If cos (50 degrees) = a, express tan (130 degrees) in terms of a.

Answer:

The probability that exactly 5 are unable to complete the  race is 0.1047

Step-by-step explanation:

We are given that 25% of all who enters a race do not complete.

30 have entered.

what is the probability that exactly 5 are unable to complete the  race?

So, We will use binomial

Formula : [tex]P(X=r) =^nC_r p^r q^{n-r}[/tex]

p is the probability of success i.e. 25% = 0.25

q is the probability of failure =  1- p  = 1-0.25 = 0.75

We are supposed to find the probability that exactly 5 are unable to complete the  race

n = 30

r = 5

[tex]P(X=5) =^{30}C_5 (0.25)^5 (0.75)^{30-5}[/tex]

[tex]P(X=5) =\frac{30!}{5!(30-5)!} \times(0.25)^5 (0.75)^{30-5}[/tex]

[tex]P(X=5) =0.1047[/tex]

Hence the probability that exactly 5 are unable to complete the  race is 0.1047

Final answer:

Explaining the impact of selling additional sandwiches on MTA's profits.

Explanation:

To prepare the schedule showing the impact on MTA's profits:

Calculate the total income from selling 800 additional sandwiches: $2.85 x 800 = $2,280.

Calculate the total cost of producing 800 sandwiches: $5 x 800 = $4,000.

Subtract the cost from the income to find the profit impact: $2,280 - $4,000 = lower profit.

Explanation (150 words): Adding 800 sandwiches at a cost of $5 each while selling at $2.85 results in a loss, impacting MTA's profits negatively. This shift reduces the overall profitability due to the increased production costs outweighing the revenue generated from the additional sandwich sales.

Answer:

You must pick at least 13 pennies to be sure of getting at least five from the same year.

Step-by-step explanation:

You have:

12 1967 pennies

7 1868 pennies

11 1971 pennies.

how many must you pick to be sure of getting at least five pennies from the same year?

This value is the multiplication of the number of different pennies by the antecessor of the number of pennies you want, added by 1.

So

You have 3 differennt pennies

You want to get at least five from the same year.

[tex]3*4 + 1 = 13[/tex]

You must pick at least 13 pennies to be sure of getting at least five from the same year.

For example, if you pick 12 pennies, you can have four from each year.

Adding three values, 13 is the smallest number that you need at least one term of the addition being equal or bigger than 5.

To be sure of getting at least five pennies from the same year, you need to pick 35 pennies in total.

To ensure that you get at least five pennies from the same year, you need to consider the worst-case scenario. In this case, the worst-case scenario is where you pick pennies from each of the three different years first before getting five from the same year. So, you need to pick the maximum number of pennies from each year first before reaching the desired goal. The maximum number of pennies you need to pick is:

12 + 7 + 11 + 5 = 35 pennies

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Answer: There were 469,532 American prison for drug related offenses in 2015.

Step-by-step explanation:

Since we have given that

Number of Americans in prison for drug related offenses in 1980 = 40,900

Rate of increment in Americans in prison from 1980 to 2015 = 1048%

So, Number of Americans who were in prison for drug related offenses in 2015 is given by

[tex]\dfrac{100+1048}{100}\times 40900\\\\\dfrac{1148}{100}\times 40900\\\\=1148\times 409\\\\=469,532[/tex]

Hence, there were 469,532 American prison for drug related offenses in 2015.

Answer:

Length = Width = 7 inches

Height = 14 inches

Step-by-step explanation:

Let's call  

x = length and width of the bottom and top

y = heigth

The bottom and top are squares, so their total area is

[tex]A_{bt}=x^2+x^2=2x^2[/tex]

the area of one side is xy. As we have 4 sides, the total area of the sides is

[tex]A_s=4xy[/tex]

The cost for the bottom and top would be

[tex]C_{bt}=\$4.2x^2=\$8x^2[/tex]

The cost for the sides would be

[tex]C_s=\$2.4xy=\$8xy[/tex]

The total cost to made the box is

[tex]C_t=\$(8x^2+8xy)[/tex]

But the volume of the box is  

[tex]V=x^2y=686in^3[/tex]

isolating

[tex]y=\frac{686}{x^2}[/tex]

Replacing this expression in the formula of the total cost

[tex]C_t(x)=8x^2+8x(\frac{686}{x^2})=8x^2+\frac{5488}{x}[/tex]

The minimum of the cost should be attained at the point x were the derivative of the cost is zero.

Taking the derivative

[tex]C'_t(x)=16x-\frac{5488}{x^2}[/tex]

If C' = 0 then

[tex]16x=\frac{5488}{x^2}\Rightarrow x^3=\frac{5488}{16}\Rightarrow x^3=343\Rightarrow x=\sqrt[3]{343}=7[/tex]

We have to check that this is actually a minimum.

To check this out, we take the second derivative

[tex]C''_t(x)=16+\frac{10976}{x^3}[/tex]

Evaluating this expression in x=7, we get C''>0, so x it is a minimum.

Now that we have x=7, we replace it on the equation of the volume to get

[tex]y=\frac{686}{49}=14in[/tex]

The dimensions of the most economical box are

Length = Width = 7 inches

Height = 14 inches

To find the dimensions of the container of least cost, assume the length of the base is x and the width of the base is also x. The height of the container is 686/(x^2). The dimensions of the container of least cost are approximately 10.33 inches by 10.33 inches by 6.68 inches.

To find the dimensions of the container of least cost, we need to minimize the cost function. Let's assume the length of the base is x, then the width of the base will also be x since it's a square. The height of the container will be 686/(x^2) since the volume is given as 686 in³. The total cost, C, is given by C = 4(2x^2) + 2(4x(686/(x^2))). Simplifying this expression gives C = 8x^2 + 5488/x. To find the minimum cost, we can take the derivative of C with respect to x and set it equal to zero:

dC/dx = 16x - 5488/x^2 = 0

Solving this equation gives x = √(343/2) ≈ 10.33. Therefore, the dimensions of the container of least cost are approximately 10.33 inches by 10.33 inches by 6.68 inches.

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Answer:

Part 1.

In 12 hours 30 units of  Humulin R insulin in 300 mL is to be infused.

So, per hour = [tex]\frac{30}{12}= 2.5[/tex] units are to be infused.

Part 2.

Now 30 units Humulin R insulin in 300 mL of normal saline (NS).

So, 2.5 units will be in : [tex]\frac{300\times2.5}{30}= 25[/tex] mL

Hence. 2.5 units Humulin R insulin in 25 mL of normal saline (NS).

Answer:

Venom C is  0.625 times as potent as venom A.

Step-by-step explanation:

We are given that venom  A is four times as potent as venom B.

Venom C is 2.5 times as potent as Venom B.

We have to find that how many times as potent is venom C is compared to venom A.

Let Venom B =x

Then Venom A=[tex]4\times x=4x[/tex]

Venom C=[tex]2.5\times x=2.5 x[/tex]

[tex]\frac{Venom\;C}{venom\;A}=\frac{2.5x}{4x}=\frac{25}{40}=\frac{5}{8}=0.625[/tex]

Hence, venom C is  0.625 times as potent as venom A.

In relative terms, venom C is 0.625 times as potent as venom A, using venom B as a common comparison base.

To solve this, we first need to establish a common comparison base. If venom A is four times as potent as venom B, let's assign a relative potency value to venom B, let's say 1. Therefore, the potency of venom A is 4. Similarly, if venom C is 2.5 times as potent as venom B, then the potency of venom C is 2.5.

To find out how much more potent venom C is compared to venom A, we divide the potency of venom C by the potency of venom A. Mathematically, it looks like this: Potency of C ÷ Potency of A = 2.5 ÷ 4 = 0.625. Therefore, venom C is 0.625 times as potent as venom A.

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Answer:

a) The patient weighs 75.15kg = 75.1kg, rounded to the nearest tenth.

b) The patient should 75.1mg a day of the drug.

c) The patient should receive 0.37mL per dose, rounded to the nearest hundreth.

Step-by-step explanation:

These problems can be solved by direct rule of three, in which we have cross multiplication.

a. How many kilograms does the patient weigh?

The problem states that patient weighs 167lb. Each lb has 0.45kg. So:

1 lb - 0.45kg

167 lb - xkg

[tex]x = 167*0.45[/tex]

[tex]x = 75.15[/tex]kg

The patient weighs 75.15kg = 75.1kg, rounded to the nearest tenth.

b. How many milligrams should the patient receive per day?

The drug has 1mg/kg. The patient weighs 75.1kg. So

1 mg - 1 kg

x mg - 75.1kg

[tex]x = 75.1[/tex]mg

The patient should 75.1mg a day of the drug.

c. How many milliliters should the patient receive per dose?

The drug is SubQ, q12h. This means that the drug is administered twice a day, so there are 2 doses. 75.1mg of the drug are administered a day. so:

2 doses - 75.1mg

1 dose - xmg

[tex]2x = 75.1[/tex]

[tex]x = \frac{75.1}{2}[/tex]

[tex]x = 37.5[/tex] SubQ, q12h

For each dose, the patient should receive 37.5mg. Each 40mg of the drug has 0.4mL. So:

40mg - 0.4ml

37.5mg - xmL

[tex]40x = 0.4*37.4[/tex]

[tex]x = \frac{0.4*37.4}{40}[/tex]

[tex]x = 0.374mL[/tex]

The patient should receive 0.37mL per dose, rounded to the nearest hundreth.

The conversion of 1.256 grams is as follows: 1,256,000 micrograms, 1256 milligrams, and 0.001256 kilograms.

To convert 1.256 grams (g) to different units of mass, you use the following conversion factors:

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1.256 grams is equal to 1,256,000 micrograms, 1,256 milligrams, and 0.001256 kilograms. These conversions use basic multiplication and division of the metric system.

To convert 1.256 grams to different units, you should be familiar with the respective conversion factors:

Steps to Convert:

Therefore, the conversions are:

Answer:

I think that what you are trying to show is:  If [tex]s[/tex] is irrational and [tex]r[/tex] is rational, then [tex]r+s[/tex]  is rational. If so, a proof can be as follows:

Step-by-step explanation:

Suppose that [tex]r+s[/tex] is a rational number. Then [tex]r[/tex] and [tex]r+s[/tex] can be written as follows

[tex]r=\frac{p_{1}}{q_{1}}, \,p_{1}\in \mathbb{Z}, q_{1}\in \mathbb{Z}, q_{1}\neq 0[/tex]

[tex]r+s=\frac{p_{2}}{q_{2}}, \,p_{2}\in \mathbb{Z}, q_{2}\in \mathbb{Z}, q_{2}\neq 0[/tex]

Hence we have that

[tex]r+s=\frac{p_{1}}{q_{1}}+s=\frac{p_{2}}{q_{2}}[/tex]

Then

[tex]s=\frac{p_{2}}{q_{2}}-\frac{p_{1}}{q_{1}}=\frac{p_{2}q_{1}-p_{1}q_{2}}{q_{1}q_{2}}\in \mathbb{Q}[/tex]

This is a contradiction because we assumed that [tex]s[/tex] is an irrational number.

Then [tex]r+s[/tex] must be an irrational number.

Answer:

[tex]m\angle B=m\angle C=120^{\circ}[/tex]

[tex]m\angle A=m\angle D=60^{\circ}[/tex]

Step-by-step explanation:

Trapezoid ABCD is isosceles trapezoid, because AB = CD (given). In isosceles trapezoid, angles adjacent to the bases are congruent, then

Since BK ⊥ AD, the triangle ABK is right triangle. In this triangle,  AB = 8, AK = 4. Note that the hypotenuse AB is twice the leg AK:

[tex]AB=2AK.[/tex]

If in the right triangle the hypotenuse is twice the leg, then the angle opposite to this leg is 30°, so,

[tex]m\angle ABK=30^{\circ}[/tex]

Since BK ⊥ AD, then BK ⊥ BC and

[tex]m\angle KBC=90^{\circ}[/tex]

Thus,

[tex]m\angle B=30^{\circ}+90^{\circ}=120^{\circ}\\ \\m\angle B=m\angle C=120^{\circ}[/tex]

Now,

[tex]m\angle A=m\angle D=180^{\circ}-120^{\circ}=60^{\circ}[/tex]

Answer:

A. A straight line intersecting two parallel lines produces corresponding equal angles.

Step-by-step explanation:

If two straight lines are parallel, and a straight line intersects them, then we get alternate angles equal, exterior angle equal to the opposite interior angle on the same side and the interior angles on the same sides equal to two right angles.

Hence, here the rule that is not true is : A. A straight line intersecting two parallel lines produces corresponding equal angles.

Final answer:

The number 4320 is an abundant number because the sum of its proper divisors exceeds 4320. All positive multiples of 6 greater than 6 are abundant because their smallest divisor, 6, is the sum of its proper divisors, and they have additional divisors that increase the sum.

Explanation:

To determine if the number 4320 is perfect, abundant, or deficient, we must compare the sum of its proper divisors (excluding the number itself) with the number. A perfect number is equal to the sum of its divisors. A number is abundant if the sum of its divisors is greater than the number, and it is deficient if the sum is less.

For 4320, the divisors are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 27, 30, 32, 36, 40, 45, 48, 54, 60, 64, 72, 80, 90, 96, 108, 120, 135, 144, 160, 180, 192, 216, 240, 270, 288, 320, 360, 432, 480, 540, 576, 720, 864, 1080, and 1440. Their sum is greater than 4320, so it is an abundant number.

Regarding positive multiples of 6 greater than 6 being abundant, the smallest divisor of such a multiple is always 6, which is already the sum of its divisors (1, 2, and 3). Since there are additional divisors beyond 1, 2, and 3, the sum of divisors must exceed the number, making it abundant.

Answer:

The solution is: [tex](x_{1}, x_{2}, x_{3}) = (1,0,-4)[/tex]

Step-by-step explanation:

The Gauss-Jordan elimination method is done by transforming the system's augmented matrix into reduced row-echelon form by means of row operations.

We have the following system:

[tex]2x_{1} - x_{2} + 3x_{3} = -10[/tex]

[tex]x_{1} - 2x_{2} + x_{3} = -3[/tex]

[tex]x_{1} - 5x_{2} + 2x_{3} = -7[/tex]

This system has the following augmented matrix:

[tex]\left[\begin{array}{ccc}2&-1&3|-10\\1&-2&1|-3\\1&-5&2| -7\end{array}\right][/tex]

To make the reductions easier, i am going to swap the first two lines. So

[tex]L1 <-> L2[/tex]

Now the matrix is:

[tex]\left[\begin{array}{ccc}1&-2&1|-3\\2&-1&3|-10\\1&-5&2| -7\end{array}\right][/tex]

Now we reduce the first row, doing the following operations

[tex]L2 = L2 - 2L1[/tex]

[tex]L3 = L3 - L1[/tex]

So, the matrix is:

[tex]\left[\begin{array}{ccc}1&-2&1|-3\\0&3&1|-4\\0&-3&1| -4\end{array}\right][/tex]

Now we divide L2 by 3

[tex]L2 = \frac{L2}{3}[/tex]

So we have

[tex]\left[\begin{array}{ccc}1&-2&1|-3\\0&1&\frac{1}{3}|\frac{-4}{3}\\0&-3&1| -4\end{array}\right][/tex]

Now we have:

[tex]L3 = 3L2 + L3[/tex]

So, now we have our row reduced matrix:

[tex]\left[\begin{array}{ccc}1&-2&1|-3\\0&1&\frac{1}{3}|\frac{-4}{3}\\0&0&2| -8\end{array}\right][/tex]

We start from the bottom line, where we have:

[tex]2x_{3} = -8[/tex]

[tex]x_{3} = \frac{-8}{2}[/tex]

[tex]x_{3} = -4[/tex]

At second line:

[tex]x_{2} + \frac{x_{3}}{3} = \frac{-4}{3}[/tex]

[tex]x_{2} - \frac{4}{3} = -\frac{4}{3}[/tex]

[tex]x_{2} = 0[/tex]

At the first line

[tex]x_{1} -2x_{2} + x_{3} = -3[/tex]

[tex]x_{1} - 4 = -3[/tex]

[tex]x_{1} = 1[/tex]

The solution is: [tex](x_{1}, x_{2}, x_{3}) = (1,0,-4)[/tex]

Answer:

Proved

Step-by-step explanation:

To prove that if n is a perfect square, then n+1 can never be a perfect square

Let n be a perfect square

[tex]n=x^2[/tex]

Let [tex]n+1 = y^2[/tex]

Subtract to get

[tex]1 = y^2-x^2 =(y+x)(y-x)[/tex]

Solution is y+x=y-x=1

This gives x=0

So only 0 and 1 are consecutive integers which are perfect squares

No other integer satisfies y+x=y-x=1

Answer:

[tex]y(x) = e^{3x} [Acos x+Bsin x][/tex]

Step-by-step explanation:

Given is a differential equation

[tex]y" - 6y' +10y=0.[/tex]

We have characteristic equation as

[tex]m^2-6m+10 =0[/tex]

The above quadratic cannot be factorised hence use formula

[tex]m=\frac{6+/-\sqrt{36-40} }{2} \\=3+i, 3-i[/tex]

Hence general solution would be

[tex]y(x) = e^{3x} [Acos x+Bsin x][/tex]

Answer:

The concentration of the drug is 50mg/mL.

Step-by-step explanation:

This problem can be solved by a rule of three.

In a rule of three problem, the first step is identifying the measures and how they are related, if their relationship is direct of inverse.

When the relationship between the measures is direct, as the value of one measure increases, the value of the other measure is going to increase too.

When the relationship between the measures is inverse, as the value of one measure increases, the value of the other measure will decrease.

In this problem, as the number of mg increases, so does the number of mL. It means the we have a direct rule of three.

Tha problem states that the antibody is available in a prefilled syringe containing 40 mg/0.8 mL. Calculate the concentration of drug on a mg/mL basis.

The problem wants to know how many mg are there in a mL of the drug.

40mg - 0.8mL

x mg - 1 mL

0.8x = 40

[tex]x = \frac{40}{0.8}[/tex]

x = 50mg.

The concentration of the drug is 50mg/mL.

Answer:   100

Step-by-step explanation:

Given : The lodhl diner offers a meal combination consisting of an appetizer, a soup, a main course, and a dessert.

There are 5 appetizers, 5 soups, 4 main courses, and 5 desserts.

Also, a dessert and a appetizer are not allowed to take together.

By Fundamental counting principal ,

Number of three-course meals with dessert and without appetizer :

[tex]5\times4\times5=100[/tex]      (1)

Number of three-course meals with appetizer and without dessert :

[tex]5\times5\times4=100[/tex]      (2)

Now, the number of meals with either dessert or appetizer :-

[tex]100+100=200[/tex]           [Add (1) and (2)]

Answer:

500

Step-by-step explanation:

5x5x4x5

we do

5x5=25

25x4=100

100x5= 500

Answer:

y = -0.9x+10.1

Step-by-step explanation:

The equation of the line is:

[tex]y=mx+b[/tex]

You have been asked to stimate m and b. To do so, first find the product between each pair of x and y and the value of x squared:

[tex]\left[\begin{array}{cccc}x&y&x*y&x^2\\1&9&9&1\\3&7&21&9\\5&7&35&25\\7&3&21&49\\9&2&18&81\end{array}\right][/tex]

Then calculate the total sum of all columns:

[tex]\left[\begin{array}{cccc}x&y&x*y&x^2\\1&9&9&1\\3&7&21&9\\5&7&35&25\\7&3&21&49\\9&2&18&81\\\bold{25}&\bold{28}&\bold{104}&\bold{165}\end{array}\right][/tex]

m can be calculated following the next equation:

[tex]m=\frac{\frac{\sum{xy}-\sum{y}}{n}}{\sum{x^2}-\frac{(\sum{x})^2}{n}}[/tex]

where n is the number of (x, y) couples (5 in our case).

Replacing the values calculated previously:

[tex]m=\frac{104-\frac{25*28}{5} }{165-\frac{25^2}{5} }=\frac{104-\frac{700}{5} }{165-\frac{625}{5} } = \frac{104-140}{165-125 } = \frac{-36}{40} = -0.9[/tex]

For b:

[tex]b=\bar{y}- m\bar{x}=\frac{\sum{y}}{n}-m\frac{\sum{x}}{n}=\frac{28}{5}-(-0.9)\frac{25}{5}= \frac{28}{5}+\frac{22.5}{5}=\frac{50.5}{5}=10.1[/tex]

In the figure attached you can see the points given and the stimated line.

Answer:

[tex] (x-4)^2+(y+5)^2=90[/tex]

Step-by-step explanation:

The equation of a circle of radius r, centered at the point (a,b) is

[tex](x-a)^2+(y-b)^2=r^2[/tex]

We already know the center is at [tex](4,-5)[/tex], we are just missing the radius. To find the radius, we can use the fact that the circle passes through the point (7,4), and so the radius is just the distance from the center to this point (see attached image). So we find the distance by using distance formula between the points (7,4) and (4,-5):

radius[tex]=\sqrt{(7-4)^2+(4-(-5))^2}=\sqrt{3^2+9^2}=\sqrt{90}[/tex]

And now that we know the radius, we can write the equation of the circle:

[tex] (x-4)^2+(y-(-5))^2=\sqrt{90}^2[/tex]

[tex] (x-4)^2+(y+5)^2=90[/tex]

Answer:

0001 0110 1110 1011 0010

Step-by-step explanation:

First decimal number = 11540

second decimal number = 5972

BCD of 0 to 9 decimal number

Decimal number                               BCD

       0                                                  0000

       1                                                    0001

       2                                                   0010

       3                                                   0011

       4                                                   0100

       5                                                   0101

       6                                                   0110

       7                                                    0111

       8                                                    1000

       9                                                    1001

Hence, we can write above two numbers in BCD form as

11540 => 0001 0001 0101 0100 0000

5972 => 0101 1001 0111 0010

So, sum of 11540 and 5972 in BCD form can be given by

       0001 0001 0101 0100 0000

        +       01 01 1001 01 11  001 0

      0001  01 10 11 10 10 11  001 0

So, the sum of 11540 and 5972 in BCD form is

                  0001 0110 1110 1011 0010

Answer:

Yes, rational numbers are closed under addition.

Step-by-step explanation:

Rational numbers are number that can be expressed in the form of fraction [tex]\frac{x}{y}[/tex], where x and y are integers and y ≠ 0.

Now, the closure property of addition of rational number states that if we add two rational number, then the sum of these two rational number will also be a rational number.

Let a and b be two rational number, then,

a+b = c, where c is the sum of and b

c is also a rational number.

Thus, rational numbers are closed under addition.

This can be explained with the help of a example.

[tex]\frac{1}{7} + \frac{2}{7}[/tex] = [tex]\frac{2}{7}[/tex]

It is clear that [tex]\frac{1}{7}, \frac{2}{7}, \frac{3}{7}[/tex] are rational numers.

Thus, rational numbers are closed under addition.

Answer:

OBSERVATIONAL

Step-by-step explanation:

Let`s see the two kind of experiments mentioned:

In this case, the patient's parent only answer a questionnairewith no intervention from the doctors into the pacients at all, so it's an observational  experiment

Answer:

Step-by-step explanation:

First you would find the area of the lot:

50 x 36 = 1800 m^2

Then you would find the area of the house:

31 x 9 = 279 m^2

Then you subtract the are of the lot minus the area of the house:

1800 - 279 = 1521 m^2

And the final answer is:

1521 square meters is left

The area of space which is left over is [tex]1521m^{2} [/tex]

Required steps shown below,

Area of lot [tex]=length*width[/tex]

                  [tex]=50*36=1800m^{2} [/tex]

Area of house[tex]=length*width[/tex]

                       [tex]=31*9=279m^{2} [/tex]

To calculate area of space is left over, subtract area of house from area of lot.

The space left over is,

                                [tex]1800-279=1521m^{2} [/tex]

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Answer: There are 90 snack boxes.

Step-by-step explanation:

Given : The number of kinds of fruits = 5

The number of kinds of herbal teas = 3

The number of kinds of flavors of wrap sandwich = 6

Then by using the fundamental principal of counting, the number of possible snack are there will be :_

[tex]5\times3\times6= 90[/tex]

Therefore, the number of possible snacks = 90

Answer:

The posterior probability that the lathe tool is properly adjusted is 94.7%

Step-by-step explanation:

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

[tex]P = \frac{P(B).P(A/B)}{P(A)}[/tex]

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In your problem we have that:

-A is the probability that the part chosen is found to be acceptable.

The problem states that the probability its lathe tool is properly adjusted is 0.8. When it happens, there is a 0.9 probability that the parts produced pass inspection. There is also a 0.2 probability of  the lathe is out of adjustment, when it happens  the probability of a good part being produced is only 0.2.

So, P(A) = P1 + P2 = 0.8*0.9 + 0.2*0.2 = 0.72 + 0.04 = 0.76

Where P1 is the probability of a good part being produced when lathe tool is properly adjusted and P2 is the probability of a good part being produced when lathe tool is not properly adjusted.

- P(B) is the the probability its lathe tool is properly adjusted. The problem states that P(B) = 0.8

P(A/B) is the probability of A happening given that B has happened. We have that A is the probability that the part chosen is found to be acceptable and B is the probability its lathe tool is properly adjusted. The problem states that when the lathe is properly adjusted, there is a 0.9 probability that the parts produced pass inspection. So P(A/B) = 0.9

So, probability of B happening, knowing that A has happened, where B is the lathe tool is properly adjusted and A is that the part randomly chosen is inspected and found to be acceptable is:

[tex]P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.8*0.9}{0.76} = \frac{0.72}{0.76} = 0.947 = 94.7%[/tex]

The posterior probability that the lathe tool is properly adjusted is 94.7%

Using Bayes' Theorem, the posterior probability that the lathe tool is properly adjusted given a part passes inspection is approximately 0.947 (94.7%). We determined this by combining the prior probabilities and the conditional probabilities to find the total probability of a part passing inspection.

To find the posterior probability, we can use Bayes' Theorem. Let's define the events:

Prior Probabilities

The probability that the lathe tool is properly adjusted is 0.8. Therefore, P(A) = 0.8 and P(A') = 0.2.

Conditional Probabilities

Applying Bayes' Theorem:

Bayes' Theorem states:

First, we need to find P(B), the total probability that a part passes inspection:

Substituting values:

Now, using Bayes' Theorem, we can calculate P(A|B):

Thus, the posterior probability that the lathe tool is properly adjusted given that a part is found to be acceptable is approximately 0.947 (94.7%).

Answer:

Y = [tex]e^{t}[/tex] +  [tex]te^{t}[/tex] + [tex]t^{2} e^{t}[/tex] + t - 18

Step-by-step explanation:

y''' − 3y'' + 3y' − y = ex − x + 21

Homogeneous solution:

First  we propose a solution:

Yh = [tex]e^{r*t}[/tex]

Y'h = [tex]r*e^{r*t}[/tex]

Y''h = [tex]r^{2}*e^{r*t}[/tex]

Y'''h = [tex]r^{3}*e^{r*t}[/tex]

Now we solve the following equation:

Y'''h - 3*Y''h + 3*Y'h - Yh = 0

[tex]r^{3}*e^{r*t}[/tex] - 3*[tex]r^{2}*e^{r*t}[/tex] + 3*[tex]r*e^{r*t}[/tex] - [tex]e^{r*t}[/tex] = 0

[tex]r^{3} - 3r^{2} + 3r - 1 = 0[/tex]

To solve the equation we must propose a solution to the  polynomial :

r = 1

To find the other r we divide the polynomial by (r-1) as you can see  

attached:

solving the equation:

(r-1)([tex]r^{2} - 2r + 1[/tex]) = 0

[tex]r^{2} - 2r + 1[/tex] = 0

r = 1

So we have 3 solution [tex]r_{1} = r_{2} =r_{3}[/tex] = 1

replacing in the main solution

Yh =  [tex]e^{t}[/tex] +  [tex]te^{t}[/tex] + [tex]t^{2} e^{t}[/tex]

The t and [tex]t^{2}[/tex] is used because we must have 3 solution  linearly independent

Particular solution:

We must propose a Yp solution:

Yp = [tex]c_{1} (t^{3} + t^{2} + t + c_{4} )e^{t} + c_{2} t + c_{3}[/tex]

Y'p = [tex]c_{1}(t^{3} + t^{2} + t + c_{4} )e^{t} + c_{1}( 3t^{2} + 2t + 1 )e^{t} + c_{2}[/tex]

Y''p = [tex]c_{1}(t^{3} + t^{2} + t + c_{4} )e^{t} + c_{1}(6t + 2)e^{t}[/tex]

Y'''p = [tex]c_{1}(t^{3} + t^{2} + t + c_{4} )e^{t} + 6c_{1}e^{t}[/tex]

Y'''p - 3*Y''p + 3*Y'p - Yp = [tex]e^{t} - t + 21[/tex]

[tex]6c_{1}e^{t} - 18c_{1} te^{t} - 6c_{1} e^{t} + 6c_{1} te^{t} + 9c_{1} t^{2} e^{t} + 3c_{1}e^{t} + 3c_{2} - c_{2} t -  c_{3}[/tex] = [tex]e^{t} - t + 21[/tex]

equalizing coefficients of the same function:

- 12c_{1} = 0

9c_{1} = 0

3c_{1} = 0

c_{1} = 0

3c_{2} - c_{3} = 21 => c_{5} = [tex]\frac{1}{3}[/tex]

-c_{2} = -1

c_{2} = 1

c_{3} = -18

Then we have:

Y = [tex]e^{t}[/tex] +  [tex]te^{t}[/tex] + [tex]t^{2} e^{t}[/tex] + t - 18

Answer:

K=6

Step-by-step explanation:

we know that

A function is: a relation from a set of inputs to a set of possible outputs where each input is related to exactly one output

so

In this problem

If the x-coordinate of the first point is equal to the x-coordinate of the second point, then the relation will not be a function

Remember that the x-coordinate is the input value and the y-coordinate is the output value

Equate the x-coordinates and solve for k

[tex]1.5k-4=-0.5k+8[/tex]

[tex]1.5k+0.5k=8+4[/tex]

[tex]2k=12[/tex]

[tex]k=6[/tex]

therefore

If the value of k is 6 the, the relation will not be a function

Final answer:

To determine when the relation A is not a function, set the first elements of the ordered pairs equal to each other. Solving for k, we find that k must equal 6 to make the relation not a function, as this value of k would result in the same input for both pairs, violating the definition of a function.

Explanation:

For a relation to be considered a function, each input value must be associated with exactly one output value. Looking at the provided relation A = {(1.5k-4, 7), (-0.5k+8, 15)}, we can see that there are two ordered pairs. For A to not be a function, the first element of both ordered pairs must be the same because this would mean that a single input is associated with two different outputs, which violates the definition of a function.

We set the first elements equal to each other to find the value of k that would make the relation not a function: 1.5k - 4 = -0.5k + 8. Solving this equation, we add 0.5k to both sides and add 4 to both sides to obtain 2k = 12, and then divide both sides by 2 to find that k = 6.

Therefore, when k equals 6, the relation A will not be a function because both ordered pairs will have the first element equal to 5, which corresponds to two different output values (7 and 15).

In summary, the raw data can only be retrieved from a dot plot, not a histogram, as the dot plot represents individual data points, whereas a histogram represents data in grouped intervals.

When comparing a dot plot to a histogram, there are several key points to consider:

A frequency table or a frequency polygon can also be useful tools for data representation in complement to histograms and dot plots, especially when comparing distributions.