Can somebody help me if you know what this is

The given information does not provide any clear indication for determining the position that should be taken on 3/01 and 3/02. Without additional information, it is not possible to make a decision. The table only displays the closing prices of the July GBP Futures Contract on different days, and it is unclear what trading strategy or what scenario is being considered. Additional information about the goals and objectives, the market conditions, and other relevant factors would be necessary to make a decision about trading positions.

hi

Let call money invested  at  7%  "X " and money at  9 % "Y" .

7% =  7/100 = 0.07  

9% = 9 /100 =  0.09  

So   X+Y = 18 000

0.07X + 0.09Y =  1560

So if  :  X+Y = 18 000

                Y =  18 000 -X

then  

        0.07X + 0.09 ( 18 000 -X ) =  1560

        0.07X +0.09 * 18 000 - 0.09X = 1560

        0.07X + 1620 -0.09X =  1560

        0.07X-0.09X = 1560-1620

           -0.02X =  -60  

                   X = -60 /-0.02

                   X = 3 000

If   X = 3 000   then  Y =  18 000 -3 000 = 15 000

Conclusion :    3 000 dollars were invested at  7 % and  15 000 at 9%

0.07* 3 000 = 210

15 000* 0.09 =1350  

Total =  210+1350 = 1 560

(a). The smallest quantity at which marginal cost is equal to 15 dollars per item is 2.76. (b). The positive value of q at which average variable cost is equal to marginal cost is \(\frac{15}{2}\). (c). The longest interval on which marginal revenue exceeds marginal cost is (1.8377, 8.1622). (d). The profit is increasing at 8.1622 at p(q) is maximum. (e). The maximum value of profit is 78.2455 dollars.

The total revenue: TR(q) = 30q

The total cost: Tc(q) = \(q^3-15 q^2+75 q+10$\)

The change in cost is referred to as the change in the cost of production when there is a need for change in the volume of production.

(a).  Marginal cost MC\(=\frac{\partial}{\partial q}(T C)$\)

\($$=3q^2-30 q+75 \text {. }$$\)

According to question.

\($$\begin{aligned}& 3 q^2-30 q+75=15 \\& 3 q^2-30 q+60=0 \\& q^2-10 q+20=0 \\& q= \frac{10 \pm \sqrt{100-80}}{2} \\&= \frac{10 \pm 2 \sqrt{5}}{2}=5 \pm \sqrt{5}\end{aligned}\)

\($$$\therefore \quad q=5+\sqrt{5} \quad$ and $q=5-\sqrt{5}$\)

we have to find smallest quantity.

\($$\therefore \quad q=5-\sqrt{5}=2.76 \text {. }$$\)

b)

\($$\begin{aligned}& F C=T C(0)=10 \\& V C(q)=T C(q)-F C \\& =q^3-15 q^2+75 q . \\& \text { AVC(q) }=\frac{V C(q)}{q}=q^2-15 q+75 .\end{aligned}$$\)

When AVC(q)=MC(q)

\($$\begin{array}{ll}\Rightarrow & q^2-15 q+7 ;=3 q^2-30 q+75 \\\Rightarrow & 2 q^2-15 q=0 .\ \Rightarrow q(2 q-15)=0 . \\\Rightarrow & q=0, \frac{15}{2} \quad \Rightarrow \quad q=\frac{15}{2}\end{array}$$\)

Therefore, the positive value of q at which average variable cost is equal to marginal cost is \(\frac{15}{2}\).

(c).

\(& M R(q)=\frac{d}{d q}(T R)=30 . \\\)

\(& M C(q)=3 q^2-30 q+75 . \\\)

Let MR(a)>MC(q)

\(& \Rightarrow \quad 30 > 3 q^2-30 q+75 \text {. } \\\)

\(& \Rightarrow \quad 3 q^2-3 p q+45 < 0 \\\)

\(& \Rightarrow \quad q^2-10 q+15 < 0 \text {. } \\\)

\(& \because \quad q=\frac{10 \pm \sqrt{100-60}}{2} \Rightarrow q=\frac{10 \pm 2 \sqrt{10}}{2} \\\)

\(& q=(5 \pm \sqrt{10}) \\\)

\(& \Rightarrow \quad(q-5+\sqrt{10})(q-5-\sqrt{10}) < 0 . \\\)

\(& \Rightarrow \(q-5+\sqrt{10}) < 0 \quad \& \quad(q-5-\sqrt{10}) > 0 \text {. } \\\)

\(& \text { or } \quad(q-5+\sqrt{10}) > 0 \quad \& \quad(q-5-\sqrt{10}) < 0 \text {. } \\\)

\(& \therefore \quad \text { other } q < 5-\sqrt{10}=1.8377 ., q > 5+\sqrt{10}=8.1622 \text {. } \\\)

\(& \Rightarrow \quad q < 1.8377 \& q > 0.1622 \text {. } \\\)

\(& \text { or } q > 5-\sqrt{10} \quad \& \quad q < 5+\sqrt{10} \text {. } \\\)

\(& \Rightarrow q > 1.8377 \quad < q < 8.1622 \text {. } \\\)

The Interval is (1.8377, 8.1622).

(d).  P(q) =TR(q)-TC(q)

\(& =30 q-\left(q^3-15 q^2+75 q+10\right) \\& =-q^3+15 q^2-75 q-10+30 q \\& =-q^3+15 q^2-45 q-10 .\)

on (1.8377,8.1622). we have to find a point such that p"(q)=0 and p"(q)<0.

\(& p^{\prime}(q)=-3 q^2+30 q-45 \\\)

\(& \Rightarrow \quad-3\left(q^2-10 q+15\right)=0 \\\)

\(& \Rightarrow \quad q^2-10 q+15=0 . \\\)

\(& \quad q=1.8377 \text { and } \quad q=8.1622 . \\\)

\(& \Rightarrow \quad p^{\prime \prime}(q)=-6 q+30 . \\\)

\(& p^{\prime \prime}(1.8377)=-6(1.8377)+30 \\\)

\(&=18.9738 \\\)

\(& p^{\prime \prime}(8.1622)=-6(8.1622)+30 . &=-18.9732 < 0 .\)

at 8.1622, p(q) is maximum.

(e).

Max profit-P(8.1622)

=78.2455 dollars

Therefore, the maximum value of profit is 78.2455 dollars.

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Answer: 90 more tickets

Step-by-step explanation:

230 - 50 = 180

180 divided by 2 = 90 tickets

Answer:

x=-3, y = 4/9

Step-by-step explanation:

This is in equation form.

a) The distance from point P to the pole  is: 6.2 m

b) The height of the Pole is: 6.2 m

The triangle attached shows us the triangle formed as a result of the given word problem about angle of elevation and distance and height.

Now, we are given that:

The angle of elevation of the top of the pole =  45°

Angle of elevation of the top of the pole = 58°.

|PQ|= 10m

a) PR is distance from point P to the pole and using trigonometric ratios, gives us:

PR/sin 58 = 10/sin(180 - 58 - 45)

PR/sin 58 = 10/sin 77

PR = (10 * sin 58)/sin 77

PR = 8.7 m

b) P O can be calculated with trigonometric ratios as:

P O = PR * cos 45

P O = 8.7 * 0.7071

P O = 6.2 m

Now, the two sides of the isosceles triangle formed are equal and as such:

R O = P O

Thus, height of pole R O = 6.2 m

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Step-by-step explanation:

499

im not sure if correct ,

We want to find the derivative of

To solve that we must remember the derivative of sin and cos

Therefore, let's use it to find f'(x)

Therefore

Now let's evaluate the derivative at the point 3π/4

Hence

Answer:

A. $4.90

Step-by-step explanation:

Answer:

9. -2

10. -1

11. -3

Step-by-step explanation:

9. 12-3X=22+2X

Collect like terms

12-22=2X+3X

-10=5X

Divide both side by 5

X= -10/5 = -2

10. X+7X-12= -20

Collect like terms

X+7X = -20+12

8X= -8

Divide both side by 8

X= -8/8 = -1

11. 7X+4-13X= -1+23

Collect like terms

-7X-13X = -1+23-4

-6X = 18

Divide both side by -6

X= 18/-6 = -3

Answer:

The 95% confidence interval for the proportion of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694)

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \(\pi\), and a confidence level of \(1-\alpha\), we have the following confidence interval of proportions.

\(\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}\)

In which

z is the zscore that has a pvalue of \(1 - \frac{\alpha}{2}\).

For this problem, we have that:

\(n = 603, \pi = \frac{142}{603} = 0.2355\)

95% confidence level

So \(\alpha = 0.05\), z is the value of Z that has a pvalue of \(1 - \frac{0.05}{2} = 0.975\), so \(Z = 1.96\).

The lower limit of this interval is:

\(\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2355 - 1.96\sqrt{\frac{0.2355*0.7645}{603}} = 0.2016\)

The upper limit of this interval is:

\(\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2355 + 1.96\sqrt{\frac{0.2355*0.7645}{603}} = 0.2694\)

The 95% confidence interval for the proportion of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694)

Answer:

36 points. Option C

Step-by-step explanation:

Jonathan scored 2/5th of the points of which his team had scored.

His team in the 1st game scored 55 points and 35 in the second.

The team combinedly scored 55+35=90 points.

Jonathan scored 2/5th of the 90 points

\(\frac{2}{5}\) * 90= \(\frac{180}{5}\)

= 36 points.

Answer:

A. 3/9 and D. 1/3

Step-by-step explanation:

First, you need to find the square roots of the fraction given.

√9=3 √81=9

Now it's 3/9.

This automatically gives you option A. Now, if you simplify 3/9, you get 1/3.

Hope this helps you out!! :)

Answer:

-3x+4

Step-by-step explanation:

The person is 36m away from the boat.

The given figure is triangular in nature. In order to determine the distance of the person from the boat, we will use the pythagoras theorem.

Given the following parameters

GH = 20m

GK = 30m

The measure of the length HK (The distance of the person from the boat,) is calculated as:

HK^2 = GH^2 + GK^2

HK^2 = 20^2+ 30^2

HK^2 = 400+900
HK^2 = 1300

HK = 36m

Hence the distance of the person from the boat is approximately 36m

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Answer:

Option 2) Sin 35 = \(\frac{5}{x}\)

Step-by-step explanation:

Sin 35 = \(\frac{opposite }{hypotenuse}\)

Where opposite = 5' and hypotenuse = x(unknown)

=> Sin 35 = \(\frac{5}{x}\)

It takes 10 minutes for the number of bacteria in the population to double.

To determine the time it takes for the number of bacteria in a population to double, we need to find the value of t when the function f(t) equals twice the initial number of bacteria.

The given function is f(t) = 60,000 * 2^(t/10).

To find the time it takes for the number of bacteria to double, we set f(t) equal to twice the initial number of bacteria, which is 2 * 60,000 = 120,000:

120,000 = 60,000 * 2^(t/10).

Next, we can simplify the equation by dividing both sides by 60,000:

2 = 2^(t/10).

Since both sides of the equation have the same base (2), we can equate the exponents:

t/10 = 1.

To solve for t, we multiply both sides by 10:

t = 10.

Therefore, it takes 10 minutes for the number of bacteria in the population to double.

This result is obtained by setting the growth rate of the bacteria population in the given function. The exponent t/10 determines the rate of growth, and when t is equal to 10, the exponent becomes 1, resulting in a doubling of the initial number of bacteria.

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To solve for x in the equation 4x3 = -5x - 21, we can follow these steps:

1. Move all the terms containing x to one side of the equation, and move the constant term to the other side. We can do this by adding 5x to both sides and then adding 21 to both sides:

   4x3 + 5x = -21

2. Factor out x from the left-hand side of the equation:

   x(4x2 + 5) = -21

3. Divide both sides by (4x2 + 5):

   x = -21 / (4x2 + 5)

So the solution for x is x = -21 / (4x2 + 5). Note that this is a rational function, which means that the value of x depends on the value of the variable x. This equation has no real solutions because the denominator is always positive, and the numerator is negative.

The height of the hill is approximately 25.73 ft. Where v0 is the initial velocity (108 ft/s), g is the acceleration due to gravity \((-32.2 ft/s^2)\),

To find the height of the hill, we can use the formula for the vertical position of an object under constant acceleration:

h = h0 + v0t + 1/2at^2

where h is the final height, h0 is the initial height, v0 is the initial velocity, t is the time, and a is the acceleration due to gravity (-32.2 ft/s^2).

In this case, we are given that the initial height h0 is 3 ft, the initial velocity v0 is 108 ft/s, and the time t is 9.5 s. We want to find the height of the hill, which we can denote as h_hill. The final height is the height of the plain, which we can denote as h_plain and assume is zero.

At the highest point of its trajectory, the arrow will have zero vertical velocity, since it will have stopped rising and just started to fall. So we can set the velocity to zero and solve for the time it takes for that to occur. Using the formula for velocity under constant acceleration:

v = v0 + at

we can solve for t when v = 0, h0 = 3 ft, v0 = 108 ft/s, and a = -32.2 ft/s^2:

0 = 108 - 32.2t

t = 108/32.2 ≈ 3.35 s

Thus, it takes the arrow approximately 3.35 s to reach the top of its trajectory.

Using the formula for the height of an object at a given time, we can find the height of the hill by subtracting the height of the arrow at the top of its trajectory from the initial height:

h_hill = h0 + v0t + 1/2at^2 - h_top

where h_top is the height of the arrow at the top of its trajectory. We can find h_top using the formula for the height of an object at the maximum height of its trajectory:

h_top = h0 + v0^2/2a

Plugging in the given values, we get:

h_top = 3 + (108^2)/(2*(-32.2)) ≈ 196.78 ft

Plugging this into the first equation, we get:

h_hill = 3 + 108(3.35) + 1/2(-32.2)(3.35)^2 - 196.78

h_hill ≈ 25.73 ft

If the arrow is launched at t0, the equation describing velocity as a function of time would be:

v(t) = v0 - gt

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The amount of Dividends of $3,000, $5,000, $34,500, and $71,000 were distributed to the 20,000 preferred and 25,000 common shareholders of Lightfoot Inc. over the first four years of operations.

To calculate the dividend per share of the preferred and common stock of Lightfoot Inc., the total amount of dividends paid out over the first four years must first be determined. This can be done by adding the given amounts of $3,000, $5,000, $34,500, and $71,000 to get a total of $113,500. To find the dividend per share for the preferred stock, the total dividend is divided by the number of shares (20,000) which gives a dividend per share of $5.68. To find the dividend per share for the common stock, the total dividend is divided by the number of shares (25,000) which gives a dividend per share of $4.54.

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The probability that the mean of a sample of 40 families will be between 17.1 and 18.1 pounds is given as follows:

0.5874 = 58.74%.

The z-score of a measure X of a variable that has mean symbolized by \(\mu\) and standard deviation symbolized by \(\sigma\) is obtained by the rule presented as follows:

\(Z = \frac{X - \mu}{\sigma}\)

The parameters for this problem are given as follows:

\(\mu = 17.2, \sigma = 2.5, n = 40, s = \frac{2.5}{\sqrt{40}} = 0.3953\)

The probability that the mean of a sample of 40 families will be between 17.1 and 18.1 pounds is the p-value of Z when X = 18.1 subtracted by the p-value of Z when X = 17.1, hence:

\(Z = \frac{X - \mu}{\sigma}\)

By the Central Limit Theorem:

\(Z = \frac{X - \mu}{s}\)

Z = (18.1 - 17.2)/0.3953

Z = 2.28

Z = 2.28 has a p-value of 0.9887.

Z = (17.1 - 17.2)/0.3953

Z = -0.25

Z = -0.25 has a p-value of 0.4013.

Hence:

0.9887 - 0.4013 = 0.5874 = 58.74%.

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At an amusement park, having 1,500 adults who entered the park during the 9:00 a.m. hour. and a 75% increase for the 10:00 a.m. hour, with each ticket costs at $49.95 the expression for the total amount collected during the two hours is:

​​49.95(1500+1500(1.75)) ; The expression is 1,500 for the first hour and 1,500 plus the additional 75 percent for the second hour

The expression for the cost is computed from the word problem as follows

At an amusement park, there were 1,500 adults who entered the park during the 9:00 a.m. hour.

= 1500

During the 10:00 a.m. hour, there was a 75% increase

= 1500 * 1.75

= 1500(1.75)

If each ticket costs $49.95, the total cost

= $49.9(1500 + 1500(1.75))

Hence, the expression is 1,500 for the first hour and 1,500 plus the additional 75 percent for the second hour

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Answer:

(a) The probability that X is at most 30 is 0.9726.

(b) The probability that X is less than 30 is 0.9554.

(c) The probability that X is between 15 and 25 (inclusive) is 0.7406.

Step-by-step explanation:

We are given that 11% of all steel shafts produced by a certain process are nonconforming but can be reworked. A random sample of 200 shafts is taken.

Let X = the number among these that are nonconforming and can be reworked

The above situation can be represented through binomial distribution such that X ~ Binom(n = 200, p = 0.11).

Here the probability of success is 11% that this much % of all steel shafts produced by a certain process are nonconforming but can be reworked.

Now, here to calculate the probability we will use normal approximation because the sample size if very large(i.e. greater than 30).

So, the new mean of X, \(\mu\) = \(n \times p\) = \(200 \times 0.11\) = 22

and the new standard deviation of X, \(\sigma\) = \(\sqrt{n \times p \times (1-p)}\)

                                                                  = \(\sqrt{200 \times 0.11 \times (1-0.11)}\)

                                                                  = 4.42

So, X ~ Normal(\(\mu =22, \sigma^{2} = 4.42^{2}\))

(a) The probability that X is at most 30 is given by = P(X < 30.5)  {using continuity correction}

        P(X < 30.5) = P( \(\frac{X-\mu}{\sigma}\) < \(\frac{30.5-22}{4.42}\) ) = P(Z < 1.92) = 0.9726

The above probability is calculated by looking at the value of x = 1.92 in the z table which has an area of 0.9726.

(b) The probability that X is less than 30 is given by = P(X \(\leq\) 29.5)    {using continuity correction}

        P(X \(\leq\) 29.5) = P( \(\frac{X-\mu}{\sigma}\) \(\leq\) \(\frac{29.5-22}{4.42}\) ) = P(Z \(\leq\) 1.70) = 0.9554

The above probability is calculated by looking at the value of x = 1.70 in the z table which has an area of 0.9554.

(c) The probability that X is between 15 and 25 (inclusive) is given by = P(15 \(\leq\) X \(\leq\) 25) = P(X < 25.5) - P(X \(\leq\) 14.5)   {using continuity correction}

       P(X < 25.5) = P( \(\frac{X-\mu}{\sigma}\) < \(\frac{25.5-22}{4.42}\) ) = P(Z < 0.79) = 0.7852

       P(X \(\leq\) 14.5) = P( \(\frac{X-\mu}{\sigma}\) \(\leq\) \(\frac{14.5-22}{4.42}\) ) = P(Z \(\leq\) -1.70) = 1 - P(Z < 1.70)

                                                          = 1 - 0.9554 = 0.0446

The above probability is calculated by looking at the value of x = 0.79 and x = 1.70 in the z table which has an area of 0.7852 and 0.9554.

Therefore, P(15 \(\leq\) X \(\leq\) 25) = 0.7852 - 0.0446 = 0.7406.

Answer:

24 3/4 - 15 1/6

Step-by-step explanation:

1/6 x 4 = 4/24        3/4 x 6 = 18/24    

24-15 = 9

9 14/24 or 9 7/12

Cost of white shirt = 2.50 $

Cost of blue shirt = 1.50 $

We have asked to find percentage difference\(.\)

_________________________________

\(:\implies\sf\:\dfrac{cost_{blue}-cost_{white}}{cost_{blue}+cost_{white}}\times 100\)

\(:\implies\sf\:\dfrac{2.50-1.50}{2.50+1.50}\times 100\)

\(:\implies\sf\:\dfrac{1}{4}\times 100\)

\(:\implies\:\boxed{\bf{\gray{25\:\%}}}\)

Answer:

X2 -X1 / Y2 - Y1

Step-by-step explanation:

14- 4 / (-13) - (-4)

10/-9

Answer:

x = 47

Step-by-step explanation:

x + 121 = 4x - 20

-3x + 121 = - 20

-3x = -141

x = 47

So, the answer is x = 47

Answers to both the given expressions are the same that is 25 and both answers are positive.

So,

Similarly,

Hence, we can say that the answers to both the expressions are the same and both have positive answers.

Therefore, the answers to both the given expressions are the same that is 25 and both answers are positive.

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9514 1404 393

Answer:

  the equation comes from setting expressions for the perimeters equal to each other

Step-by-step explanation:

Each of the relations is described in terms of the shortest side of the triangle, which is what we're asked to find. It makes sense to let a variable represent that length.

  shortest triangle side: s

  each of the other two sides: s+1

  perimeter of the triangles: s +(s+1) +(s+1) = 3s+2

__

  side length of square: s-2

  perimeter of square : 4(s -2) = 4s -8

__

The two perimeters are equal, so we have ...

  3s +2 = 4s -8 . . . . . your equation

  10 = s . . . . . . . . . . . add 8-3s