Suppose that 3% of all athletes are using the endurance-enhancing hormone EPO (you should be able to simply compute the percentage of all athletes that are not using EPO). For our purposes, a “positive” test result is one that indicates presence of EPO in an athlete’s bloodstream. The probability of a positive result, given the presence of EPO is .99. The probability of a negative result, when EPO is not present, is .90. What is the probability that a randomly selected athlete tests positive for EPO? 0.0297

Answer:

3/15 or 0.2

Step-by-step explanation:

Answer:

20m/s

Step-by-step explanation:

This can be solved using the acceleration / velocity equations.

Specifically,

v² = u² + 2as

Where

v = final velocity = what we need to find

u = initial velocity = given as 10.0m/s

a = acceleration = given as 0.5m/s²

s = distance = 300m

Hence,

v² = 10² + (2)(0.5) (300)

    = 100 + 300

    =400

v = √400 = 20 m/s

The shop sells 216 "I LOVE NY" T-shirts each day. Therefore, over the course of one week, the shop sells 1512 T-shirts.

The shop in Times Square is open from 9 am to 9 pm, which means the shop operates for 12 hours. Since there are 60 minutes in an hour, this shop is open for a total of 720 minutes each day.

Three "I LOVE NY" T-shirts are sold every 10 minutes. So, in 720 minutes, the number of T-shirts sold would be 720 ÷ 10 = 72 sets of three T-shirts. Therefore, 72 sets x 3 shirts = 216 T-shirts are sold per day.

Finally, to calculate the weekly total, it is necessary to multiply the daily total by 7 (the number of days in a week). So, 216 T-shirts x 7 days = 1512 T-shirts sold in one week.

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Answer:

Quotient will be 1.25

Step-by-step explanation:

First we convert decimal numbers to fractions. So write down the decimal divided by 1 and then multiply both top and bottom with 10 for every number after decimal point.

Here we found for  .4375  = [tex]\frac{4375}{10000}[/tex]

and .35 =   [tex]\frac{35}{100}[/tex]

Now we divide both the numbers as

= [tex]\frac{\frac{4375}{1000} }{\frac{35}{100} }[/tex]

= [tex]\frac{4375}{1000}[/tex] × [tex]\frac{100}{35}[/tex]

= [tex]\frac{125}{100}[/tex]

= 1.25

Quotient will be 1.25

Final answer:

The quotient of 0.4375 divided by 0.35 is 1.25, which rounded to the tenths place is 1.3.

Explanation:

The student is asking to find the quotient of two decimal numbers, which is a basic arithmetic operation involving division. The numbers are 0.4375 and 0.35. To find the quotient, simply divide 0.4375 by 0.35.

Using a calculator or performing the division manually, you would proceed as follows:

Adjust the decimals by multiplying both numbers by 100 to make them whole numbers, resulting in 43.75 divided by 35.

Perform the division to get the preliminary result: 43.75 / 35 = 1.25.

Since we need to round the final answer to the tenths place based on the least precise number given (35.5 g), round 1.25 to one decimal place, which is 1.3 (1.25 rounds up because the next digit, 5, is equal to or greater than 5).

Therefore, the quotient of 0.4375 divided by 0.35, rounded to the tenths place, is 1.3.

Answer:

C. 153.84 mL

Step-by-step explanation:

Let's say x is the volume of 75% solution and y is the volume of 10% solution.

Sum of the volumes:

x + y = 500

Sum of the alcohol amounts:

0.75x + 0.10y = 0.30(500)

0.75x + 0.10y = 150

Solve the system of equations using either substitution or elimination.  I'll use substitution.

y = 500 - x

0.75x + 0.10 (500 - x) = 150

0.75x + 50 - 0.10x = 150

0.65x = 100

x = 153.84

You need 153.84 mL of 75% solution.

"153.84 mL" of 75% alcohol should be added. A further explanation is provided below.

Let,

then,

→ [tex]x+y = 500[/tex]

         [tex]y = (500-x)[/tex]

now,

→ [tex]75(x)+10(500-x) = 500\times 30[/tex]

                [tex]65x+5000=15000[/tex]

                           [tex]65x=15000-5000[/tex]

                           [tex]65x=10000[/tex]

                               [tex]x = \frac{10000}{65}[/tex]

                                  [tex]= \frac{2000}{13}[/tex]

                                  [tex]= 153.84 \ mL[/tex]

Thus the above response i.e., "option c" is correct.

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Take up to the third-order derivative:

[tex]f(x)=(-3x+15)^{3/2}[/tex]

[tex]f'(x)=\dfrac32(-3x+15)^{1/2}(-3)=-\dfrac92(-3x+15)^{1/2}[/tex]

[tex]f''(x)=-\dfrac94(-3x+15)^{-1/2}(-3)=\dfrac{27}4(-3x+15)^{-1/2}[/tex]

[tex]f'''(x)=-\dfrac{27}8(-3x+15)^{-3/2}(-3)=\dfrac{81}8(-3x+15)^{-3/2}[/tex]

Evaluate each derivative at [tex]x=a=2[/tex]:

[tex]f(2)=9^{3/2}=27[/tex]

[tex]f'(2)=-\dfrac929^{1/2}=-\dfrac{27}2[/tex]

[tex]f''(2)=\dfrac{27}4\dfrac1{9^{1/2}}=\dfrac94[/tex]

[tex]f'''(2)=\dfrac{81}8\dfrac1{9^{3/2}}=\dfrac38[/tex]

Then the Taylor polynomial is

[tex]P_3(x)=f(2)+f'(2)(x-2)+\dfrac{f''(2)}2(x-2)^2+\dfrac{f'''(2)}6(x-2)^3[/tex]

[tex]P_3(x)=27-\dfrac{27}2(x-2)+\dfrac98(x-2)^2+\dfrac1{16}(x-2)^3[/tex]

Answer:

2nd answer.

Step-by-step explanation:

see attached.

Answer with Step-by-step explanation:

We have to find the solution of the equation:

[tex]x^4+6x^2+5=0[/tex]

Let u=x²

Then, above equation is transformed to:

[tex]u^2+6u+5=0[/tex]

it could also be written as:

[tex]u^2+5u+u+5=0[/tex]

u(u+5)+1(u+5)=0

(u+1)(u+5)=0

either  u+1=0 or u+5=0

either u= -1 or u= -5

Putting u=x²

x² = -1 or x² = -5

On taking square root both sides

x= ± i  or  x= ± i√5

Hence, roots of the equation [tex]x^4+6x^2+5=0[/tex] are:

i , -i , i√5 and -i√5

Answer:

  16

Step-by-step explanation:

Moment of inertia of a disk is proportional to its mass and to the square of its radius. For two disks with the same mass, the larger one will have a moment of inertia that is (4R/R)^2 = 16 times that of the smaller one.

It will take 16 smaller disks to make the systems have the same moment of inertia.

Answer:

a. The two events are dependent.

b. [tex]P(A\cap B)[/tex]= [tex]\frac{1}{220}[/tex].

Step-by-step explanation:

Given

Total coins =220

Number of Indian pennies= 6

A: When one of the 220 coins is randomly selected, it is one of the Indian pennies.

Therefore , the probability of getting an  Indian pennies=[tex]\frac{6}{220 }[/tex]

By using formula of probability=[tex]\frac{Number \; of\; favourable\; cases}{total\; number \; of \;cases}[/tex]

Probability of getting an  Indian pennies=[tex]\frac{3}{110}[/tex]

B: When another one of the 220 coins is randomly selected( with replacement) , It is also one of the Indian pennies.

Therefore, probability of getting an Indian pennies=[tex]\frac{6}{220}[/tex]

Probability of getting an Indian pennies =[tex]\frac{3}{110}[/tex]

[tex]A\cap B[/tex]: 1

[tex]P(A\cap B)=\frac{1}{220}[/tex]

If two events are independent. Then

[tex]P(A\cap B)= P(A)\times p(B)[/tex]

P(A).P(B)= [tex]\frac{3}{110} \times \frac{3}{110}[/tex]=[tex]\frac{9}{12100}[/tex]

Hence, [tex]P(A\cap B)\neq P(A).P(B)[/tex]

Therefore, the two events are dependent.

b. Probability that events A and B both occur

Number of favourable cases when both events A and B occur=1

Total coins=220

Probability=[tex]\frac{Number \; of\; favourable \; cases}{Total\; number\; of\; cases}[/tex]

[tex]P(A\cap B)=\frac{1}{220}[/tex]

Answer:

Step-by-step explanation:

so u do 2+2 4=243==32===3=424=4=234=234=32=43=4=34

Answer:

[tex]\frac{dh}{dt}=\frac{160}{169\pi }  ft/min[/tex]

Step-by-step explanation:

This is a classic related rates problem.  Gotta love calculus!

Start out with the formula for the volume of a cone, which is

[tex]V=\frac{1}{3}\pi r^2h[/tex]

and with what we know, which is [tex]\frac{dV}{dt}=40[/tex]

and the fact that the diameter = height (we will come back to that in a bit).

We need to find [tex]\frac{dh}{dt}[/tex] when h = 13

The thing we need to notice now is that there is no information given to us that involves the radius.  It does, however, give us a height.  We need to replace the r with something in terms of h.  Let's work on that first.

We know that d = h.  Because d = 2r, we can say that 2r = h, and solving for r gives us that [tex]r=\frac{h}{2}[/tex].

Now we can rewrite the formula with that replacement:

[tex]V=\frac{1}{3}\pi  (\frac{h}{2})^2h[/tex]

Simplify that all the way down to

[tex]V=\frac{1}{12}\pi  h^3[/tex]

The derivative of that function with respect to time is

[tex]\frac{dV}{dt}=\frac{1}{12}\pi(3h^2)\frac{dh}{dt}[/tex]

Filling in what we have gives us this:

[tex]40=\frac{1}{12}\pi (3)(13)^2\frac{dh}{dt}[/tex]

Solve that for the rate of change of the height:

[tex]\frac{dh}{dt}=\frac{160}{169\pi } \frac{ft}{min}[/tex]

or in decimal form:

[tex]\frac{dh}{dt}=.95\pi  \frac{ft}{min}[/tex]

This involves relationship between rates using Calculus.

dh/dt = 0.3 ft/min

Volumetric rate; dv/dt = 40 ft³/min

height of pile; h = 13 ft

We are not given the diameter here but as we are dealing with a right circular cone, we will assume that the diameter is equal to the height.

Thus; diameter; d = 13 ft

radius; r = h/2 = d/2 = 13/2

radius; r= 6.5 ft

V = ¹/₃πr²h

We want to find how fast the height is increasing and this is dh/dt.

Thus, we will need to express r in the volume formula in terms of h;

V = ¹/₃π(h/2)²h

V = ¹/₃π(h²/4)h

V = ¹/₁₂πh³

dV/dt = 3(¹/₁₂πh²)dh/dt

dV/dt = ¹/₄πh²(dh/dt)

Plugging in the relevant values, we have;

40 = ¹/₄π × 13² × (dh/dt)

dh/dt = (40 × 4)/(π × 13²)

dh/dt = 0.3 ft/min

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Answer:

20

Step-by-step explanation:

Answer:

160/1001, 175/1001

Step-by-step explanation:

i) There are:

₈C₁ ways to choose 1 new camera from 8 new cameras

₆C₃ ways to choose 3 refurbished cameras from 8 refurbished cameras

₁₄C₄ ways to choose 4 cameras from 14 cameras

The probability is:

P = ₈C₁ ₆C₃ / ₁₄C₄

P = 8×20 / 1001

P = 160 / 1001

P ≈ 0.160

ii) At most one new camera means either one new camera or no new cameras.  We already found the probability of one new camera.  The probability of no new cameras is the same as the probability of choosing 4 refurbished cameras:

P = ₆C₄ / ₁₄C₄

P = 15 / 1001

So the total probability is:

P = 160/1001 + 15/1001

P = 175/1001

P ≈ 0.175

To find the probability that one new camera will be selected, use the binomial coefficient and calculate the probability of selecting one new camera and three cameras that are not new. To find the probability of at most one new camera, calculate the probabilities of selecting zero new cameras and one new camera and add them together.

To find the probability that one new camera will be selected, we need to calculate the probability of selecting one new camera and three cameras that are not new. The total number of ways to select four cameras from 14 without replacement is given by the binomial coefficient 14 choose 4, which is equal to 14!/(4!(14-4)!). The probability of selecting one new camera is given by the product of the probability of selecting one new camera (8/14) and the probability of selecting three cameras that are not new (6/13 * 5/12 * 4/11). To find the probability that at most one new camera will be selected, we need to calculate the probabilities of selecting zero new cameras and one new camera and add them together.

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Answer:

The equation of tangent line is [tex]y=5x-33  [/tex]

Step-by-step explanation:

We need to find out the equation of tangent line.

Given :- g(6)=−3  and  g'(6)=  5

If  g(6)=−3

then the point on the line for the required tangent is  (6,−3)

If  g'(6)=  5

then the slope of the tangent at that point is  45

The tangent line can be specified by the slope-point form of the equation:

[tex](y-y_1)=m(x-x_1)[/tex]

which in this case is

[tex](y-(-3))=5(x-6)[/tex]

[tex](y+3)=(5x-30)[/tex]

subtract both the sides by 3,

[tex]y+3-3=5x-30-3[/tex]

[tex]y=5x-33[/tex]

Therefore, the equation of tangent line is [tex]y=5x-33[/tex]

Final answer:

The equation of the tangent line to the graph of y = g(x) at the point where x = 6 is y = 5x - 33, using the point-slope form and the given point (6, -3) with the slope of 5.

Explanation:

To find the equation of the tangent line to the graph at a particular point, we use the point-slope form of a line, given by y - y1 = m(x - x1), where (x1, y1) is the point on the graph and m is the slope at that point. Given that g(6) = -3 and g'(6) = 5, we can substitute these values into the point-slope form to get the equation of the tangent line. The equation is then y + 3 = 5(x - 6), which simplifies to y = 5x - 33.

[tex]\bf \sqrt{xy}=y\implies \left( xy \right)^{\frac{1}{2}}=y\implies \stackrel{\textit{chain rule~\hfill }}{\cfrac{1}{2}(xy)^{-\frac{1}{2}}\stackrel{\textit{product rule}}{\left(y+x\cfrac{dy}{dx} \right)}}=\cfrac{dy}{dx} \\\\\\ \cfrac{1}{2\sqrt{xy}}\left(y+x\cfrac{dy}{dx} \right)=\cfrac{dy}{dx}\implies \cfrac{y}{2\sqrt{xy}}+\cfrac{x}{2\sqrt{xy}}\cdot \cfrac{dy}{dx}=\cfrac{dy}{dx}[/tex]

[tex]\bf \cfrac{x}{2\sqrt{xy}}\cdot \cfrac{dy}{dx}=\cfrac{dy}{dx}-\cfrac{y}{2\sqrt{xy}} \implies \cfrac{x}{2\sqrt{xy}}\cdot \cfrac{dy}{dx}-\cfrac{dy}{dx}=-\cfrac{y}{2\sqrt{xy}} \\\\\\ \stackrel{\textit{common factor}}{\cfrac{dy}{dx}\left( \cfrac{x}{2\sqrt{xy}}-1 \right)}=-\cfrac{y}{2\sqrt{xy}} \implies \cfrac{dy}{dx}=-\cfrac{y}{\left( \frac{x}{2\sqrt{xy}}-1 \right)2\sqrt{xy}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \cfrac{dy}{dx}=-\cfrac{y}{x-2\sqrt{xy}}~\hfill[/tex]

Answer:

D

Step-by-step explanation:

[tex]\[\frac{x}{x^{2} +3x+2} -\frac{1}{(x+2)(x+1)} =\frac{x}{x^2+3x+2} -\frac{1}{x(x+1)+2(x+1)} =\frac{x}{x^{2}+3x+2 } -\frac{1}{x^{2} +2x+x+2} =\frac{x}{x^{2} +3x+2} -\frac{1}{x^{2} +3x+2} =\frac{x-1}{x^{2} +3x+2} \][/tex]

Answer:

he is correct

Step-by-step explanation:

dddddddddddddddddddddddddddd

Answer:

After the 6th day, beginning with the 7th day, he is walking less than 0.5 km.

Step-by-step explanation:

To find a percent of a number, change the percent to a decimal and multiply by the number. To find 90% of a number, change 90% to a decimal and multiply by the number.

90% = 0.9

He first walks 1 km. The next day, he walks 90% of 1 km. To find 90% of 1 km, multiply 0.9 by 1 km. It is 0.9 km. For the next day, he walks 90% of 0.9 km, which is 0.9 * 0.9 km = 0.81 km. To find how much he walks each day, multiply what he walked on the previous day by 0.9.

Now you can find out how much he walks each day until you see he walks less than 0.5 km.

Day 1: 1 km

Day 2: 1 km * 0.9 = 0.9 km

Day 3: 0.9 km * 0.9 = 0.81 km

Day 4: 0.81 km * 0.9 = 0.729 km

Day 4: 0.729 km * 0.9 = 0.6561

Day 5: 0.6561 km * 0.9 = 0.59049 km

Day 6: 0.59049 km * 0.9 = 0.531441 km

Day 7: 0.531441 km * 0.9 = 0.4782969 km

On the 6th day, he is still walking more than 0.5 km, but by the 7th day, he is walking less than 0.5 km.

Answer: After the 6th day, beginning with the 7th day, he is walking less than 0.5 km.

After the 6th day, beginning with the 7th day, he exists walking less than 0.5 km.

To find a percent of a number, change the percent to a decimal and multiply by the number. To discover 90% of a number, change 90% to a decimal and multiply by the number.

90% = 0.9

He first walks 1 km. The next day, he walks 90% of 1 km.

To discover 90% of 1 km, multiply 0.9 by 1 km. It is 0.9 km.

For the subsequent day, he walks 90% of 0.9 km, which exists

0.9 [tex]*[/tex] 0.9 km = 0.81 km.

To find how much he walks each day, multiply what he walked on the last day by 0.9.

Now find out how much he walks each day until sees he walks less than 0.5 km.

Day 1: 1 km

Day 2: 1 km [tex]*[/tex] 0.9 = 0.9 km

Day 3: 0.9 km [tex]*[/tex] 0.9 = 0.81 km

Day 4: 0.81 km [tex]*[/tex] 0.9 = 0.729 km

Day 4: 0.729 km [tex]*[/tex] 0.9 = 0.6561

Day 5: 0.6561 km [tex]*[/tex] 0.9 = 0.59049 km

Day 6: 0.59049 km [tex]*[/tex] 0.9 = 0.531441 km

Day 7: 0.531441 km [tex]*[/tex] 0.9 = 0.4782969 km

On the 6th day, he stands still walking more than 0.5 km, but by the 7th day, he exists walking less than 0.5 km.

Answer: After the 6th day, beginning with the 7th day, he exists walking less than 0.5 km.

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Answer: There is a probability of 40% of getting a school or district uses digital content and uses it as part of their curriculum.

Step-by-step explanation:

Since we have given that

Probability that schools or districts in a country use digital content = 80%

Probability that schools uses digital content as a part of their curriculum out of 80% = [tex]\dfrac{5}{10}[/tex]

So, the probability that a selected school or district uses digital content and uses it as  a part of their curriculum is given by

[tex]\dfrac{80}{100}\times \dfrac{5}{10}\\\\=0.8\times 0.5\\\\=0.4\\\\=40\%[/tex]

Hence, there is a probability of 40% of getting a school or district uses digital content and uses it as part of their curriculum.

The probability that a randomly selected school or district uses digital content and uses it as part of their curriculum is 40%.

To find the probability that a randomly selected school or district uses digital content and uses it as part of their curriculum, we need to multiply the probabilities of these events occurring.

Given that 80% of K-12 schools or districts use digital content and 5 out of 10 of these schools use it as part of their curriculum, we can calculate the probability as:

P(Uses digital content and uses it as part of curriculum) = P(Uses digital content) x P(Uses it as part of curriculum | Uses digital content)

Substituting the values, we have:

P(Uses digital content and uses it as part of curriculum) = 0.80 x 0.50 = 0.40 or 40%

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First, lets convert them into multipliers:

The multiplier for a:

                             25% increase = 1.25

                             40% decrease = 0.6

                             40% increase  = 1.4

Now to work out the overall percentage change, we just times all of the multipliers together, and convert it back to a percentage:

1.25 x 0.6 x 1.4 = 1.05    

So the overall multiplier is 1.05

And a multiplier of 1.05 = a 5% increase.

That means that the percentage change is + 5%

_________________________________________

Answer:

The percentage change in shares from the start of 2013 to the end of 2015 is:

+ 5%

_______________________________________

Note: if you haven't been taught multipliers - then ask and I'll try my best to explain!

Answer:

Step-by-step explanation:

Let's start by making up as many teams as we can with the 32 student. Given that each team is different, we can make 10 teams of 3 each. (we still have 23 more teams to make).

The last two people make a team of only 2. No matter which student from the 30 other students is picked, the team of two and the one the student is coming from will have one student in common. Though there are more borrowings that take place (many more), the results remain as stated. At least 2 teams will have 1 person in common.

The method is called the pigeon hole method.

By applying the Pigeonhole Principle in combinatorics, in a scenario where 32 students are assigned to 33 teams of 3 students each, there must exist two teams that share exactly one student.

This problem can be solved by using the principles of Combinatorics and the Pigeonhole Principle. The Pigeonhole Principle states that if you try to distribute n items into m containers and n > m, then at least one container must contain more than one item.

In the given scenario, we have 32 students that are being assigned to 33 teams, with each team consisting of 3 students. That means a total of 96 (3 x 32) places in teams.

If each student is a 'pigeon' and each 'place' in a team is a 'pigeonhole', the Pigeonhole Principle tells us that at least two pigeons must share at least one pigeonhole. Since each student can't be in more than one place at a time nor in the same team more than once, there must exist two teams that share exactly one student.

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Answer:

Nico invest [tex]\$2,100[/tex] at 7% and [tex]x=\$900[/tex] at 3%

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 Principal amount of money to be invested

r is the rate of interest  

t is Number of Time Periods

in this problem we have

At 7%

[tex]t=1\ years\\ P=\$x\\r=0.07[/tex]

substitute in the formula above

[tex]I1=x(0.07*1)[/tex]

[tex]I1=0.07x[/tex]

At 3%

[tex]t=1\ years\\ P=\$(x-1,200)\\r=0.03[/tex]

substitute in the formula above

[tex]I2=(x-1,200)(0.03*1)[/tex]

[tex]I2=0.03x-36[/tex]

The total interest is equal to

I=I1+I2

I=$174

substitute

[tex]174=0.07x+0.03x-36[/tex]

[tex]0.10x=174+36[/tex]

[tex]0.10x=210[/tex]

[tex]x=\$2,100[/tex]

[tex]x-1,200=2,100-1,200=\$900[/tex]

therefore

Nico invest [tex]\$2,100[/tex] at 7% and [tex]x=\$900[/tex] at 3%

Answer:  0.3936

Step-by-step explanation:

Given: Mean : [tex]\mu =1387 \text{ grams}[/tex]

Standard deviation : [tex]\sigma = 161 \text{ grams}[/tex]

The formula to calculate z is given by :-

[tex]z=\dfrac{x-\mu}{\sigma}[/tex]

For x= 1,425 grams

[tex]z=\dfrac{1425-1387}{161}=0.23602484472\approx0.27[/tex]

The P Value =[tex]P(X>1425)=P(z>0.27)=1-0.6064198=0.3935802\approx0.3936[/tex]

Hence, the  probability that a randomly selected broiler weighs more than 1,425 grams =0.3936

Final Answer:

There is approximately a 40.66% chance that a randomly selected broiler weighs more than 1,425 grams.

Explanation:

To solve this problem, you will need to apply the properties of the normal distribution. We want to find out the probability that a broiler weighs more than 1,425 grams.
Given:
- Mean (μ) = 1387 grams
- Standard deviation (σ) = 161 grams
- X = 1425 grams (the value we're interested in)

Step 1: First, we compute the z-score for the weight of 1425 grams. The z-score is a measure of how many standard deviations an element is from the mean. It can be calculated using the formula:

[tex]\[ z = \frac{(X - \mu)}{\sigma} \][/tex]

where X is the value for which we're finding the probability, μ is the mean, and σ is the standard deviation.

Step 2: Insert the values into the formula to compute the z-score for 1425 grams:

[tex]\[ z = \frac{(1425 - 1387)}{161} \\\\\[ z = \frac{38}{161} \\\\\[ z \approx 0.236 \][/tex]
Step 3: Once we have the z-score, we can use the z-table (a standard normal distribution table) to find out the probability of a z-score being less than 0.236. However, since we want the probability that the broiler weighs more than 1425 grams, we are interested in the probability of a z-score being greater than 0.236.

Step 4: Look up the corresponding probability for z = 0.236 on the z-table. The z-table gives us the area under the normal curve to the left of the z-score.

Let's assume the z-table gives us a probability of P(Z < 0.236). The value would typically be around 0.5934, which means there is a 59.34% chance that a random broiler will weigh less than 1425 grams.

Step 5: To find the probability that a broiler weighs more than 1425 grams, we subtract the value found in the z-table from 1 because the total area under the curve equals 1 (which corresponds to the probability of all possible outcomes).

[tex]\[ P(Z > 0.236) = 1 - P(Z < 0.236) \\\\\[ P(Z > 0.236) = 1 - 0.5934 \\\\\[ P(Z > 0.236) \approx 0.4066 \][/tex]

[tex]f(x_1,\ldots,x_n)=x_1+\cdots+x_n=\displaystyle\sum_{i=1}^nx_i[/tex]

[tex]{x_1}^2+\cdots+{x_n}^2=\displaystyle\sum_{i=1}^n{x_i}^2=4[/tex]

The Lagrangian is

[tex]L(x_1,\ldots,x_n,\lambda)=\displaystyle\sum_{i=1}^nx_i+\lambda\left(\sum_{i=1}^n{x_i}^2-4\right)[/tex]

with partial derivatives (all set equal to 0)

[tex]L_{x_i}=1+2\lambda x_i=0\implies x_i=-\dfrac1{2\lambda}[/tex]

for [tex]1\le i\le n[/tex], and

[tex]L_\lambda=\displaystyle\sum_{i=1}^n{x_i}^2-4=0[/tex]

Substituting each [tex]x_i[/tex] into the second sum gives

[tex]\displaystyle\sum_{i=1}^n\left(-\frac1{2\lambda}\right)^2=4\implies\dfrac n{4\lambda^2}=4\implies\lambda=\pm\frac{\sqrt n}4[/tex]

Then we get two critical points,

[tex]x_i=-\dfrac1{2\frac{\sqrt n}4}=-\dfrac2{\sqrt n}[/tex]

or

[tex]x_i=-\dfrac1{2\left(-\frac{\sqrt n}4\right)}=\dfrac2{\sqrt n}[/tex]

At these points we get a value of [tex]f(x_1,\cdots,x_n)=\pm2\sqrt n[/tex], i.e. a maximum value of [tex]2\sqrt n[/tex] and a minimum value of [tex]-2\sqrt n[/tex].

Answer:IF each vendor has his own price or (ppower) so far every single vendor will have his own price.

Step-by-step explanation:

The graph show\ing the demand (D) and supply (S = MC) curves in the market for hot dogs indicate: Competitive market.

In a market were their is competition, when demand and supply curves intersect this indicate market equilibrium.

Based on the graph the market equilibrium price will be $1.50 per hot dog while on the other hand the market equilibrium quantity will be 250 hot dogs which  is the point were demand and supply intersect.

Inconclusion the market for hot dogs indicate: Competitive market.

Learn more about competitve market here:https://brainly.com/question/25717627

Jessica has a total of 22,275 different possible schedules to choose from for her next semester given the number of sections for each class and assuming there are no time conflicts.

Jessica is creating her semester schedule and there are 15 sections of English 101, 9 sections of Spanish 102, 11 sections of Biology 102, and 15 sections of College Algebra. To figure out how many different possible schedules are available, we need to multiply the number of sections for each class.

Therefore, the total number of different possible schedules Jessica can choose is calculated as follows:

15 (English 101) * 9 (Spanish 102) * 11 (Biology 102) * 15 (College Algebra) = 22,275 possible schedules.

This is under the assumption that there are no time conflicts between the different classes.

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Answer:

  r = 11

Step-by-step explanation:

Substituting the given value for C, we have ...

  22€ = 2€·r

Dividing by the coefficient of r, we get

  22€/(2€) = r = 11

To find the value of 'r' using the equation C = 2€r; C = 22€, you simply need to substitute the given value of C into the equation and solve for r. In this case, r equals 11.

Given : C = 2€r; C = 22€

To find the value of r, substitute C = 22€ into the equation:

22€ = 2€r

Divide both sides by 2€ to isolate r:
r = 11

Answer:

22 minutes after 7:00 P.M. they will be closest.

Step-by-step explanation:

A boat heading south travelling for t hours at the rate of 20 km/h, so the distance x = 20t

The another boat will reach the dock after travelling another 1-t hours at the rate of 15 km/h, so the distance =

y = 15 - 15t

D = d²  = x² + y²

D = (20t)² + (15 - 15t)²

dD/dt = -2(15² )( 1-t ) +2 × 20² × t

dD/dt = 2 (15² + 20²) × t -2 ( 15 )² = 0

t = [tex]\frac{2(15)^{2}}{(2\times15^{2}+2\times20^{2})}[/tex]

t = 0.36 hours = 0.36 × 60 = 21.6 minutes ≈ 22 minutes

Therefore, the distance is minimized 22 minutes after 7 pm.

The two boats were closest together 12 minutes after 7:00 PM.

To find the time when the two boats were closest together, we can first determine the position of each boat at 8:00 PM. The boat traveling south will have traveled for 1 hour at a speed of 20 km/h, so it would be 20 km south of the dock. The boat traveling east will have traveled for 1 hour at a speed of 15 km/h, so it would be 15 km east of the dock. We can then calculate the distance between the two boats by using the Pythagorean theorem. The distance is the square root of the sum of the squares of the distances traveled south and east, which is approximately 25 km. Since both boats started at the dock at 7:00 PM, to find the time when they were closest together, we can subtract the time traveled by the boat heading south until it reaches the closest point to the other boat from 60 minutes. The boat heading south will have traveled (20/25) * 60 minutes, which is 48 minutes. So, the two boats were closest together 12 minutes after 7:00 PM.

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Step-by-step explanation:

[tex]x-\text{the number}\\\\\text{Nine times the difference of a number and 3}:\\\\\boxed{9\times(x-3)=9(x-3)}[/tex]

[tex]9(x-3)\qquad\text{use the distributive property}\ a(b+c)=ab+ac\\\\=9x+(9)(-3)\\\\=9x-27[/tex]

Check the pictures below.

if we knew the roots/solutions of the equation, we can set h(s) = 0 and solve for "s" to find out how many seconds is it when the height is 0.

if you notice in the first picture, when f(x) = 0, is when the parabola hits a root/solution or the ground, for David he'll be hitting the water surface, and the equation that has both of those roots/solutions conspicuous is

h(s) = -4.9(s - 2)(s + 1).

Answer:

450 km

Step-by-step explanation:

Let's say Va is the speed of the car from city A, Ta is the time it spent traveling, and Da is the distance it traveled.

Similarly, Vb is the speed of the car from city B, Tb is the time it spent traveling, and Db is the distance it traveled.

Given:

Va = Vb - 10

Ta₁ = Tb₁ = 5

Ta₂ = Tb₂ + 4.5

Db₂ = 150

Find:

D = Da₁ + Db₁ = Da₂ + Db₂

Distance = rate × time

In the first scenario:

Da₁ = Va Ta₁

Da₁ = (Vb - 10) (5)

Da₁ = 5Vb - 50

Db₁ = Vb Tb₁

Db₁ = Vb (5)

Db₁ = 5Vb

So:

D = Da₁ + Db₁

D = 10Vb - 50

In the second scenario:

Da₂ = Va Ta₂

Da₂ = (Vb - 10) (Tb₂ + 4.5)

Da₂ = Vb Tb₂ + 4.5Vb - 10Tb₂ - 45

Db₂ = Vb Tb₂

150 = Vb Tb₂

Substituting:

Da₂ = 150 + 4.5Vb - 10Tb₂ - 45

Da₂ = 105 + 4.5Vb - 10Tb₂

Da₂ = 105 + 4.5Vb - 10 (150 / Vb)

Da₂ = 105 + 4.5Vb - (1500 / Vb)

So:

D = Da₂ + Db₂

D = 105 + 4.5Vb - (1500 / Vb) + 150

D = 255 + 4.5Vb - (1500 / Vb)

Setting this equal to the equation we found for D from the first scenario:

10Vb - 50 = 255 + 4.5Vb - (1500 / Vb)

5.5Vb - 305 = -1500 / Vb

5.5Vb² - 305Vb = -1500

5.5Vb² - 305Vb + 1500 = 0

11Vb² - 610Vb + 3000 = 0

(Vb - 50) (11Vb - 60) = 0

Vb = 50, 5.45

Since Vb > 10, Vb = 50 km/hr.

So the distance between the cities is:

D = 10Vb - 50

D = 10(50) - 50

D = 450 km