There are 47 contestants at a national dog show. How many different ways can contestants fill the first place, second place, and third place positions?

Complete question is;

A machine used to fill gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of 128 ounces and a standard deviation of 0.20 ounce. You randomly select 35 cans and carefully measure the contents. The sample mean of the cans is 127.9 ounces. Does the machine need to be? reset? Explain your reasoning.

(yes/no)?, it is (very unlikely/ likely) that you would have randomly sampled 35 cans with a mean equal to 127.9 ?ounces, because it (lies/ does not lie) within the range of a usual? event, namely within (1 standard deviation, 2 standard deviations 3 standard deviations) of the mean of the sample means.

Answer:

Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.

Step-by-step explanation:

We are given;

Mean: μ = 128

Standard deviation; σ = 0.2

n = 35

Now, formula for standard error of mean is given as;

se = σ/√n

se = 0.2/√35

se = 0.0338

Normally, the range of values should be within 2 standard deviations of mean. In this case, normal range of values will be;

μ ± 2se = 128 ± 0.0338

This gives; 127.9662, 128.0338

So, Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.

De acordo com a disponibilidade da unidade, há apenas a seguinte dosagem: 1g/2mL - ou seja, uma grama de dipirona a cada 2mL

O enunciado está meio mal formulado, pois é dito que foram prescritos 500mg de dipirona e é essa quantidade de farmaco que a criança tem que tomar. Deseja-se saber quantos mL deverao ser administrados.

Fazendo a classica regra de 3, podemos chegar no volume desejado:

(atentar que 500mg = 0,5g)

     g               mL

     1    ---------   2

    0,5  ---------  X    

1 . X = 0,5 . 2

Step-by-step explanation:

Using Sine Rule

[tex] \frac{ \sin(a) }{ |a| } = \frac{ \sin(b) }{ |b| } = \frac{ \sin(c) }{ |c| } [/tex]

[tex] \frac{ \sin(42) }{5} = \frac{ \sin(38) }{a} [/tex]

[tex]a = \frac{5( \sin(38))}{ \sin(42) } [/tex]

[tex]a = 4.6[/tex]

[tex] \frac{ \sin(42) }{5} = \frac{ \sin(100) }{b} [/tex]

[tex]b= \frac{5( \sin(100))}{ \sin(42) } [/tex]

[tex]b = 7.4[/tex]

Answer:

  x = 1.00995066776

  x = 2.52925492433

Step-by-step explanation:

This sort of equation is best solved using a graphing calculator. For that purpose, I like to rewrite the equation as a function whose zeros we're seeking. Here, that becomes ...

  [tex]f(x)=\log{(x)}-\log{(x-1)}^2-2\log{(x-1)}[/tex]

The attached graph shows zeros at

  x = 1.00995066776 and 2.52925492433

_____

Comment on the equation

Note that we have taken the middle term to be the square of the log, rather than the log of a square. For the latter interpretation, see mberisso's answer at https://brainly.com/question/17210068

Comment on the answer refinement

We have used Newton's method iteration to refine the solutions to this equation. The solution near 1.00995 requires the initial guess be very close for that method to work properly. Fortunately, the 1.01 value shown on the graph is sufficient for the purpose.

Answer:

1/8

Step-by-step explanation:

3/8-1/8-1/8=1/8

Step-by-step explanation:

utilise the formula a+(n-1)d

a is the first number while d is common difference

Answer:

22

Step-by-step explanation:

Using the formular, Un = a + (n - 1)d

Where n = 10; a = -23; d = 5

U10 = -23 + (9)* 5

U10 = -23 + 45 = 22

Answer:

The two numbers are 1 and 2.

Step-by-step explanation:

Let the two numbers be a and b.

One number is twice another, so let's let b=2a.

Their reciprocals are 3/2. Thus:

[tex]\frac{1}{a}+\frac{1}{b} =\frac{3}{2}[/tex]

Substitute and solve for a:

[tex]\frac{1}{a}+\frac{1}{2a} =\frac{3}{2}\\[/tex]

Combine the fractions by forming a common denominator by multiplying the left term by 2:

[tex]\frac{2}{2a} +\frac{1}{2a}=\frac{3}{2}[/tex]

Combine and cross-multiply:

[tex]3/2a=3/2\\6a=6\\a=1\\b=2(1)=2[/tex]

Thus, the two numbers are 1 and 2.

Answer:

A) C1 = 0.00187 m = 0.187 cm,  C2 = 0.0062 m = 0.62 cm

B)  A sample of how the graph looks like is attached below ( periodic sine wave )

C) w = [tex]\sqrt[4]{3}[/tex] is when the amplitude of the forced response is maximum

Step-by-step explanation:

Given data :

mass = 5kg

length of spring = 10 cm = 0.1 m

f(t) = 10sin(t) N

viscous force = 2 N

speed of mass = 4 cm/s = 0.04 m/s

initial velocity = 3 cm/s = 0.03 m/s

Formulating initial value problem

y = viscous force / speed = 2 N / 0.04 m/s = 50 N sec/m

spring constant = mg/ Length of spring = (5 * 9.8) / 0.1 = 490 N/m

f(t) = 10sin(t/2) N

using the initial conditions of u(0) = 0 m and u"(0) = 0.03 m/s to express the equation of motion

the equation of motion = 5u" + 50u' + 490u = 10sin(t/2)

A) finding the solution of the initial value

attached below is the solution and

B) attached is a periodic sine wave replica of how the grapgh of the steady state solution looks like

C attached below

volume of a cone

.

.

.

volume of sphere

.

.

number of spheres that can be made......

.

.

hence a hemisphere can be formed

Answer:

1/3(a+b)

hope it helps :>

Find the powers [tex]a=\sqrt{2}+\sqrt{3}[/tex]

$a^{2}=5+2 \sqrt{6}$

$a^{3}=11 \sqrt{2}+9 \sqrt{3}$

The cubic term gives us a clue, we can use a linear combination to eliminate the root 3 term $a^{3}-9 a=2 \sqrt{2}$ Square $\left(a^{3}-9 a\right)^{2}=8$ which gives one solution. Expand we have $a^{6}-18 a^{4}-81 a^{2}=8$ Hence the polynomial $x^{6}-18 x^{4}-81 x^{2}-8$ will have a as a solution.

Note this is not the simplest solution as $x^{6}-18 x^{4}-81 x^{2}-8=\left(x^{2}-8\right)\left(x^{4}-10 x^{2}+1\right)$

so fits with the other answers.

Answer:

[tex]y^3 -6y-6[/tex]

Answer:

Change due is 3.50

Step-by-step explanation:

20.00-16.50 is 3.50

Answer: $3.50

Step-by-step explanation:

You subtract 20 from 16.50.

Answer:

z = 83/( -c+6-t)

Step-by-step explanation:

-cz + 6z = tz + 83

Subtract tz from each side

-cz + 6z -tz= tz-tz + 83

-cz + 6z - tz = 83

Factor out z

z( -c+6-t) = 83

Divide each side by ( -c+6-t)

z( -c+6-t)/( -c+6-t)  = 83/( -c+6-t)

z = 83/( -c+6-t)

Answer:

6.02

six point zero two

Step-by-step explanation:

Answer:

602 / 100= 6,02

Step-by-step explanation:

602 to divide 100 = 6,02

Answer:

mean=87

median=87

Step-by-step explanation:

mean=sum of test score/number of subject

mean=79+91+93+85+86+88/6

mean=522/6

mean=87

Literal meaning of median is medium.

To find the number which lies in the medium, we must rearrange the number in ascending.

79, 91, 93, 85, 86, 88

79, 85, 86, 88, 91, 93

86+88/2=87

Hope this helps ;) ❤❤❤

Let me know if there is an error in my answer.

Assuming the cube is closed, you can use the divergence theorem:

[tex]\displaystyle\iint_S\vec F\cdot\mathrm dS=\iiint_T\mathrm{div}\vec F\,\mathrm dV[/tex]

where [tex]S[/tex] is the surface of the cube and [tex]T[/tex] is the region bounded by [tex]S[/tex].

We have

[tex]\mathrm{div}\vec F=\dfrac{\partial(y+z)}{\partial x}+\dfrac{\partial(x+z)}{\partial y}+\dfrac{\partial(x+y)}{\partial z}=0[/tex]

so the flux is 0.

Answer:

  [tex]\dfrac{3}{(y+2)^3}[/tex]

Step-by-step explanation:

Maybe you want this written using math symbols. It will be ...

  [tex]\boxed{\dfrac{3}{(y+2)^3}}[/tex]

Answer:

x=9/2

Step-by-step explanation:

Let's solve your equation step-by-step.

5x+4(−x−2)=−5x+2(x−1)+12

Step 1: Simplify both sides of the equation.

5x+4(−x−2)=−5x+2(x−1)+12

5x+(4)(−x)+(4)(−2)=−5x+(2)(x)+(2)(−1)+12 (Distribute)

5x+−4x+−8=−5x+2x+−2+12

(5x+−4x)+(−8)=(−5x+2x)+(−2+12) (Combine Like Terms)

x+−8=−3x+10

x−8=−3x+10

Step 2: Add 3x to both sides.

x−8+3x=−3x+10+3x

4x−8=10

Step 3: Add 8 to both sides.

4x−8+8=10+8

4x=18

Step 4: Divide both sides by 4.

4x/4=18/4

x=9/2

Answer:

300.05 miles

Step-by-step explanation:

initial fee= $39.99

final bill = $ 100

cost =$ 0.20 per mile

remaining amount = $ 60.01

solution,

she drive = remaining amount / cost

=60.01/0.20

=300.05 miles

Answer:

500 miles

Step-by-step explanation:

Let us use cross multiplication to find the unknown amount.

Given:

1) Cost for 1 mile=$0.20

2)Cost for x miles=$100

Solution:

No of miles                             Cost

1) 1                                             $0.20

2)x                                             $100

By cross multiplying,

100 x 1= 0.20x

x=100/0.20

x=500 miles

Thank you!

Answer:

this? hope it helps ........

Answer:

The answer is area=32pi-64 and the perimeter is 8pi

Step-by-step explanation:

Answer:

A.the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]

B. β  = 0.0122

C. β  = 0.0000

Step-by-step explanation:

Given that:

Mean = 100

standard deviation = 2

sample size = 9

The null and the alternative hypothesis can be computed as follows:

[tex]\mathtt{H_o: \mu = 100}[/tex]

[tex]\mathtt{H_1: \mu \neq 100}[/tex]

A. If the acceptance region is defined as [tex]98.5 < \overline x > 101.5[/tex] , find the type I error probability [tex]\alpha[/tex] .

Assuming the critical region lies within [tex]\overline x < 98.5[/tex] or [tex]\overline x > 101.5[/tex], for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is [tex]\mu = 100[/tex]

∴

[tex]\mathtt{\alpha = P( type \ 1 \ error ) = P( reject \ H_o)}[/tex]

[tex]\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5 )}[/tex]

when  [tex]\mu = 100[/tex]

[tex]\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }[/tex]

[tex]\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z > \dfrac{1.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\alpha = P ( Z <-2.25 ) + P(Z > 2.25) }[/tex]

[tex]\mathtt{\alpha = P ( Z <-2.25 ) +( 1- P(Z < 2.25) })[/tex]

From the standard normal distribution tables

[tex]\mathtt{\alpha = 0.0122+( 1- 0.9878) })[/tex]

[tex]\mathtt{\alpha = 0.0122+( 0.0122) })[/tex]

[tex]\mathbf{\alpha = 0.0244 }[/tex]

Thus, the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]

B. Find beta for the case where the true mean heat evolved is 103.

The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis [tex]\mathtt{H_o}[/tex]

Thus;

β = P( type II error) - P( fail to reject [tex]\mathtt{H_o}[/tex] )

∴

[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]

Given that [tex]\mu = 103[/tex]

[tex]\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }[/tex]

[tex]\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }[/tex]

[tex]\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}[/tex]

From standard normal distribution table

β  = 0.0122 - 0.0000

β  = 0.0122

C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?

[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]

Given that [tex]\mu = 105[/tex]

[tex]\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }[/tex]

[tex]\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }[/tex]

[tex]\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}[/tex]

From standard normal distribution table

β  = 0.0000 - 0.0000

β  = 0.0000

The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.

Complete Question

On the uploaded image is a similar question that will explain the given question

Answer:

The value of k is  [tex]k = 214285.7[/tex]

The percentage  of the oil that will be cleaned is [tex]x = 80.77\%[/tex]

Step-by-step explanation:

From the question we are told that

   The  cost of cleaning up the spillage is  [tex]C = \frac{ k x }{100 - x }[/tex]  [tex]x \le x \le 100[/tex]

     The  cost of cleaning x =  70% of the oil is  [tex]C = \$500,000[/tex]

Now at  [tex]C = \$500,000[/tex] we have  

       [tex]\$ 500000 = \frac{ k * 70 }{100 - 70 }[/tex]

       [tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]

      [tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]

      [tex]k = 214285.7[/tex]

Now  When  [tex]C = \$900,000[/tex]

       [tex]x = 80.77\%[/tex]

the graph has 12 segments so angle enclosed by each segment is [tex] {2\pi\over 12}=\frac{\pi}6[/tex]

anti-clockwise is taken as positive, so if you want positive q, you need to rotate 8 segments [tex] q=8\frac,{\pi}6=\frac{4\pi}3 [/tex] , and and 8 circles or units so r=8

and for a negative angle, you need to rotate clockwise

Which is 4 segments from the horizontal line. so [tex]q=-\frac{2\pi}3[/tex] and r will be same, 8 units.

[not sure about -r so I won't include it in answer]

Answer:

Points : ( 8, - 2π/3 ), ( - 8, π/3 ), ( - 8, - 5π/3 )

Step-by-step explanation:

For the first two cases, ( r, θ ) r would be > 0, where r is the directed distance from the pole, and theta is the directed angle from the positive x - axis.

So when r is positive, we can tell that this point is 8 units from the pole, so r is going to be 8 in either case,

( 8, 240° ) - because r is positive, theta would have to be an angle with which it's terminal side passes through this point. As you can see that would be 2 / 3rd of 90 degrees more than a 180 degree angle,or 60 + 180 = 240 degrees.

( 8, - 120° ) - now theta will be the negative side of 360 - 240, or in other words - 120

Now let's consider the second two cases, where r is < 0. Of course the point will still be 8 units from the pole. Again for r < 0 the point will lay on the ray pointing in the opposite direction of the terminal side of theta.

( - 8, 60° ) - theta will now be 2 / 3rd of 90 degrees, or 60 degrees, for - r. Respectively the remaining degrees will be negative, 360 - 60 = 300, - 300. Thus our second point for - r will be ( - 8, - 300° )

_________________________________

So we have the points ( 8, 240° ), ( 8, - 120° ), ( - 8, 60° ), and ( - 8, - 300° ). However we only want 3 cases, so we have points ( 8, - 120° ), ( - 8, 60° ), and ( - 8, - 300° ). Let's convert the degrees into radians,

Points : ( 8, - 2π/3 ), ( - 8, π/3 ), ( - 8, - 5π/3 )

Answer:

[tex]L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ][/tex]

Step-by-step explanation:

Given that:

[tex]f(t) = 12 cos (t- \dfrac{\pi}{6})[/tex]

recall that:

cos (A-B) = cos AcosB + sin A sin B

∴

[tex]f(t) = 12 [cos\ t \ cos \dfrac{\pi}{6}+ sin \ t \ sin \dfrac{\pi}{6}][/tex]

[tex]f(t) = 12 [cos \ t \ \dfrac{3}{2}+ sin \ t \ sin \dfrac{1}{2}][/tex]

[tex]f(t) = 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t)[/tex]

[tex]L(f(t)) = L ( 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t) ][/tex]

[tex]L(f(t)) = 6 \sqrt{3} \ L [cos \ (t) ] + 6\ L [ sin \ (t) ][/tex]

[tex]L(f(t)) = 6 \sqrt{3} \dfrac{S}{S^2 + 1^2}+ 6 \dfrac{1}{S^2 +1^2}[/tex]

[tex]L(f(t)) = \dfrac{6 \sqrt{3} +6 }{S^2+1}[/tex]

[tex]L(f(t)) = \dfrac{6( \sqrt{3} \ S +1 }{S^2+1}[/tex]

[tex]L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ][/tex]

Answer:

csctheta= [tex]\frac{\sqrt{13} }{3}[/tex]

Step-by-step explanation:

answer is provided on top

The value of the [tex]\rm cosec \theta = \frac{\sqrt{13} }{3}[/tex]. Cosec is found as the ratio of the hypotenuse and the perpendicular.

The field of mathematics is concerned with the relationships between triangles' sides and angles, as well as the related functions of any angle

The given data in the problem is;

[tex]\rm cot \theta = \frac{2}{3}[/tex]

The [tex]cot \theta[/tex] is found as;

[tex]\rm cot \theta = \frac{B}{P} \\\\ \rm cot \theta = \frac{2}{3} \\\\ B=2 \\\\ P=3 \\\\[/tex]

From the phythogorous theorem;

[tex]\rm H=\sqrt{P^2+B^2} \\\\ \rm H=\sqrt{2^2+3^2} \\\\ H=\sqrt{13} \\\\[/tex]

The value of the cosec is found as;

[tex]\rm cosec \theta = \frac{H}{P} \\\ \rm cosec \theta = \frac{\sqrt{13} }{3}[/tex]

Hence the value of the [tex]\rm cosec \theta = \frac{\sqrt{13} }{3}[/tex].

To learn more about the trigonometry refer to the link;

https://brainly.com/question/26719838

Answer:

The first picture's answer would be (6, 21)

Step-by-step explanation:

You have to find the points on the 8th and the 9th day, and then you would add them together, and then divide by two finding the average, which would be 24 and 18, so when added, you get 42, divided by 2 you get 21. You look on the graph for the point with 21, and you find it is on 6.

Answer:

A: 19

Step-by-step explanation:

For this, we can complete the square. We first look at the first 2 terms,

t^2 and -6t.

We know that [tex](t-3)^2[/tex] will include terms.

[tex](t-3)^2 = t^2 - 6t + 9[/tex]

But [tex](t-3)^2[/tex] will also add 9, so we can subtract 9. Putting this into the equation, we get:

[tex]m(t) = (t-3)^2 - 9 +28[/tex]

[tex]m(t) = (t-3)^2 +19[/tex]

Using the trivial inequality, which states that a square of a real number must be positive, we can say that in order to have the minimum number of members, we need to make (t-3) = 0. Luckily, 3 months is in our domain, which means that the minimum amount of members is 19.

Answer:

Step-by-step explanation:

A school bag contains 12 pens of which 5 are red and the other are black. 4 pens are selected from the bag.

(a) How many different samples of size 4 pens are possible?

12C4=12!/(4!*8!)=495

(b) How many samples have 3 red pens and 1 black pen?

5C3*7C1

5C3=5!/(3!*2!)=10

7C1=7!/(1!*6!)=7

=>5C3*7C1=10*7=70

(c) How many samples of size 4 contain at least two red pens?

(5C2*7C2)+(5C3*7C1)+(5C4*7C0)

5C2=5!/(2!*3!)=10

7C2=7!/(2!*5!)=21

5C3=5!/(3!*2!)=10

7C1=7!/(1!*6!)=7

5C4=5!/(4!*1!)=5

7C0=7!/(0!*7!)=1

=>(5C2*7C2)+(5C3*7C1)+(5C4*7C0)=285

(d) How many samples of size 4 contain at most one black pen?

(7C1*5C3)+(7C0*5C4)

7C1=7!/(1!*6!)=7

7C0=7!/(0!*7!)=1

5C3=5!/(3!*2!)=10

5C4=5!/(4!*1!)=5

=>(7C1*5C3)+(7C0*5C4)=(7*10)+(1*5)=75

Answer:

Step-by-step explanation:

5x - 15 = 5(-4x - 3)

Multiply the terms in the bracket

5x - 15 = - 20x - 15

Group like terms

Send the constants to the right side of the line and those with variables to the left side

That's

5x + 20x = - 15 + 15

Simplify

25x = 0

Divide both sides by 25

We have the final answer as

Hope this helps you

Answer:

x=0

Step-by-step explanation:

5x - 15 = 5(-4x - 3)

To find the solution to this equation, we have to get x by itself on one side of the equation.

First, distribute the 5 on the right side. Multiply each term by 5.

5x - 15= (5*-4x) + (5*-3)

5x-15 = -20x + (5*-3)

5x-15= -20x -15

Next, add 20x to both sides of the equation.

(5x+20x) -15 = (-20x+20x) -15

(5+20x) -15 = -15

25x -15=-15

Next, add 15 to both sides of the equation.

25x -15 +15 = -15+15

25x= -15+15

25x=0

Finally, divide both sides of the equation by 25.

25x/25=0/25

x= 0/25

x= 0

The solution to this equation is x=0

Answer:

work pictured and shown

Answer:

Last one

Step-by-step explanation:

● [ ( 3^2 × 5^0) / 4 ]^2

5^0 is 1 since any number that has a null power is equal to 1.

●[ (3^2 ×1 ) / 4 ]^2

● (9/4)^2

● 81 / 16