{asymptotes}{y=\pm\frac{3}{4}x} \end{cases}\implies \stackrel{\textit{using the positive asymptote}}{0+\cfrac{6}{b}(x-0)=\cfrac{3}{4}x}\implies \cfrac{6x}{b}=\cfrac{3x}{4}\implies 24x=3xb \\\\\\ \cfrac{24x}{3x}=b\implies 8=b \\\\[-0.35em] ~\dotfill\\\\ \cfrac{(y- 0)^2}{ 6^2}-\cfrac{(x- 0)^2}{ 8^2}=1\implies \cfrac{y^2}{36}-\cfrac{x^2}{64}=1[/tex]

Answer:

company B

Step-by-step explanation:

average salary of Company A(μ) = $51,500

standard deviation of Company A (σ)=  $2,175.

average salary of Company B(μ) = $46,820

standard deviation of Company B(σ) =$5,920

desired salary(x) = $55,000

z-score for company A = [tex]\dfrac{x-\mu}{\sigma}[/tex]

                                    = [tex]\dfrac{55000-51500}{ 2175} = 1.61[/tex]

z-score for company A = [tex]\dfrac{x-\mu}{\sigma}[/tex]

                                    =  [tex]\dfrac{55000-46820}{ 5920} = 1.38[/tex]

higher the value of z less chances of getting the desired salary hence company B has value of z is less so, the chances of getting desired salary is more in company B.

Jason has to calculate z-score to compare his desired salary with the average salaries at two different companies. The z-score for Company A is 1.61 and for Company B is 1.38. Hence, the desired wage of $55,000 is 1.61 and 1.38 standard deviations away from the mean salaries at Company A and Company B, respectively.

The z-score is a measure of how many standard deviations an observation or datum is from the mean. To calculate Jason's z-score at each company, we would subtract the mean salary at that company from $55,000 and then divide by the standard deviation for that company.

For Company A: [tex]Z_A = ($55,000 - $51,500) / $2,175 = 1.61.[/tex]

For Company B: [tex]Z_B = ($55,000 - $46,820) / $5,920 = 1.38.[/tex]

Therefore, Jason's desired salary of $55,000 is 1.61 standard deviations away from the mean at Company A and 1.38 standard deviations away from the mean at Company B.

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Answer: 0.7257

Step-by-step explanation:

Given : The weights of steers in a herd are distributed normally.

[tex]\mu= 1100\text{ lbs }[/tex]

Standard deviation : [tex]\sigma=300 \text{ lbs }[/tex]

Let x be the weight of the randomly selected steer .

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

[tex]z=\dfrac{920-1100}{300}=-0.6[/tex]

The the probability that the weight of a randomly selected steer is greater than 920 lbs using standardized normal distribution table  :

[tex]P(x>920)=P(z>-0.6)=1-P(z<-0.6)\\\\=1-0.2742531=0.7257469\approx0.7257[/tex]    

Hence, the probability that the weight of a randomly selected steer is greater than 920lbs =0.7257

Answer:

The correct option is d.

Step-by-step explanation:

The approximate population P, in thousands, for a species of frogs in a particular rain forest, x years after 1999 is given by the formula

[tex]P=0.672x^2-0.046x+3[/tex]

We need to find the year it which the population reach 182 frogs.

Substitute P=182 in the given formula.

[tex]182=0.672x^2-0.046x+3[/tex]

Subtract 182 from both the sides.

[tex]0=0.672x^2-0.046x+3-182[/tex]

[tex]0=0.672x^2-0.046x-179[/tex]

Multiply both sides by 1000 to remove decimals.

[tex]0=672x^2-46x-179000[/tex]

Quadratic formula:

[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

Substitute a=672, b=-46 and c=-179000 in the quadratic formula.

[tex]x=\frac{-\left(-46\right)\pm\sqrt{\left(-46\right)^2-4\cdot \:672\left(-179000\right)}}{2\cdot \:672}[/tex]

On simplification we get

[tex]x=\frac{-\left(-46\right)+\sqrt{\left(-46\right)^2-4\cdot \:672\left(-179000\right)}}{2\cdot \:672}\approx 16.355[/tex]

[tex]x=\frac{-\left(-46\right)-\sqrt{\left(-46\right)^2-4\cdot \:672\left(-179000\right)}}{2\cdot \:672}\approx -16.287[/tex]

The value of x can not be negative because x is number of years after 1999.

x=16.35 in means is 17th year after 1999 the population reach 182 frogs.

[tex]1999+17=2016[/tex]

The population reach 182 frogs in 2016. Therefore the correct option is d.

Answer:

[tex][\frac{\pi}{2},\frac{3\pi}{2}][/tex]

Step-by-step explanation:

Let me first state that I am assuming your function is

[tex]f(x)=7cos^2(x)-14sin(x)[/tex]

If this is incorrect, then disregard this whole answer/explanation.

In order to find where the function is increasing or decreasing, we need to first find the first derivative, set it equal to 0, and then factor to find the values that cause the derivative to equal 0.  This is where you expect to find a max or a min value in the function itself.  But this function is not going to be easily solved for 0 once we find the derivative unless we make it in terms of either sin or cos right now, before taking the first derivative.  

Let [tex]cos^2(x)=1-sin^2(x)[/tex]

This is a Pythagorean trig identity, and I'm assuming that if you're in calculus solving for the intervals of increasing and decreasing values that you have, at one time, used trig identities.

Rewriting:

[tex]f(x)=7(1-sin^2(x))-14sin(x)[/tex] which simplifies to

[tex]f(x)=7-7sin^2(x)-14sin(x)[/tex] and in order of descending values of x:

[tex]f(x)=-7sin^2(x)-14sin(x)+7[/tex]

Now we can find the derivative.  For the first term, let u = sin(x), therefore,

[tex]f(u)=u^2[/tex], u' = cos(x), and f'(u) = 2u.  The derivative is found by multiplying f'(u) by u', which comes out to 2sin(x)cos(x)

The derivative for the next 2 terms are simple, so the derivative of the function is

[tex]f'(x)=-7[2sin(x)cos(x)]-14cos(x)[/tex] which simplifies down to

[tex]f'(x)=-14sin(x)cos(x)-14cos(x)[/tex]

We will set that equal to zero and solve for the values that cause that derivative to equal 0.  But first we can simplify it a bit.  You can factor out a -14cos(x):

[tex]f'(x)=-14cos(x)(sin(x)+1)[/tex]

By the Zero Product Property, either

-14cos(x) = 0 or sin(x) + 1 = 0

Solving the first one for cos(x):

cos(x) = 0

Solving the second one for sin(x):

sin(x) = -1

We now look to the unit circle to see where, exactly the cos(x) = 0.  Those values are

[tex]\frac{\pi}{2},\frac{3\pi}{2}[/tex]

The value where the sin is -1 is found at

[tex]\frac{3\pi}{2}[/tex]

We set up a table (at least that's what I advise my students to do!), separating the intervals in ascending order, starting at 0 and ending at 2pi.

Those intervals are

0 < x < [tex]\frac{\pi}{2}[/tex], [tex]\frac{\pi}{2}<x<\frac{3\pi}{2}[/tex], and [tex]\frac{3\pi}{2}<x<2\pi[/tex]

Now pick a value that falls within each interval and evaluate the derivative at that value and determine the sign (+ or -) that results.  You don't care what the value is, only the sign that it carries.  For the first interval I chose

[tex]f'(\frac{\pi}{4})=-[/tex] so the function is decreasing here (not what you wanted, so let's move on to the next interval).

For the next interval I chose:

[tex]f'(\pi)=+[/tex] so the function is increasing here.

For the last interval I chose:

[tex]f'(\frac{7\pi}{4})=-[/tex]

It appears that the only place this function is increasing is on the interval

[tex][\frac{\pi}{2},\frac{3\pi}{2}][/tex]

Final answer:

The interval on which the function f(x) = 7cos^2(x) - 14sin(x) is increasing is DNE.

Explanation:

To find the interval on which the function f(x) = 7cos^2(x) - 14sin(x) is increasing, we need to determine where the derivative of the function is positive. The derivative of f(x) can be found using the chain rule, which gives us f'(x) = -14cos(x) - 28sin(x)cos(x). To find where f'(x) > 0, we need to solve the inequality -14cos(x) - 28sin(x)cos(x) > 0.

We can simplify this inequality to cos(x)(-14 - 28sin(x)) > 0. Since cos(x) is positive on the interval 0 ≤ x ≤ 2π and -14 - 28sin(x) is negative on the interval 0 ≤ x ≤ 2π, the product of these two terms will be negative. Therefore, there are no values of x on the interval 0 ≤ x ≤ 2π where f(x) is increasing.

a. This recurrence is of order 2.

b. We're looking for a function [tex]A(x)[/tex] such that

[tex]A(x)=\displaystyle\sum_{n=0}^\infty a_nx^n[/tex]

Take the recurrence,

[tex]\begin{cases}a_0=a_0\\a_1=a_1\\a_n-5a_{n-1}+6a_{n-2}=0&\text{for }n\ge2\end{cases}[/tex]

Multiply both sides by [tex]x^{n-2}[/tex] and sum over all integers [tex]n\ge2[/tex]:

[tex]\displaystyle\sum_{n=2}^\infty a_nx^{n-2}-5\sum_{n=2}^\infty a_{n-1}x^{n-2}+6\sum_{n=2}^\infty a_{n-2}x^{n-2}=0[/tex]

Pull out powers of [tex]x[/tex] so that each summand takes the form [tex]a_kx^k[/tex]:

[tex]\displaystyle\frac1{x^2}\sum_{n=2}^\infty a_nx^n-\frac5x\sum_{n=2}^\infty a_{n-1}x^{n-1}+6\sum_{n=2}^\infty a_{n-2}x^{n-2}=0[/tex]

Now shift the indices and add/subtract terms as needed to get everything in terms of [tex]A(x)[/tex]:

[tex]\displaystyle\frac1{x^2}\left(\sum_{n=0}^\infty a_nx^n-a_0-a_1x\right)-\frac5x\left(\sum_{n=0}^\infty a_nx^n-a_0\right)+6\sum_{n=0}^\infty a_nx^n=0[/tex]

[tex]\displaystyle\frac{A(x)-a_0-a_1x}{x^2}-\frac{5(A(x)-a_0)}x+6A(x)=0[/tex]

Solve for [tex]A(x)[/tex]:

[tex]A(x)=\dfrac{a_0+(a_1-5a_0)x}{1-5x+6x^2}\implies\boxed{A(x)=\dfrac{a_0+(a_1-5a_0)x}{(1-3x)(1-2x)}}[/tex]

c. Splitting [tex]A(x)[/tex] into partial fractions gives

[tex]A(x)=\dfrac{2a_0-a_1}{1-3x}+\dfrac{3a_0-a_1}{1-2x}[/tex]

Recall that for [tex]|x|<1[/tex], we have

[tex]\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n[/tex]

so that for [tex]|3x|<1[/tex] and [tex]|2x|<1[/tex], or simply [tex]|x|<\dfrac13[/tex], we have

[tex]A(x)=\displaystyle\sum_{n=0}^\infty\bigg((2a_0-a_1)3^n+(3a_0-a_1)2^n\bigg)x^n[/tex]

which means the solution to the recurrence is

[tex]\boxed{a_n=(2a_0-a_1)3^n+(3a_0-a_1)2^n}[/tex]

d. I guess you mean [tex]a_0=2[/tex] and [tex]a_1=5[/tex], in which case

[tex]\boxed{\begin{cases}a_0=2\\a_1=5\\a_2=13\\a_3=35\\a_4=97\\a_5=275\end{cases}}[/tex]

e. We already know the general solution in terms of [tex]a_0[/tex] and [tex]a_1[/tex], so just plug them in:

[tex]\boxed{a_n=2^n+3^n}[/tex]

Answer: Option(c) Richard Lester is correct.

Step-by-step explanation:

Both the films were directed by Richard Lester.

A Hard Day's night was a scripted comic farce and its main focus on  Beatlemania and the band's hectic touring lifestyle. It is a black and white movie.

Help! film also directed by Richard Lester. And this film was shot in various exotic locations. Help! was the first Beatles film that is filmed in colour.  

Answer:

B. 78 and 82.

Step-by-step explanation:

We have been given that the average test score of the class was an 80 and the standard deviation was 2. We are asked to find two values between which 68% of class will score.

We know that in a normal distribution approximately 68% of the data falls within one standard deviation of the mean.

So 68% scores will lie within one standard deviation below and above mean that is:

[tex](\mu-\sigma,\mu+\sigma)[/tex]

Upon substituting our given values, we will get:

[tex](80-2,80+2)[/tex]

[tex](78,82)[/tex]

Therefore, about 68% of the class would score between 78 and 81 and option B is the correct choice.

Answer: The store sold 288 tapes last month.

Step-by-step explanation:

Let the number of CDs be x , then the number of tapes is given by the expression : 4x

Also, the total quantity of these two items sold was 360.

Now, we have the following equation :-

[tex]x+4x=360\\\\\Rightarrow\ 5x=360\\\\\Rightarrow\ x=\dfrac{360}{5}\\\\\Rightarrow\ x=72[/tex]

The number of CDs sold in last month = 72

The number of tapes sold in last month =[tex]4\times72=288[/tex]

Hence, the store sold 288 tapes last month.

Answer:

  (a) increasing: (-ln(2)/3, ∞); decreasing: (-∞, -ln(2)/3)

  (b) minimum: (-ln(2)/3, (9/8)∛2) ≈ (-0.21305, 1.41741); maximum: DNE

  (c) inflection point: DNE; concave up: (-∞, ∞); concave down: DNE

Step-by-step explanation:

The first derivative of f(x) = e^(8x) +e^(-x) is ...

  f'(x) = 8e^(8x) -e^(-x)

This is zero at the function minimum, where ...

  8e^(8x) -e^(-x) = 0

  8e^(9x) -1 = 0 . . . . . . multiply by e^x

  e^(9x) = 1/8 . . . . . . .  add 1, divide by 8

  9x = ln(2^-3) . . . . . . take the natural log

  x.min = (-3/9)ln(2) = -ln(2)/3 . . . divide by the coefficient of x, simplify

This value of x is the location of the minimum.

__

The function value there is ...

  f(-ln(2)/3) = e^(8(-ln(2)/3)) + e^(-(-ln(2)/3))

  = 2^(-8/3) +2^(1/3) = 2^(1/3)(2^-3 +1)

  f(x.min) = (9/8)2^(1/3) . . . . . minimum value of the function

__

A graph shows the first derivative to have positive slope everywhere, so the curve is always concave upward. There is no point of inflection. The minimum point found above is the place where the function transitions from decreasing to increasing.

To find the intervals on which f(x) = [tex]e^8^x + e^(^-^x^)[/tex] is increasing and decreasing, analyze the first derivative. The function is increasing on (-1/8, ∞) and decreasing on (-∞, -1/8). The local minimum is at x = -1/8, and the function is concave up on (-∞, -1/8) and concave up on (-1/8, ∞).

To find the intervals on which the function f(x) = [tex]e^8^x + e^(^-^x^)[/tex] is increasing and decreasing, we need to analyze the first derivative of the function. The first derivative is f'(x) = [tex]8e^8^x - e^(^-^x^)[/tex]. We set this derivative equal to zero and solve for x to find the critical points. There is one critical point at x = -1/8. We can test intervals to the left and right of this critical point to determine the behavior of the function. The function is decreasing on (-∞, -1/8) and increasing on (-1/8, ∞). Therefore, the function is increasing on the interval (-1/8, ∞) and decreasing on the interval (-∞, -1/8).

To find the local minimum and maximum values of f, we analyze the second derivative of the function. The second derivative is f''(x) =[tex]64e^8^x + e^(^-^x^)[/tex]. We evaluate this second derivative at the critical point x = -1/8. The second derivative at x = -1/8 is positive, so the function has a local minimum at x = -1/8.

The inflection point of the function can be found by analyzing the points where the concavity changes. The second derivative changes sign at x = -1/8. Therefore, the inflection point of the function is (-1/8, f(-1/8)). To find the intervals on which the function is concave up and concave down, we analyze the sign of the second derivative. The second derivative is positive on (-∞, -1/8) and positive on (-1/8, ∞), meaning the function is concave up on (-∞, -1/8) and concave up on (-1/8, ∞).

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Answer:

-1

Step-by-step explanation:

Move all terms containing  x

to the left side of the equation.

Tap for fewer steps...

Subtract  4x from both sides of the equation.

2 x−10−4x= −8 y=4x−8

Subtract  4 x  from  2 x − 2 x− 10= − 8 y = 4 x − 8

Move all terms not containing  x  to the right side of the equation.

Tap for more steps...

− 2 x = 2 y = 4 x − 8

Divide each term by  − 2  and simplify.

Tap for fewer steps...

Divide each term in  − 2 x= 2  by  − 2 .

− 2 x − 2 = 2 − 2 y = 4 x − 8

Simplify the left side of the equation by cancelling the common factors.

Tap for fewer steps...

Reduce the expression by cancelling the common factors.

Tap for more steps...

− ( − 1 ⋅ x ) = 2 2 y = 4 x − 8

Rewrite  

− 1 ⋅ x  as  - x . x = 2 − 2 y = 4 x − 8

Divide  2  by  − 2 .

x = − 1 y = 4 x − 8

Replace all occurrences of  x  with the solution found by solving the last equation for  x . In this case, the value substituted is  − 1 . x= − 1 y = 4 ( − 1 ) − 8

Simplify  4 ( − 1 ) − 8 .

Tap for fewer steps...

Multiply  4 by  − 1 .

x = − 1 y = − 4 − 8

Subtract  8  from  - 4 .

x = − 1 y = − 12

The solution to the system of equations can be represented as a point.

( − 1 , − 12 )

The result can be shown in multiple forms.

Point Form:  ( − 1 , − 12 )

Equation Form:  x  =− 1  ,y = − 12

Answer: 2.68

Step-by-step explanation:

Claim : The proportion of peas with yellow pods is equal to 0.25​

i.e. p=0.25

Sample size : [tex]460[/tex]

Proportion of peas with yellow pods in sample  :

[tex]P=\dfrac{91}{460}=0.19782608695\approx0.20[/tex]

Now, the test statistic for the population proportion is given by :-

[tex]z=\dfrac{p-P}{\sqrt{\dfrac{P(1-P)}{n}}}[/tex]

[tex]\Rightarrow\ z=\dfrac{0.25-0.20}{\sqrt{\dfrac{0.20(1-0.20)}{460}}}\Rightarrow\ z=2.68095132369\approx2.68[/tex]

Hence, the value of the test statistic is 2.68

The test statistic for the given population proportion and sample data is approximately -2.57. This result was calculated using the Z test formula for testing population proportions, with a sample proportion of 0.1978, an expected proportion of 0.25, and a sample size of 460.

To calculate the test statistic, we'll use the formula for Z: Z = (p' - p0) / sqrt[(p0(1 - p0)) / n]

Plugging these values into the formula, we get: Z = (0.1978 - 0.25) / sqrt[(0.25 * 0.75) / 460] ≈ -2.57

The estimated proportion p' in this formula represents the proportion of peas with yellow pods within our sample. We use this, along with the hypothesized proportion p0 and the sample size n, to calculate the test statistic, which gives us an idea of how far our observed data is from the expected hypothesis.

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Answer:

Probability: 0.1825 or 18.25%  .......  Unlikely

Step-by-step explanation:

Hello, great question. These types are questions are the beginning steps for learning more advanced Probability problems.

To start of we need to calculate the total amount of respondents that took the survey. We do this by adding all the answers together.

169 + 12,527 + 2,834 = 15,530 total people

Now that we know the total amount of people we can calculate the probability of each response by dividing the amount of people that had that response by the total amount of people that took the survey.

Never Used: [tex]\frac{169}{15,530} = 0.0109 = 1.09%[/tex]

Sometimes Used: [tex]\frac{12,527}{15,530} = 0.8066 = 80.66%[/tex]

Frequently used: [tex]\frac{2834}{15,530} = 0.1825 = 18.25%[/tex]

So we can see that the probability of a randomly selected person using a credit card frequently is 0.1825 or 18.25%

Unlikely

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Final answer:

The probability that a randomly selected person uses a credit card frequently is calculated as 2834 divided by the total of 15530, resulting in 0.1825 (rounded to four decimal places). Since this is less than 20%, it is considered 'Unlikely' for someone to frequently use a credit card.

Explanation:

To calculate the probability that a randomly selected person uses a credit card frequently, we need to use the basic probability formula, which is the number of favorable outcomes divided by the total number of outcomes. In this case, the number of people who use a credit card frequently is the favorable outcome, and the total number of respondents is the sum of all categories of credit card usage.

Number of people who use a credit card frequently: 2834

Total number of respondents: 169 (never) + 12527 (sometimes) + 2834 (frequently) = 15530

Probability of frequent use: 2834 / 15530 = 0.1825 (rounded to 4 decimal places)

Now, let's interpret the result. A probability of 0.1825, when rounded, is about 18.25%. This number is less than 20%, which is generally considered the benchmark for something to be considered "likely". Therefore, it is Unlikely for someone to use a credit card frequently.

Answer with explanation:

The System of equations which we have to solve by Gauss Jordan Method:

  [tex]1.\rightarrow 2x_{1}-6x_{2}-2x_{3}=14, 2.\rightarrow 3x_{1}+4x_{2}-7x_{3}= 16, 3.\rightarrow 3x_{1}-6x_{2}+9x_{3}=21[/tex]

Writing it in the form of Augmented Matrix=3 Rows and 4 Columns:

  [tex]\left[\begin{array}{cccc}2&-6&-2&14\\3&4&-7&16\\3&-6&9&21\end{array}\right]\\\\R_{1}=\frac{R_{1}}{2},R_{3}=\frac{R_{3}}{3}\\\\ \left[\begin{array}{cccc}1&-3&-1&7\\3&4&-7&16\\1&-2&3&7\end{array}\right]\\\\R_{3}\rightarrow R_{3}-R_{1}\\\\\left[\begin{array}{cccc}1&-3&-1&7\\3&4&-7&16\\0&1&4&0\end{array}\right]\\\\R_{2}\rightarrow R_{2}-3R_{1}\\\\\left[\begin{array}{cccc}1&-3&-1&7\\0&13&-4&-5\\0&1&4&0\end{array}\right][/tex]

 [tex]R_{3}\rightarrow R_{2}+R_{3}\\\\\left[\begin{array}{cccc}1&-3&-1&7\\0&13&-4&-5\\0&14&0&-5\end{array}\right]\\\\\rightarrow14 x_{2}= -5\\\\x_{2}=\frac{-5}{14}\\\\\rightarrow 13 x_{2}-4x_{3}=-5\\\\ \frac{-65}{14}-4 x_{3}=-5\\\\-4x_{3}=-5+\frac{65}{14}\\\\x_{3}=\frac{5}{56}\\\\x_{1}-3x_{2}-x_{3}=7\\\\x_{1}+\frac{15}{14}-\frac{5}{56}=7\\\\x_{1}+\frac{55}{56}=7\\\\x_{1}=7-\frac{55}{56}\\\\x_{1}=\frac{337}{56}[/tex]

Solution set

  [tex]=(\frac{337}{56},\frac{-5}{14},\frac{5}{56})[/tex]

Set

[tex]\begin{cases}x=r\cos\theta\\y=r\sin\theta\end{cases}\implies\mathrm dA=r\,\mathrm dr\,\mathrm d\theta[/tex]

The region [tex]R[/tex] is given in polar coordinates by the set

[tex]R=\left\{(r,\theta)\mid2\le r\le3,0\le\theta\le\dfrac\pi2\right\}[/tex]

So we have

[tex]\displaystyle\iint_R\sin(x^2+y^2)\,\mathrm dA=\int_0^{\pi/2}\int_2^3r\sin(r^2)\,\mathrm dr\,\mathrm d\theta=\boxed{\frac\pi4(\cos4-\cos9)}[/tex]

The Cartesian coordinates are converted into polar coordinates so that the integral sin(x2 + y2) dA R becomes the integral sin(r2) r dr dθ. But this specific integral can't be solved analytically, yet using polar coordinates can simplify other integration issues related to circular regions or distances from the origin.

To evaluate the given integral using polar coordinates, one must firstly translate the Cartesian coordinates (x,y) into polar coordinates (r,θ), so that x is replaced with rcosθ and y with rsinθ. Consequently, the integral sin(x2 + y2) dA R becomes the integral sin(r2) r dr dθ, with r varying from 2 to 3, and θ from 0 to π/2 (since we are only dealing with the first quadrant).

However, this integral becomes very complex and is not feasible to solve analytically. You would need to use a numeric method to get an approximate answer. In the context of other problems, switching to polar coordinates can simplify the integration, especially when dealing with circular regions or equations related to distances from the origin.

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By the chain rule,

[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}[/tex]

Then for all [tex]t[/tex] the first derivative has a value of 7.

By the product rule,

[tex]\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac{\mathrm dy}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}\right]=\dfrac{\mathrm d\left(\frac{\mathrm dy}{\mathrm dt}\right)}{\mathrm dx}\dfrac{\mathrm dt}{\mathrm dx}+\dfrac{\mathrm dy}{\mathrm dt}\dfrac{\mathrm d^2t}{\mathrm dx^2}[/tex]

but [tex]t=x\implies\dfrac{\mathrm dt}{\mathrm dx}=1\implies\dfrac{\mathrm d^2t}{\mathrm dx^2}=0[/tex], so we're left with

[tex]\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d\left(\frac{\mathrm dy}{\mathrm dt}\right)}{\mathrm dx}[/tex]

By the chain rule,

[tex]\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d\left(\frac{\mathrm dy}{\mathrm dt}\right)}{\mathrm dx}=\dfrac{\mathrm d\left(\frac{\mathrm dy}{\mathrm dt}\right)}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}=\dfrac{\mathrm d^2y}{\mathrm dt^2}[/tex]

but [tex]y=7t-2\implies\dfrac{\mathrm dy}{\mathrm dt}=7\implies\dfrac{\mathrm d^2y}{\mathrm dt^2}=0[/tex] so the second derivative is 0 for all [tex]t[/tex].

The derivative dy/dx for the given parametric equations is 7, and the second derivative d2y/dx2 is zero. The slope at t = 9 is 7, and the curve does not exhibit concavity since it is linear.

To find dy/dx for the parametric equations x = t and y = 7t - 2, we need to compute the derivatives of both x and y with respect to t and then use the chain rule to find dy/dx as dy/dt divided by dx/dt. Since the derivative of x with respect to t is 1, and the derivative of y with respect to t is 7 (as the derivatives of the constants -2 and 1 are zero), dy/dx equals 7/1, which is 7. To find the second derivative d2y/dx2, we note that since dx/dt is constant (equals 1), the second derivative is zero. Therefore, the concavity of the curve does not change and is neither concave up nor down.

At the given value of the parameter t = 9, the slope of the tangent line is 7, as it is for all values of t. Since the second derivative is zero, the curve is linear and does not exhibit concavity at any point, including t = 9.

Answer:

2/7

Step-by-step explanation:

For the president position, there are 4 females from a total of 7 people.

For the vice president position, there are 3 males from 6 people left over.

So the probability is 4/7 × 3/6 = 2/7.

Answer: [tex]\dfrac{2}{7}[/tex]

Step-by-step explanation:

Given : Number of males = 3

Number of females = 4

The number of ways to select a female is elected president and a male is elected vice president :-

[tex]^3P_1\times ^4C_1=\dfrac{3!}{(3-1)!}\times\dfrac{4!}{(4-1)!}=3\times4=12[/tex]

The total number of ways to select 2 people from 7 :_

[tex]^7P_2=\dfrac{7!}{(7-2)!}=42[/tex]

Now, the probability that a female is elected president and a male is elected vice president will be :-

[tex]\dfrac{12}{42}=\dfrac{2}{7}[/tex]

Answer:

1) So the students do not make the same mistakes

2) So the students can see the importance of their jobs, to save lives

Approximately 2.28% of pediatricians work more than 72 hours per week according to the table.

To find the percentage of pediatricians who work more than 72 hours per week, we need to calculate the z-score for this value and then use a standard normal distribution table to find the corresponding percentage.

Calculate the z-score using the formula:

[tex]z = (x - u) / \alpha[/tex]

where x is the value (72 hours), u is the mean (48 hours), and a is the standard deviation (12 hours).

Substitute the values into the formula: z = (72 - 48) / 12 = 2.

Using a standard normal distribution table, find the percentage of values that are greater than 2.

Based on the table, approximately 2.28% of pediatricians work more than 72 hours per week.

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I'll abbreviate the definite integral with the notation,

[tex]I(f(x),a,b)=\int_a^bf(x)\,\mathrm dx[/tex]

We're given

Recall that the definite integral is additive on the interval [tex][a,b][/tex], meaning for some [tex]c\in[a,b][/tex] we have

[tex]I(f,a,b)=I(f,a,c)+I(f,c,b)[/tex]

The definite integral is also linear in the sense that

[tex]I(kf+\ell g,a,b)=kIf(a,b)+\ell I(g,a,b)[/tex]

for some constant scalars [tex]k,\ell[/tex].

Also, if [tex]a\ge b[/tex], then

[tex]I(f,a,b)=-I(f,b,a)[/tex]

a. [tex]I(2f,1,5)=2I(f,1,5)=2(I(f,1,8)-I(f,5,8))=2(9-4)=\boxed{10}[/tex]

b. [tex]I(f-g,1,8)=I(f,1,8)-I(g,1,8)=9-5=\boxed{4}[/tex]

c. [tex]I(f-g,1,5)=I(f,1,5)-I(g,1,5)=\dfrac{I(2f,1,5)}2-I(g,1,5)=10-3=\boxed{7}[/tex]

d. [tex]I(g-f,5,8)=I(g,5,8)-I(f,5,8)=(I(g,1,8)-I(g,1,5))-I(f,5,8)=(5-3)-4=\boxed{-2}[/tex]

e. [tex]I(7g,5,8)=7I(g,5,8)=7(5-3)=\boxed{14}[/tex]

f. [tex]I(3f,5,1)=3I(f,5,1)=-3I(f,1,5)=-\dfrac32I(2f,1,5)=-\dfrac32(10)=\boxed{-15}[/tex]

In this integral calculus problem, we leverage properties of definite integrals to compute the values of various expressions. Key steps usually involve substituting given integral values and multiplying by constant factors when required

To solve the problem, we first need to consider the properties of integral calculus, specifically those of definite integrals. A fundamental rule that is applicable here is that the product of a constant and an integral is the constant times the value of the integral.

So for problem a, integral^5_1 2f(x) dx = 2* integral^5_1 f(x) dx = 2 * 5 = 10.

Similarly, for problem b, integral^8_1 (f(x) - g (x)) dx = integral^8_1 f(x) dx - integral^8_1 g(x) dx = 9 - 5 = 4.

Following through similar steps of substitutions, we obtain the following solutions:

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Answer:

  linear function growth rate: 8.5

Step-by-step explanation:

The growth rate of the linear function is the coefficient of t: 8.5. (It is a constant.)

__

The growth rate of g(t) is its derivative: g'(t) = (1/8)(160e^(t/8)) = 20e^(t/8). Then the relative growth rate is ...

  g'(t)/g(t) = (20e^(t/8))/(160e^(t/8)) = 20/160 = 1/8

It is a constant.

Answer:

  (a)  20% to Longmont; 80% to Denver

  (b)  20% to Longmont; 80% to Denver

Step-by-step explanation:

(a) The bus to Longmont is the first bus to arrive, only between 8:31 and 8:37, and again between 9:01 and 9:07. That is, for a total of 12 minutes every hour, the Longmont bus is the first to arrive. The probability of going to Longmont is 12/60 = 1/5 = 20%.

__

(b) Same as for (a). As long as passenger arrival times are uniform within an hour, the probability is the same.

Answer:a-396

b-420

Step-by-step explanation:

[tex]\alpha[/tex] =0.1

Margin of Error=0.04

Level of significance is z[tex]\left ( 0.1\right )=1.64[/tex]

Previous estimate[tex]\left ( p\right ) =0.38[/tex]

sample size is given by:

n=[tex]\left (\frac{Z_{\frac{\alpha }{2}}}{E}\right )p\left ( 1-p\right )[/tex]

n=[tex]\frac{1.64}{0.04}^{2}0.38\left ( 1-0.38\right )=396.0436\approx 396[/tex]

[tex]\left ( b\right )[/tex]Does not use prior estimate

Assume

[tex]\alpha [/tex]=0.1

Margin of Error=0.04

Level of significance is z[tex]\left ( 0.1\right )=1.64[/tex]

Population proportion[tex]\left ( p\right )[/tex]=0.5

n=[tex]\left (\frac{Z_{\frac{\alpha }{2}}}{E}\right )p\left ( 1-p\right )[/tex]

n=[tex]\frac{1.64}{0.04}^{2}0.5\left ( 1-0.5\right )[/tex]

n=420.25[tex]\approx 420[/tex]

Answer: The range is 13.  The mean is 10. The standard deviation is 4.62 .

Step-by-step explanation:

The given data : 3, 6, 9, 11, 15, 16

Total number of data values : n = 6

The mean of data is given by :-

[tex]\overline{x}=\dfrac{\sum^6_{i=1}x_i}{n}\\\\\Rightarrow\overline{x}=\dfrac{60}{6}=10[/tex]

The standard deviation is given by :-

[tex]\sqrt{\dfrac{1}{n}(\sum^6_{i=1}(x_i-\overline{x})^2)}\\\\=\sqrt{\dfrac{1}{6}(\sum^6_{i=1}(x_i-10)^2)}\\\\=\sqrt{\dfrac{1}{6}\times(49+16+1+1+25+36)}=4.61880215352\approx4.62[/tex]

The range of the data : Maximum value -Minimum value

[tex]=16-3=13[/tex]

Answer:

2.5 ml for a dosage of 20 mg.

Step-by-step explanation:

Hello, great question. These types are questions are the beginning steps for learning more advanced Ratio problems.

Since this is basically a ratio problem we can use the simple Rule of Three property to solve this problem. The Rule of Three property can be seen in the photo below. Now we just plug in the values and solve for x.

 8 mg.   ⇒   1 ml.

20 mg.   ⇒   x

[tex]\frac{20mg*1ml}{8mg} = 2.5ml[/tex]

Now we can see that we should administer a 2.5 ml for a dosage of 20 mg.

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