A survey was conducted among 78 patients admitted to a hospital cardiac unit during a​ two-week period. The data of the survey are shown below. Let B equals the set of patients with high blood pressure. Let C equals the set of patients with high cholesterol levels. Let S equals the set of patients who smoke cigarettes.​n(B) equals 36 ​n(B intersect​ S) equals 10 ​n(C) equals 34 ​n(B intersect​ C) equals 12 ​n(S) equals 30 ​n(B intersect C intersect​S) equals 5 ​n[(B intersect​ C) union​ (B intersect​ S) union​ (C intersect​ S)] equals 21

Answer:

Option A. y= -1/5x + 5

Step-by-step explanation:

step 1

Find the slope of the given line

we have

y=5x-2

The slope m is

m=5

step 2

Find the slope of a line perpendicular to the given line

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is -1)

m1*m2=-1

we have

m1=5

(5)*m2=-1

m2=-1/5

therefore

The equation of a line perpendicular to the given line could be

y= -1/5x + 5

Answer:

Inverse Laplace of [tex]\frac{1}{(S+4)^2}[/tex] will be [tex]te^{-4t}[/tex]

Step-by-step explanation:

We have to find the inverse Laplace transform of [tex]H(S)=\frac{1}{(S+4)^2}[/tex]

We know that of [tex]\frac{1}{s+4}[/tex] is [tex]e^{-4t}[/tex]

As in H(s) there is square of [tex]s+4[/tex]

So i inverse Laplace there will be multiplication of t

So the inverse Laplace of [tex]\frac{1}{(s+4)^2}[/tex]  will be [tex]te^{-4t}[/tex]

[tex]L^{-1}\frac{1}{(S+4)^2}=te^{-4t}[/tex]

Assume a standard deck of 52 cards with 4 suits of 13 cards each.

a. There are [tex]\dbinom{13}8\dbinom{39}2[/tex] ways of being dealt a hand consisting of 8 hearts and 2 non-hearts, so the probability of being dealt such a hand is

[tex]\dfrac{\dbinom{13}8\dbinom{39}2}{\dbinom{52}{10}}\approx0.0000602823[/tex]

b. This time, the non-hearts specifically belong to the suit of diamonds, for which there are [tex]\dbinom{13}2[/tex] ways of getting drawn, so the probability is

[tex]\dfrac{\dbinom{13}8\dbinom{13}2}{\dbinom{52}{10}}\approx0.0000063455[/tex]

a. The probability P of being dealt exactly 8 hearts is:

[tex]\[ P = \frac{k}{n} = \frac{\binom{13}{8} \times \binom{39}{2}}{\binom{52}{10}} \][/tex]

b. The probability  P'  of being dealt exactly 8 hearts given the first two cards were diamonds is:

[tex]\[ P' = \frac{k'}{n'} = \frac{\binom{8}{6} \times \binom{42}{2}}{\binom{50}{8}} \][/tex]

To solve this problem using combinatorics, we can calculate the probability by considering the total number of possible outcomes and the number of favorable outcomes.

Let's denote:

-  n  as the total number of ways to deal 10 cards from a standard deck (52 cards).

-  k  as the number of ways to deal exactly 8 hearts and 2 non-hearts from the remaining 44 cards in the deck.

a. To calculate the probability a player is dealt exactly 8 hearts:

Total number of ways to choose 8 hearts out of 13 hearts:

[tex]\[ \binom{13}{8} \][/tex]

Total number of ways to choose 2 non-hearts out of 39 non-hearts:

[tex]\[ \binom{39}{2} \][/tex]

Therefore, the number of favorable outcomes is:

[tex]\[ k = \binom{13}{8} \times \binom{39}{2} \][/tex]

The total number of ways to deal 10 cards from a deck of 52 cards is:

[tex]\[ n = \binom{52}{10} \][/tex]

So, the probability P of being dealt exactly 8 hearts is:

[tex]\[ P = \frac{k}{n} = \frac{\binom{13}{8} \times \binom{39}{2}}{\binom{52}{10}} \][/tex]

b. To calculate the probability a player is dealt exactly 8 hearts if the first two cards dealt were diamonds:

If the first two cards are diamonds, then there are 50 cards remaining, out of which 8 are hearts and 42 are non-hearts.

Total number of ways to choose 6 more hearts out of the remaining 8 hearts:

[tex]\[ \binom{8}{6} \][/tex]

Total number of ways to choose 2 non-hearts out of the remaining 42 non-hearts:

[tex]\[ \binom{42}{2} \][/tex]

Therefore, the number of favorable outcomes is:

[tex]\[ k' = \binom{8}{6} \times \binom{42}{2} \][/tex]

The total number of ways to deal 8 cards from the remaining 50 cards is:

[tex]\[ n' = \binom{50}{8} \][/tex]

So, the probability  P'  of being dealt exactly 8 hearts given the first two cards were diamonds is:

[tex]\[ P' = \frac{k'}{n'} = \frac{\binom{8}{6} \times \binom{42}{2}}{\binom{50}{8}} \][/tex]

Answer:

Step-by-step explanation:

The question is:

The amount of red blood cells in a blood sample is equal to the total amount in the sample minus the amount of plasma. What is the total amount of blood drawn? Red blood cells = 45% Plasma = 5.5 ml

Solution:

We have given:

The amount of red blood cells in a blood sample is equal to the total amount in the sample minus the amount of plasma. We need to find how many ml is 1% of blood.

The equation we get is:

Red blood cells = total sample - amount of plasma

45% = 100% - 5.5 ml

Combine the percentages:

5.5 ml = 100%-45%

5.5 ml = 55%

0.1 ml = 1%

1% = 0.1 ml

Now you can find out red blood cells volume or total sample volume.

Red blood cells volume = 45% x 0.1ml/%

= 4.5ml

Total sample volume =100% x 0.1ml/%

= 10ml .

Answer:

Thousandths: 7

Ten-thousandths: 4

Tenths: 3

Step-by-step explanation:

The given number is 59.3274659

The place values of a numner is given in the followwing format:

Tens  Ones.Tenths  Hundredths  Thousandths Tenthousandths  and so on

We can fit our number

Tens  Ones.Tenths  Hundredths  Thousandths Tenthousandths HT M

  5          9 .     3                 2                     7                       4        

Thousandths: 7

Ten-thousandths: 4

Tenths: 3

Answer:

[tex]x=\frac{-31\pm \sqrt{1000.2}}{4}[/tex]

Step-by-step explanation:

Given quadratic equation,

[tex]2x^2+31x-4.9=0[/tex]

Since, by the quadratic formula,

The solution of a quadratic equation [tex]ax^2+bx+c=0[/tex] is,

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

Here, a = 2, b = 31, c = -4.9,

Thus, by the quadratic formula,

[tex]x=\frac{-31\pm \sqrt{31^2-4\times 2\times -4.9}}{2\times 2}[/tex]

[tex]=\frac{-31\pm \sqrt{961+39.2}}{4}[/tex]

[tex]=\frac{-31\pm \sqrt{1000.2}}{4}[/tex]

[tex]\implies x = \frac{-31+\sqrt{1000.2}}{4}\text{ or }x=\frac{-31- \sqrt{1000.2}}{4}[/tex]

Answer: The negations are:

1. The train is not late and my watch is not fast.

2. Dogs don't bark or cats don't meow.

Step-by-step explanation:

Hi!

De Morgan's laws for two propositions P and Q are:

1. ¬(P ∨ Q) = (¬P) ∧ (¬Q)

2. ¬(P ∧ Q) = (¬P) ∨ (¬Q)

where the symbols are,

¬ = not

∨ = or

∧ = and

1. In this case the proposition is P ∨  Q, with

P = "the train is late"

Q = "my watch is fast"

Then by law 1:  ¬(P ∨ Q) =  "The train is not late and my watch is not fast"

2. In this case the proposition is P  ∧ Q, with

P = "dogs bark"

Q = "cats meow"

Then by law 2: ¬(P ∧ Q) = "Dogs don't bark or cats don't meow"

Answer:

2

Step-by-step explanation:

The  difference in the cost for a one-night stay with and without breakfast is:

                             $ 23

Total cost to stay one night at sleepyhead Motel without breakfast is: $119.

and cost to stay one night at sleepyhead Motel with breakfast is: $142.

Hence, the difference in the cost for a one-night stay with and without breakfast is calculated by:

               $ (142-119)

                 =   $ 23

Hence, the answer is: $ 23

Let [tex]q[/tex] be the unknown amount of the chemical originally in the pond, so [tex]Q(0)=q[/tex].

a. The incoming water introduces the chemical at a rate of

[tex]Q'_{\rm in}=\left(0.1\dfrac{\rm kg}{\rm L}\right)\left(4\dfrac{\rm L}{\rm hr}\right)=\dfrac25\dfrac{\rm kg}{\rm hr}[/tex]

and the mixture flows out at a rate of

[tex]Q'_{\rm out}=\left(\dfrac Q{2760}\dfrac{\rm kg}{\rm L}\right)\left(4\dfrac{\rm L}{\rm hr}\right)=\dfrac Q{690}\dfrac{\rm kg}{\rm hr}[/tex]

so that the net rate of change (in kg/hr) of the chemical in the pond is given by the differential equation,

[tex]\boxed{Q'=\dfrac25-\dfrac Q{690}}[/tex]

b. The ODE is linear; multiplying both sides by [tex]e^{t/690}[/tex] gives

[tex]e^{t/690}Q'+\dfrac{e^{t/690}}{690}Q=\dfrac{2e^{t/690}}5[/tex]

Condense the left side into the derivative of a product:

[tex]\left(e^{t/690}Q\right)'=\dfrac{2e^{t/690}}5[/tex]

Integrate both sides to get

[tex]e^{t/690}Q=276e^{t/690}+C[/tex]

and solve for [tex]Q[/tex] to get

[tex]Q=276+Ce^{-t/690}[/tex]

The pond starts with [tex]q[/tex] kg of the chemical, so when [tex]t=0[/tex] we have

[tex]q=276+C\implies C=q-276[/tex]

so that the amount of chemical in the water at time [tex]t[/tex] is

[tex]Q(t)=276+(q-276)e^{-t/690}[/tex]

As [tex]t\to\infty[/tex], the exponential term will converge to 0, leaving a fixed amount of 276 kg of the chemical in the pond.

The differential equation for the amount of chemical in the pond is dQ/dt = 0.01 kg/L * 4 L/h. After a long time, the amount of chemical in the pond is Q = 0.04 kg/h * t.

(a)

To write a differential equation for the amount of chemical in the pond at any time, we need to consider the rate of change of the chemical in the pond. The chemical flows into the pond at a rate of 0.01 kg/L and flows out at the same rate, so the rate of change of the chemical Q(t) in the pond is 0.01 kg/L multiplied by the rate of change of the volume of water in the pond, which is 4 L/h. Therefore, the differential equation for the amount of chemical in the pond is:

dQ/dt = 0.01 kg/L * 4 L/h

(b)

To determine how much chemical will be in the pond after a long time, we can solve the differential equation.

We can rewrite the differential equation as:

dQ = 0.01 kg/L * 4 L/h * dt

Integrating both sides:

∫dQ = ∫0.01 kg/L * 4 L/h dt

Q = 0.04 kg/h * t + C

Where C is a constant of integration. Given that the amount of chemical in the pond is initially 0 (since the pond starts with only pure water), we can substitute Q = 0 and solve for C:

0 = 0.04 kg/h * 0 + C

C = 0

Therefore, the amount of chemical in the pond after a long time is given by:

Q = 0.04 kg/h * t

https://brainly.com/question/33433874

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Answer:

No

.3 = .30

.32 > .30

Step-by-step explanation:

Answer:

There are 52 dimes and 18 nickles

Step-by-step explanation:

Lets call x= number of dimes and y= number of nickles

then we have the first equation  

(1)   x + y = 70

As a nickel is worth 0.05 US$ and a dime is worth 0.10 US$, we have the second equation

(2)   0.10x + 0.05y = 6.10

We then have a linear system of 2 equations and 2 unknowns  

(1) x +y = 70

(2) 0.10x + 0.05y = 6.10

In order to solve the system by the elimination method, we have to multiply on of the equations by a suitable number to eliminate one unknown when adding the two equations.  

There are several ways of doing this. We could, for example, multiply (1) by -0.05  and then add it to (2)

(1) -0.05x – 0.05y = (-0.05)70

(2) 0.10x + 0.05y = 6.10

That is to say,

(1) -0.05x – 0.05y = -3.5

(2) 0.10x + 0.05y = 6.10

Adding (1) and (2) we get

-0.05x = -2.6 => x = (-2.6)/(-0.05) = 52 => x = 52

So we have 52 dimes.

Substituting this value in equation 1, we obtain

  y = 70 - x = 70 – 52 = 18

Then we have 18 nickels

Answer:

the equation of the least-squares line for the data is: [tex]\hat Y=9.3+1.3x[/tex]

Step-by-step explanation:

In a simple linear regression model, such as, [tex]\hat Y=b_0+b_1x[/tex], the coefficients bo and b1 are estimated through the method of least squares by the use of the equations:

[tex]b_1=\frac{S{xy}}{S_x^2}\\\\b_0=\bar{y}+b_1 \bar{x}[/tex]

For the data provided you have to:

[tex]S_{xy}=\frac{\sum {(x_i-\bar x)(y_i-\bar y)}}{n-1}=3.25\\\\S_x^2=\frac{\sum {(x_i-\bar x)^2}}{n-1}=2.5\\\\\bar y=5.4[/tex], thus:

[tex]b_1=\frac{3.25}{2,5}=1.3\\\\b_0=5.4+1.3(3.0)=9.3[/tex]

the equation of the least-squares line for the data is:

[tex]Y=9.3+1.3x[/tex]

Answer:

y = 2

Step-by-step explanation:

Answer:

y=2

Step-by-step explanation:

Step-by-step explanation:

As the hint says, for any function [tex]f:\Omega\to\mathcal{P}(\Omega)[/tex], we can think of the set [tex] X=\{ \omega\in\Omega : \omega \notin f(\omega)\}[/tex] (which is the set of all those elements of [tex]\Omega[/tex] which don't belong to their image). So [tex]X[/tex] is made of elements of [tex]\Omega[/tex], and so it belongs to [tex]\mathcal{P}(\Omega)[/tex].

Now, this set [tex]X[/tex] is NOT the image of any element in [tex] \Omega[/tex], since if there was some [tex]a\in\Omega[/tex] such that [tex]f(a)=X[/tex], then the following would happen:

If [tex]a\in X=f(a)[/tex], then by definition of the set [tex]X[/tex], [tex]a\notin f(a)[/tex], so we're getting that [tex]a\in f(a)[/tex] and also [tex] a\notin f(a)[/tex], which is a contradiction.

On the other hand, if [tex]a\notin f(a)[/tex], then by definition of the set [tex]X[/tex], we would get that [tex]a\in X=f(a)[/tex], so we're getting that [tex]a\in f(a)[/tex] and also [tex] a\notin f(a)[/tex], which is a contradiction again.

So in any case, the assumption that this set [tex]X[/tex] is the image of some element in [tex]\Omega[/tex] leads us to a contradiction, therefore this set [tex]X[/tex] is NOT the image of any element in [tex]\Omega[/tex], and so there cannot be a bijection from [tex]\Omega[/tex] to [tex]\mathcal{P}(\Omega)[/tex], and so the two sets cannot have the same cardinality.

Step-by-step explanation:

a) Give two pairs which are in the relation [tex]\equiv \mod 4[/tex] and two pairs that are not.

As stated before, a pair [tex](x,y)\in \mathbb{Z}\times\mathbb{Z}[/tex] is equal mod m (written [tex]x\equiv y\mod m[/tex]) if [tex]m\mid (x-y)[/tex]. Then:

b) Show the [tex]\equiv \mod m[/tex] is an equivalence relation.

An equivalence relation is a binary relation that is reflexive, symmetric and  transitive.

By definition [tex]\equiv \mod m[/tex] is a binary relation. Observe that:

[tex]m\mid [(y-x)+(z-y)] \implies m\mid (z-x) \implies x\equiv z \mod m[/tex].

In conclusion, [tex]\equiv \mod m[/tex] defines an equivalence relation.

Answer:

The linear equation relating price to demand is [tex]p=-0.05x+360.2[/tex]

Step-by-step explanation:

A Linear Demand Function expresses demand p (the number of items demanded) as a function of the unit price x (the price per item).

From the information given we know two facts:

Let x be the price and p the number of cards sold.

The fact 1. is the slope of the function because is the change in demand per unit change in price.

We can use fact 2. to find the value of b in the equation

[tex]p=-0.05x+b\\b=p+0.05x[/tex]

[tex]b=360+0.05\cdot 4\\b=360+0.2\\b=360.2[/tex]

The linear equation relating price to demand is [tex]p=-0.05x+360.2[/tex]

Answer:

Every line in [tex]\mathbb{R}^{3}[/tex] is a function of the form [tex]\gamma (t)={\bf p}+t {\bf v} [/tex], where [tex]{\bf p}[/tex] is point where the line passes and [tex]{\bf v}[/tex] is a nonzero vector which is called the direction vector of the line. Then, if we derive the function [tex]\gamma[/tex] we obtain [tex]\gamma'(t)={\bf v} \neq (0,0,0)[/tex], so [tex]\gamma(t)={\bf p}+t {\bf v}[/tex] is a regular curve.

Step-by-step explanation:

Every line in [tex]\mathbb{R}^{3}[/tex] can be parametrized by

[tex]\gamma (t)={\bf p}+t{\bf v}=(p_{1},p_{2},p_{3})+t(v_{1},v_{2},v_{3})=(p_{1}+tv_{1},p_{2}+tv_{2},p_{3}+tp_{3})[/tex], where [tex]t\in \mathbb{R}[/tex]. To derivate the function [tex]\gamma [/tex] we only need to derive each component. Then we have that

[tex]\gamma'(t)=(\frac{d}{dt}(p_{1}+tv_{1}),\frac{d}{dt}(p_{2}+tv_{2}),\frac{d}{dt}(p_{3}+tv_{3}))=(v_{1},v_{2},v_{3})={\bf v}\neq (0,0,0).[/tex]

Now, remember that a a parametrized curve is said to be regular if [tex]\gamma'\neq 0[/tex] for all [tex]t[/tex].

Answer:

The possible combinations are (4 & 6) ,(8 &3), (2 &12), (24 &1)

Step-by-step explanation:

The area of a flower bed is 24 square feet.

Now, the factors of 24 are 2 x 2 x 2 x 3.

Hence, if the other sides were whole number dimensions, then the possible combinations will be (4 & 6) ,(8 &3), (2 &12), (24 &1)

Answer:

5

Step-by-step explanation:

We have to see the pattern, we start with -8, the second number is -2, the distance between -8 and -2 is 6 units. Now from -2 to 2 the distance is 4 units. From 2 to 4 distance is 2 units. Then we can conclude that with each step the distance is divided by 2, then the next number is 5 because the distance between 4 and 5 is 1 unit.

Area = Length time width.

The area of the shaded area is Length x width of the entire shape, minus the length x width of the unshaded area.

Full area: 27 x 10 = 270 square m.

Unshaded area: 17 x 5 = 85 square m.

Area of shaded region: 270 - 85 = 185 square m.

Answer:

Distributive property

Subtraction property of equality

Division property of equality

Step-by-step explanation:

Given

-6(x-4)=42

-6*x+-4*-6=42-------------------distributive property

-6x+24-24=42-24-----------------subtraction property of equality

-6x=18

-6x/-6=18/-6--------------------------division property of equality

x= -3

Answer: Fraction of all eligible voters voted for Shelly is [tex]\dfrac{3}{20}[/tex] .

Morgan received the most votes .

Step-by-step explanation:

Given : The fraction of all eligible voters cast votes =[tex]\dfrac{1}{2}[/tex]

The fraction of votes received by Shelly = [tex]\dfrac{3}{10}[/tex]        (1)

Now, the fraction of all eligible voters voted for Shelly is given by :_

[tex]\dfrac{3}{10}\times\dfrac{1}{2}=\dfrac{3}{20}[/tex]

Thus, the fraction of all eligible voters voted for Shelly is [tex]\dfrac{3}{20}[/tex] .

The fraction of vote received by Morgan= [tex]\dfrac{5}{8}[/tex]          (2)

To compare the fractions given in (1) and (2), we need to find least common multiple of 10 and 8 .

LCM (10, 8)=40

Now, make denominator 40 (to make equivalent fraction)  in (1), (2) we get

Fraction of all eligible voters voted for Shelly = [tex]\dfrac{3\times4}{10\times4}=\dfrac{12}{40}[/tex]

Fraction of all eligible voters voted for Morgan =[tex]\dfrac{5\times5}{8\times5}=\dfrac{25}{40}[/tex]

Since, [tex]\dfrac{25}{40}>\dfrac{12}{40}[/tex]  [By comparing numerators]

Therefore, Morgan received the most votes .

Answer:

The solutions are: [tex]x=4,\:x=-4,\:x=3i,\:x=-3i[/tex]

Step-by-step explanation:

Consider the provided equation.

[tex]x^4-7x^2-144=0[/tex]

Substitute [tex]u=x^2\mathrm{\:and\:}u^2=x^4[/tex]

[tex]u^2-7u-144=0[/tex]

[tex]u^2-16u+9u-144=0[/tex]

[tex](u-16)(u+9)=0[/tex]

[tex]u=16,\:u=-9[/tex]

Substitute back [tex]\:u=x^2[/tex] and solve for x.

[tex]x^2=16\\x=\sqrt{16}\\ \quad x=4,\:x=-4[/tex]

Or

[tex]x^2=-9\\x=\sqrt{-9}\\ \quad x=3i,\:x=-3i[/tex]

Hence, the solutions are: [tex]x=4,\:x=-4,\:x=3i,\:x=-3i[/tex]

Check:

Substitute x=4 in provided equation.

[tex]4^4-7(4)^2-144=0[/tex]

[tex]256-112-144=0[/tex]

[tex]0=0[/tex]

Which is true.

Substitute x=-4 in provided equation.

[tex](-4)^4-7(-4)^2-144=0[/tex]

[tex]256-112-144=0[/tex]

[tex]0=0[/tex]

Which is true.

Substitute x=3i in provided equation.

[tex](3i)^4-7(3i)^2-144=0[/tex]

[tex]81+63-144=0[/tex]

[tex]0=0[/tex]

Which is true.

Substitute x=-3i in provided equation.

[tex](-3i)^4-7(-3i)^2-144=0[/tex]

[tex]81+63-144=0[/tex]

[tex]0=0[/tex]

Which is true.

Hi here´s a way to solve it

Let m and n be odd integers. Then,  we can express m as 2r + 1 and n as 2s + 1, where r and s are integers.

This means that any odd number can be written as the sum of some even integer and one.

Substituting, we have that m + n = (2r + 1) + 2s + 1 = 2r + 2s + 2.

As we defined r and s as integers, 2r + 2s + 2 is also an integer.

Now It is clear that 2r + 2s + 2 is an integer divisible by 2 becasue we have 2 in each of the integers.

Therefore,  2r + 2s + 2 = m + n is even.

So, the sum of two odd integers is even.

Answer:  Option 'c' is correct.

Step-by-step explanation:

Since we have given that

Number of ounces steaks = 4.5 oz

Number of guests = 180

Percentage of waste = 25%

Number of ounces of fresh steak is given by

[tex]4.5\times 180\\\\=810\ oz[/tex]

Let the number of ounces of raw steak be x.

According to question,

[tex]\dfrac{100-25}{100}\times x=810\\\\\dfrac{75}{100}\times x=810\\\\0.75\times x=810\\\\x=\dfrac{810}{0.75}\\\\x=1080\ oz[/tex]

As we need in pounds, and we know that

1 pound = 16 ounces

So, Number of pounds of raw steak that Chef could order is given by

[tex]\dfrac{1080}{16}=67.5\ lbs[/tex]

Hence, Option 'c' is correct.

To determine how many pounds of raw steak Chef should order for 180 guests with 25% waste, calculate the weight per guest and multiply by the number of guests. Convert the weight to pounds by dividing by 16. The answer is approximately B. 63.3 lbs.

To determine how many pounds of raw steak Chef should order, we need to calculate the total weight of steak required for 180 guests, taking into account the 25% waste.

First, we calculate the weight of steak for one guest by multiplying 4.5 oz by 1.25 (to account for the waste), which equals 5.625 oz.

Then, we multiply the weight per guest by the number of guests to get the total weight of steak required: 5.625 oz x 180 = 1012.5 oz.

Finally, we convert the ounces to pounds by dividing by 16, since there are 16 ounces in a pound.

Thus, Chef should order 1012.5 oz ÷ 16 = 63.28125 lbs, which rounds to approximately 63.3 lbs (option B).

Answer:

88

Step-by-step explanation:

We are given that in arithmetic sequence , the nth term [tex]a_n[/tex] is given by the formula

[tex]A_n=a_1+(n-1)d[/tex]

Where [tex]a_1=first term[/tex]

d=Common difference

In an geometric sequence, the nth term is given by

[tex]a_n=a_1r^{n-1}[/tex]

Where r= Common ratio

1,4,7,10,..

We have to find 30th term.

[tex]a_1=1,a_2=4,a_3=7,a_4=10[/tex]

[tex]d=a_2-a_1=4-1=3[/tex]

[tex]d=a_3-a_2=7-4=3[/tex]

[tex]d=a_4-a_3=10-7=3[/tex]

[tex]r_1=\frac{a_2}{a_1}=\frac{4}{1}=4[/tex]

[tex]r_2=\frac{a_3}{a_2}=\frac{7}{4}[/tex]

[tex]r_1\neq r_2[/tex]

Therefore, given sequence is an arithmetic sequence because the difference between consecutive terms is constant.

Substitute n=30 , d=3 a=1 in the given formula of arithmetic sequence

Then, we get

[tex]a_{30}=1+(30-1)(3)=1+29(3)=1+87=88[/tex]

Hence, the 30th term of sequence is 88.

Answer: 0.002718

Step-by-step explanation:

Given : The population mean annual salary for environmental compliance specialists is about ​$62,000.

i.e. [tex]\mu=62000[/tex]  

Sample size : n= 32

[tex]\sigma=6200[/tex]

Let x be the random variable that represents the annual salary for environmental compliance specialists.

Using formula [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex], the z-value corresponds to x= 59000 will be :

[tex]z=\dfrac{59000-62000}{\dfrac{6200}{\sqrt{32}}}\approx\dfrac{-3000}{\dfrac{6200}{5.6568}}=-2.73716129032\approx-2.78[/tex]

Now, by using the standard normal z-table , the probability that the mean salary of the sample is less than ​$59,000 :-

[tex]P(z<-2.78)=0.002718[/tex]

Hence, the probability that the mean salary of the sample is less than ​$59,000= 0.002718

Answer:

0.00125 Km

Step-by-step explanation:

Data provided in the question:

Speed of the horse = 9 Km/h

Duration for which the Lysera's eyes were shut = 0.50 seconds

now,

1 hour = 3600 seconds

or

1 second = [tex]\frac{\textup{1}}{\textup{3600}}\ textup{hours}[/tex]

Thus,

0.50 seconds =  [tex]\frac{\textup{0.50}}{\textup{3600}}\ textup{hours}[/tex]

Also,

Distance = speed × Time

on substituting the values, we get

Distance = 9 × [tex]\frac{\textup{0.50}}{\textup{3600}}[/tex]

or

Distance = 0.00125 Km

Answer:

[tex]10.897[/tex]

Step-by-step explanation:

An baseball player’s batting average decreases from 0.312 to 0.278 .

Let [tex]x_0[/tex] be the initial  baseball player’s batting average and [tex]x_1[/tex] be the final baseball player’s batting average .

Initial value [tex]\left ( x_0 \right )[/tex] = 0.312

Final value [tex]\left ( x_1 \right )[/tex] = 0.278

So, change in value =Final value - Initial value =  [tex]x_1-x_0[/tex] = [tex]0.278-0.312=-0.034[/tex]

Therefore , decrease in value = 0.034

We know that  percent decreased = ( decrease in value × 100 ) ÷ Initial value

i.e. , [tex]\frac{0.034}{0.312}\times 100=\frac{3400}{312}=10.897[/tex]

Or we can say percentage change in  baseball player’s batting average = [tex]-10.897 \%[/tex]

Answer:

It is sufficient to prove that  [tex] p_1\implies p_2, p_2\implies p_3, p_3\implies p_1[/tex]

Step-by-step explanation:

The propositions [tex] p_1,p_2,p_3[/tex] being equivalent means they should always have the same truth value. If one of them is true, then all of them must be true. And if one of them is false, then all of them must be false.

Suppose we've proven that [tex] p_1\implies p_2, p_2\implies p_3, p_3\implies p_1[/tex] (call these first, second and third implications).

If [tex]p_1[/tex] was true, then by the first implication that we proved, it would follow that [tex]p_2[/tex] is also true. And then by the second implication that we prove it would follow then that [tex]p_3 [/tex] is also true. Therefore the three of them would be true. Notice the reasoning would have been the same if we had started assuming that the one that was true was either [tex]p_2~or~p_3[/tex]. So one of them being true makes all of them be true.

On the other hand, if [tex]p_1[/tex] was false, then by the third implication that we proved, it would follow that [tex]p_3[/tex] has to be false (otherwise [tex]p_1[/tex] would have to be true, which would be a contradiction). And then, since [tex]p_3[/tex] is false, by the second implication that we proved it would follow that [tex] p_2[/tex] is false (otherwise [tex] p_3[/tex] would have to be true, which would be a contradiction). Therefore the three of them would be false. Notice the reasoning would have been the same if we had started assuming that the one that was false was either [tex]p_2~or~p_3[/tex]. So one of them being false makes all of them be false.

So, the three propositions always have the same truth value, and so they're all equivalent.