The area under the normal curve between the mean and a score for which z= - 1.20 is

a. .2563
b. .7563
c. .1138
d. .3862
e. .3849

Answer:

=10km/h

Step-by-step explanation:

Let motor boat speed be represented by x and current y

The speed upstream = Motor boats speed - rate of current

=x-y

The net speed down stream = Motor boats speed + rate of current

=x+y

Let us find the speed upstream =distance/ time taken

=150km/5hrs

=30km/h

Speed down stream= 150km/3h

=50 km/h

The problem forms simultaneous equations.

x-y=30

x+y=50

Using elimination method we solve the equations.

Add the two equations to eliminate y.

2x=80

x=40

Current, y= 50-x

=10km/h

Answer:

1) [tex]40\ \frac{km}{h}[/tex]

2) [tex]10\ \frac{km}{h}[/tex]

Step-by-step explanation:

 Let' call "b" the speed of the motorboat and "c" the speed of the current.

We know that:

[tex]V=\frac{d}{t}[/tex]

Where "V" is the speed, "d" is distance and "t" is time.

Then:

[tex]d=V*t[/tex]

We know that distance traveled upstream is 150 km and the time is 5 hours. Then, we set up the folllowing equation:

 [tex]5(b-c)=150[/tex]  (Remember that in the trip upstream the speed of the river is opposite to the motorboat)

For the return trip:

 [tex]3(b+c)=150[/tex]  

 By solving the system of equations, we get:

- Make both equations equal to each other and solve for "c".

[tex]5(b-c)=3(b+c)\\\\5b-5c=3b+3c\\\\5b-3b=3c+5c\\\\2b=8c\\\\c=\frac{b}{4}[/tex]

- Substitute "c" into the any original equation and solve for "b":

[tex]5(b-\frac{b}{4})=150\\\\\frac{3}{4}b=30\\\\b=40\ \frac{km}{h}[/tex]

- Substitute "b" into [tex]c=\frac{b}{4}[/tex]:

[tex]c=\frac{40}{4}\\\\c=10\ \frac{km}{h}[/tex]

Answer:  (A U B) n (B n C') = {5, 8, 11}.

Step-by-step explanation:  We are given the following sets :

U = {4, 5, 6, 7, 8, 9, 10, 11},

A = {5, 7, 9},

B = {4, 5, 8, 11}

and

C = {4, 6, 10}.

We are to find the following :

(A U B) n (B n C')

We know that for any two sets A and B,

A ∪ B contains all the elements present in set A or set B or both,

A ∩ B contains all the elements present in both A and B,

A - B contains all those elements which are present in A but not B

and

A' contains all the elements present in the universal set U but not A.

We will be suing the following rule of set of theory :

A ∩ B' = A - B.

Therefore, we have

[tex](A\cup B)\cap(B\cap C')\\\\=(A\cup B)\cap (B-C)\\\\=(\{5,7,9\}\cup\{4,5,8,11\})\cap (\{4,5,8,11\}-\{4,6,10\})\\\\=\{4,5,7,8,9,11\}\cap\{5,8,11\}\\\\=\{5,8,11\}.[/tex]

Thus,  (A U B) n (B n C') = {5, 8, 11}.

Final answer:

To solve the set operation (A U B) n (B n C'), we first find the union of A and B, then the complement of C, and the intersection of B and C'. The final step is to intersect the results of (A U B) and (B n C'), which gives us {4, 5, 8, 11}.

Explanation:

The question involves operations on sets, specifically union, intersection, and complement. We have a universal set U and subsets A, B, and C. The objective is to find the result of (A U B) n (B n C'), which involves set union (U), set intersection (n), and the complement of a set (').

First, let's find the union of sets A and B: A U B = {s, 7, 9, 4, 5, 8, 11}.

Next, we need to find the complement of set C, which is C' = {4, 5, 7, 8, 9, 11} as these are the elements of U that are not in C.

Then, identify the intersection of sets B and C': B n C' = {4, 5, 8, 11}, because these elements are common to both B and C'.

Finally, we find the intersection of the two results: (A U B) n (B n C') = {4, 5, 8, 11}.

Answer:

so principal amount is $7862.07

Step-by-step explanation:

Given data

rate = 7*1/4 % = 29/4 %

interest = $190

time = 4 months = 4/12 year

to find out

principal

solution

we know the simple interest formula  i.e.

interest = ( principal × rate  × time ) /100             ..................1

now put all value rate time and interest in equation 1 we get interest here

interest = ( principal × rate  × time ) /100  

190 =  ( principal × 29/4  × 4/12 ) /100  

principal = 190 × 12  ×100 /   29

principal = 7862.068966

so principal amount is $7862.07

Answer:

Maximum velocity, v = 50 ft/s

Step-by-step explanation:

Given

1000[tex]\frac{dv}{dt}=5000-100v[/tex]    -----------(1)

Dividing (1) by 1000, we get

[tex]\frac{dv}{dt}=5-\frac{v}{10}[/tex]

[tex]\frac{dv}{dt}+\frac{v}{10}=5[/tex]     -----------------(2)

Now we can solve the above equation using method of integrating factors

[tex]u(t)=e^{\int \frac{1}{10}dt}[/tex]

[tex]u(t)=e^{\frac{1}{10}t}[/tex]

Now multiplying each side of (2) by integrating factor,

[tex]e^{\frac{1}{10}t}(\frac{dv}{dt})+\frac{v}{10}e^{\frac{1}{10}t}=5e^{\frac{1}{10}t}[/tex]

Combining the LHS into one differential we get,

[tex]\frac{d}{dt}\left ( e^{\frac{1}{10}t}v \right ) = \int 5e^{\frac{1}{10}t}.dt[/tex]

[tex]e^{\frac{1}{10}t}v = 50e^{\frac{1}{10}t}[/tex] + c

v(t)=50+ce

Appltying the initial condition v(0)=0, we get

[tex]0=50+ce^{-\frac{1}{10}(0)}[/tex]

0=50+c

c=-50

So the particular solution is

[tex]v(t)=50-50e^{-\frac{1}{10}t}[/tex]

[tex]v(t)=50\left (1-e^{-\frac{1}{10}t}\right)[/tex]

Therefore, the maximum velocity is 50 ft/s

The forces balance out at 50 ft/s, which is the maximum velocity the boat can achieve.

To determine the maximum velocity of the boat, we need to consider the forces acting on it. The equation given is:

1000  [tex]\frac{dv}{dt}[/tex] = 5000 - 100v

At maximum velocity, the acceleration of the boat will be zero, which means  [tex]\frac{dv}{dt}[/tex] = 0. Therefore, setting the left-hand side of the equation to zero, we get:

0 = 5000 - 100v

Solving for v, we have:

Thus, the maximum velocity of the boat is 50 ft/s. This is where the force provided by the motor equals the resistive force from the water.

Answer:

Largest possible rank =6

Minimum possible rank=0

Step-by-step explanation:

We are given that A is a matrix of order [tex] 6\times 9 [/tex]

We have to find the maximum possible rank and minimum possible rank of A.

In given matrix we have 6 rows and 9 columns.

Rank: Rank is defined as the number of non zero rows or columns of matrix and any row or column is not a linear  combination of other two or more rows or columns

Rank of matrix of order [tex] m\times n [/tex]

Where m< n

Then largest possible rank is m

We are given a matrix of order[tex]6\times 9[/tex] where 6< 9

Therefore, the largest possible rank is 6.

The minimum possible rank  is zero because a  given matrix can be zero matrix therefore, the rank of zero matrix is zero.

Largest possible rank =6

Minimum possible rank=0

Answer: The value of b = 0.99

The probability that a day requires more water than is stored in city reservoirs is 0.161.

Step-by-step explanation:

Given : Water use in the summer is normally distributed with

[tex]\mu=310.4\text{ million gallons per day}[/tex]

Standard deviation : [tex]\sigma=40 \text{ million gallons per day}[/tex]

Let x be the combined storage capacity requires by the reservoir on a random day.

Z-score : [tex]\dfrac{x-\mu}{\sigma}[/tex]

[tex]z=\dfrac{350-310.4}{40}=0.99[/tex]

The probability that a day requires more water than is stored in city reservoirs is  :

[tex]P(x>350)=P(z>0.99)=1-P(z<0.99)\\\\=1-0.8389129=0.1610871\approx0.161[/tex]    

Hence, the probability that a day requires more water than is stored in city reservoirs is 0.161

Answer:

m<1 = 45°  

m<2 = 76°

m<3 = 80°

Step-by-step explanation:

Points to remember

1). Vertically opposite angles are equal

2). Sum of angles of a triangles is 180°

To find the measures of given angles

From the figure we get,

m<1 = 45°   [Vertically opposite angles]

By using angle sum property <2 + 59 + 45 = 180

m<2 = 180 - 104

m<2 = 76°

Also m<1 + m<3 + 55 = 180

m<3 = 180 - (m<1 + 55)

 = 180 - (45 + 55)

 = 180 - 100

m<3 = 80°

Answer:

  1a. y = 5x +3

  1b. y = -1/7x + 2

  1c. y = 2x +3

  2a. y = -1/2x +3; slope = -1/2, y-intercept = +3

  2b. y = 3/2x +6; slope = 3/2, y-intercept = +6

Step-by-step explanation:

Apparently the "y-intercept form" referred to is the one more commonly called "slope-intercept form." That form is ...

  y = mx + b . . . . . . . . m is the slope; b is the y-intercept

You arrive at this form by solving each equation for y. You do that the same way you solve any equation for any variable: undo what is done to the variable of interest.

In these general form equations, the y-variable term has a value multiplying y and some other terms added to that. First of all, you subtract the added terms (from both sides of the equation). This transforms ...

  ax +by +c = 0

into

  by = -ax -c

Next, you divide by the constant that is multiplying y. Of course all terms on both sides of the equation are divided by that, so you now have ...

  y = (-a/b)x -c/b

The slope is -a/b, and the y-intercept is -c/b.

___

1a. 5x -y +3 = 0

  -y = -5x -3 . . . . . subtract non-y terms

  y = 5x +3 . . . . . . divide by -1

__

1b. x +7y -14 = 0

  7y = -x +14 . . . . . subtract non-y terms

  y = -1/7x +2 . . . . .divide by 2

__

1c. 6x -3y +9 = 0

  -3y = -6x -9 . . . . subtract terms not containing y

  y = 2x +3 . . . . . . divide by -3

__

2a. x +2y -6 = 0

  2y = -x +6 . . . . . subtract non-y terms

  y = -1/2x +3 . . . . divide by 2 -- this is graphed as the red line below

__

2b. 3x -2y +12 = 0

  -2y = -3x -12 . . . . subtract non-y terms

  y = 3/2x +6 . . . . . divide by -2 -- this is graphed as the blue line below

_____

Comment on the 2nd problem

Of course, the y-intercept (constant term in slope-intercept form) is the point on the y-axis where the line crosses. The slope tells you the ratio of "rise" to "run". That is, a slope of -1/2 means the line drops one unit for each 2 units it goes to the right. A slope of 3/2 means the line increases (rises) by 3 units for each 2 units it goes to the right.

To approximate the volume with 8 boxes, we have to split up the interval of integration for each variable into 2 subintervals, [0, 1] and [1, 2]. Each box will have midpoint [tex]m_{i,j,k}[/tex] that is one of all the possible 3-tuples with coordinates either 1/2 or 3/2. That is, we're sampling [tex]f(x,y,z)=\cos(xyz)[/tex] at the 8 points,

(1/2, 1/2, 1/2)

(1/2, 1/2, 3/2)

(1/2, 3/2, 1/2)

(3/2, 1/2, 1/2)

(1/2, 3/2, 3/2)

(3/2, 1/2, 3/2)

(3/2, 3/2, 1/2)

(3/2, 3/2, 3/2)

which are captured by the sequence

[tex]m_{i,j,k}=\left(\dfrac{2i-1}2,\dfrac{2j-1}2,\dfrac{2k-1}2\right)[/tex]

with each of [tex]i,j,k[/tex] being either 1 or 2.

Then the integral of [tex]f(x,y,z)[/tex] over [tex]B[/tex] is approximated by the Riemann sum,

[tex]\displaystyle\iiint_B\cos(xyz)\,\mathrm dV\approx\sum_{i=1}^2\sum_{j=1}^2\sum_{k=1}^2\cos m_{i,j,k}\left(\frac{2-0}2\right)^2[/tex]

[tex]=\displaystyle\sum_{i=1}^2\sum_{j=1}^2\sum_{k=1}^2\cos\frac{(2i-1)(2j-1)(2k-1)}8[/tex]

[tex]=\cos\dfrac18+3\cos\dfrac38+3\cos\dfrac98+\cos\dfrac{27}8\approx\boxed{4.104}[/tex]

(compare to the actual value of about 4.159)

The Midpoint Rule for triple integrals approximates the integral of a function by taking the function's value at the midpoint of sub-boxes into which the domain is divided. In this problem, B is divided into eight 1x1x1 sub-boxes, and cos(xyz) is evaluated at each sub-box's center.

The Midpoint Rule for triple integrals is a method for approximating the value of a triple integral over a box B by dividing the box into smaller sub-boxes and evaluating the function at the midpoint of each sub-box.

For the given integral of cos(xyz) over the box B defined by the ranges 0 ≤ x ≤ 2, 0 ≤ y ≤ 2, and 0 ≤ z ≤ 2, we are asked to divide B into eight sub-boxes of equal size.

This involves calculating the midpoints for each sub-box (xi, yj, zk) and then evaluating the function at these points.

The size of each sub-box would be 1 x 1 x 1 since the original box dimensions are 2 x 2 x 2.

The midpoints for the sub-boxes will be at (0.5, 0.5, 0.5), (0.5, 0.5, 1.5), (0.5, 1.5, 0.5), (0.5, 1.5, 1.5), (1.5, 0.5, 0.5), (1.5, 0.5, 1.5), (1.5, 1.5, 0.5), and (1.5, 1.5, 1.5).

The volume of each sub-box is 1, so the Riemann sum becomes the sum of cos(xyz) evaluated at these midpoints times the volume of each sub-box.

After calculating the Riemann sum, it provides an approximate value for the triple integral.

The actual calculation would require substituting each midpoint into the function cos(xyz) to get the respective value for each sub-box, then summing these values and finally multiplying by the volume of a single sub-box to get an approximation for the entire integral.

Answer: There is 49.28% of cumulative markup for the department at the end of April.

Step-by-step explanation:

Since we have given that

Price of opening inventory = $170,000

Mark up rate = 48%

Amount of mark up is given by

[tex]\dfrac{48}{100}\times 170000\\\\=\$81600[/tex]

Price of additional merchandise = $80000

Mark up rate = 52%

Amount of mark up is given by

[tex]\dfrac{52}{100}\times 80000\\\\=\$41600[/tex]

So, total mark up would be

$81600 + $41600 = 123200

So, the cumulative markup percentage for the department at the end of April is given by

[tex]\dfrac{123200}{170000+80000}\times 100=\dfrac{123200}{250000}\times 100=49.28\%[/tex]

Hence, there is 49.28% of cumulative markup for the department at the end of April.

Answer:

30 gallons of 20% acid solution should be mixed.

Step-by-step explanation:

Let x gallons of a 20% acid solution was mixed with 30 gallons of a 40% solution, to obtain a mixture of 30% acid solution.

Therefore, final volume of the solution will be (x + 30) gallons.

Now concept to solve this question is

20%.(x) + 40%.(30) = 30%.(x + 30)

0.20(x) + 0.40(30) = 0.30(x + 30)

0.20x + 12 = 0.30x + 9

0.30x - 0.20x = 12 - 9

.10x = 3

x = [tex]\frac{3}{0.1}[/tex]

x = 30 gallons

Therefore, 30 gallons of the 20% acid solution should be mixed.

Answer:

The method for selecting a stratified sample is to order a population in some way and then select members of the population at regular intervals. - FALSE.

This can be re written as :The method for selecting a systematic sample is to order a population in some way and then select members of the population at regular intervals.

Explanation for stratified sampling:

In stratified sampling, the members of a population are divided into two or more strata with similar characteristics and then a random sample is selected from each strata. This way ensures that members of each group within a population will be sampled.

Explanation for cluster sampling:

The method for selecting a cluster sample is to order a population in some way and then select members of the population at regular intervals.

In cluster sampling, the population is divided into​​ clusters, and all of the members of one or more​ clusters are selected.

Final answer:

The statement provided is false as it describes the method of systematic sampling instead of stratified sampling. Stratified sampling requires dividing the population into strata and randomly selecting individuals from each stratum, while systematic sampling selects members at regular intervals.

Explanation:

The statement is false. The correct method for selecting a stratified sample is to divide the population into groups known as strata, and then use simple random sampling to identify a proportionate number of individuals from each stratum. The statement describes the method for selecting a systematic sample, not a stratified sample. A systematic sample is obtained by ordering a population and then selecting members at regular intervals, not by stratifying them.

On the other hand, a cluster sample involves dividing the population into clusters (groups), often geographically defined, and then randomly selecting some of these clusters. All the members from these selected clusters are included in the sample. Hence, the method of selecting individuals at regular intervals is not used in cluster sampling either.

Answer Step-by-step explanation:

Given statement if a-b is even and b-c is even then a-c is even .

Let p: a-b and b-c are even

q: a-c is even.

Converse: If a-c is even then a-b and b-c are both even.

Inverse:If a-b and b-c  are not both even then a-c is not even.

If a= Even number

b= Even number

c=Even number

If a-c  is even then a-b and b-c are both even..Hence, the converse statement is true.

If a=Odd number

b=Odd number

c= Odd number

If a-c is even then a-b and b-c are both even number .Hence, the converse  statement is true.

If a=Even number

b= Even number

c= Odd number

a-b  and b-c are both odd not even number but a-c is even number

a=8,b=6 c=3

a-b=8-6=2

b-c=6-3=3

a-c=8-3=5

If a-c is odd then a-b even but b-c is odd .Hence , the converse statement is false.But the inverse statement is true.

If a= Odd number

b=Even number

c= Even number

If a-b is odd and b-c is even then a-c is odd not even . Hence, the inverse statement is true.

If a= Odd number

b=Eve number

c=Odd number

a=9,b=6,c=5

a-b=9-6=3

b-c=6-5=1

a-c=9-5=4

Here, a-b and b-c are not both even but a-c is even .Hence, the inverse statement is false.

Answer:

Step-by-step explanation:

Let p : It is raining

Truth value of this statement is True.

q : I will give you a ride home after class

Truth value of the statement q is True.

So for the statement p→q (If it is raining, then I will give you a ride home)

truth value will be True.

Therefore, when statement p is True and q is False, p→q will be False.

Answer:

a) bijective

b) neither surjective or injective

c) injective

Step-by-step explanation:

Since we are looking from real numbers to real numbers, we want the following things

1) We want every real number y to get it. (surjective)

2) We want every y that gets hit to be hit only once. (injective).

---

If we have both things then the function is bijective.

a) f(x)=5x+4 this is a line with a positive slope.

That means it is increasing left to right.  Every increasing or even decreasing line is going to hit every real number y.  This is a bijection.

b) g(x)=2x^2-2 is a parabola. Parabola functions always have y's that get hit more than once and not all y's get hit because the parabola is either open up or down starting from the vertex.  This function is neither injective or surjective.

c) h(x)=1+(2/x)  x is not 0.

1+(2/x) is never 1 because 2/x is never 0 for any x. This means the real number y=1 will never be hit and is therefore not surjective. This function is injective because every 1 that is hit is only hit once. If you want use the horizontal line test to see this.

To find the net worth of the Crabby Apple restaurant at the beginning of January before it lost $2500, we add back the loss to the net worth at the end of the month, resulting in an initial net worth of $2900.

The student is asking about calculating the net worth of a restaurant before a financial loss. To find the net worth at the beginning of the month, we would add the loss sustained during that month back to the net worth at the end of the month. Since the restaurant lost $2500 in January, and had a net worth of $400 at the end of January, we can calculate its net worth at the beginning of January by the following calculation:

Net Worth at Beginning of Month = Net Worth at End of Month + Loss During Month

Net Worth at Beginning of Month = $400 + $2500

Net Worth at Beginning of Month = $2900

Therefore, the net worth of the Crabby Apple restaurant at the beginning of January was $2900.

The net worth of the Crabby Apple restaurant at the beginning of January was $2900.

To determine the net worth of the Crabby Apple restaurant at the beginning of January, we need to account for the loss incurred during the month. Given that the net worth at the end of January is $400 and the restaurant lost $2500 during the month, we can set up the following equation:

[tex]\[\text{Net worth at the beginning of January} =[/tex][tex]\text{Net worth at the end of January} + \text{Loss during January}\][/tex]

Substitute the given values into the equation:

[tex]\[\text{Net worth at the beginning of January} = \$400 + \$2500\][/tex]

[tex]\[\text{Net worth at the beginning of January} = \$2900\][/tex]

Answer: 0.125

Step-by-step explanation:

Given: A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed in interval (50,52).

∴ The probability density function of X will be :-

[tex]f(x)=\dfrac{1}{b-a}=\dfrac{1}{52-50}=\dfrac{1}{2}[/tex]

The required probability will be:-

[tex]P(51.25<x<51.5)=\int^{51.5}_{51.25}f(x)\ dx\\\\=\dfrac{1}{2}\int^{51.5}_{51.25}\ dx\\\\=\dfrac{1}{2}[x]^{51.5}_{51.25}\\\\=\dfrac{1}{2}(51.5-51.25)=\dfrac{0.25}{2}=0.125[/tex]

Hence, the probability that a given class period runs between 51.25 and 51.5 minutes =0.125

Answer:

[tex]y'=\frac{5}{x} \cdot \cos(\ln(2x^5))[/tex]

Step-by-step explanation:

[tex]y=\sin(\ln(2x^5))[/tex]

We are going to use chain rule.

The most inside function is [tex]y=2x^5[/tex] which gives us [tex]y'=10x^4[/tex].

The next inside function going out is [tex]y=\ln(x)[/tex] which gives us [tex]y'=\frac{1}{x}[/tex].

The most outside function is [tex]y=\sin(x)[/tex] which gives us [tex]y'=\cos(x)[/tex].

[tex]y'=10x^4 \cdot \frac{1}{2x^5} \cdot \cos(\ln(2x^5))[/tex]

[tex]y'=\frac{5}{x} \cdot \cos(\ln(2x^5))[/tex]

Answer:

The probability is 0.1444.

Step-by-step explanation:

Let X be the event that college business students cheated on final exams,

Since, the probability that student cheats on exam, p = 75 % = 0.75,

So, the probability that student does not cheat on exam, q = 1 - p = 0.25,

The binomial distribution formula,

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

Where, [tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]

The probability that exactly 30 of the students monitored cheat on the final exam out of 40 students is,

[tex]P(X=30)=^{40}C_{30} 0.75^{30} 0.25^{40-30}[/tex]

[tex]=847660528\times 0.75^{30} \times 0.25^{10}[/tex]

[tex]=0.144364346356[/tex]

[tex]\approx 0.1444[/tex]

The correct probability that exactly 30 of the 40 students monitored cheat on the final exam is 0.0256.

To solve this problem, we can use the binomial probability formula, which is given by:

[tex]\[ P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k} \][/tex]

 Given:

- ( n = 40 ) (the number of students monitored),

- ( k = 30 ) (the number of students who cheat),

- ( p = 0.75 ) (the probability of a student cheating on the final exam).

We can plug these values into the binomial probability formula:

[tex]\[ P(X = 30) = \binom{40}{30} \cdot (0.75)^{30} \cdot (1-0.75)^{40-30} \][/tex]

First, calculate the binomial coefficient:

[tex]\[ \binom{40}{30} = \frac{40!}{30!(40-30)!} \][/tex]

Next, calculate the powers of p and  (1-p) :

[tex]\[ (0.75)^{30} \] \[ (0.25)^{10} \][/tex]

Now, multiply these values together:

[tex]\[ P(X = 30) = \binom{40}{30} \cdot (0.75)^{30} \cdot (0.25)^{10} \][/tex]

Using a calculator or a software tool, we can compute the exact value of the probability:

[tex]\[ P(X = 30) \approx 0.0256 \][/tex]

Therefore, the probability that exactly 30 of the 40 students monitored cheat on the final exam is approximately 0.0256 when rounded to four decimal places.

Answer:

Step-by-step explanation:

∠C and ∠F are vertical angles, so are congruent. Then any angle complementary to one of those will also be complementary to the other.

Likewise, ∠B and ∠E are vertical angles and congruent. Any angle complementary to one of them will also be complementary to the other. Here, ∠E and ∠F are listed as complementary, so we know ∠B and ∠F will be also.

Answer:2 and 4

Step-by-step explanation:

^B and ^F

^F and ^E

The scientist can determine the amount of Solution A and Solution B required by setting up and solving a system of two linear equations representing the total solution volume and the total salt amount.

Lets let the amount of Solution A the scientist will use be x and the amount of Solution B she will also use be y. We know that x + y = 110 ounces because her final mix should be 110 ounces. Also, we know that 0.4x + 0.65y = 0.55*(x+y) = 60.5 because the amount of salt from Solution A and Solution B should add up to the amount of salt in the final mixture. Solving this system of linear equations to obtain the values for x and y, gives the required amounts of Solution A and Solution B needed.

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Answer:

The correct option is D. a = -2 and b = 4.

Step-by-step explanation:

Consider the provided equation:

[tex]y=a(x-2)^2+b\ \text{and}\ y=5[/tex]

The vertex form of a quadratic is:

[tex]y= a(x-h)^2+k[/tex]

Where, (h,k) is the vertex and the quadratic opens up if 'a' is positive and opens down if 'a' is negative.

Now consider the provided option A. a = 1 and b = -4.

Since the value of a is positive the graph opens up and having vertex (2,-4). Thus graph will intersect the line y = 5.

Refer the figure 1:

Now consider the option B. a = 2 and b = 5.

Since the value of a is positive the graph opens up and having vertex (2,5). Thus graph will intersect the line y = 5.

Refer the figure 2:

Now consider the option C. a = -1 and b = 6.

Since the value of a is negative the graph opens down and having vertex (2,6). Thus graph will intersect the line y = 5.

Refer the figure 3:

Now consider the option D. a = -2 and b = 4.

Since the value of a is negative the graph opens down and having vertex (2,4). Thus graph will not intersect the line y = 5.

Refer the figure 4:

Hence, the correct option is D. a = -2 and b = 4.

Answer:

[tex]36.93\ ft[/tex]

Step-by-step explanation:

step 1

Find the circumference of the circle

The circumference is equal to

[tex]C=\pi D[/tex]

we have

[tex]D=25.2\ ft[/tex]

substitute

[tex]C=25.2\pi\ ft[/tex]

step 2

we know that

The circumference of a circle subtends a central angle of 360 degrees

so

using proportion

Find out the arc length by a central angle of 168 degrees

[tex]\frac{25.2\pi}{360}=\frac{x}{168}\\\\x=25.2\pi*168/360\\\\x=(25.2*3.14)*168/360\\\\x=36.93\ ft[/tex]

Answer:

A. It is false that the bus has an engine and the people have money.

Step-by-step explanation:

Given statement is -

The bus does not have an engine or the people do not have money.

The De Morgan's laws states that the complement of the union of two given sets is the intersection of their complements.

Also, the complement of the intersection of these two sets is the union of their complements.

negative (p ∨ q)⇔ p ∧ negative q

So, here option A is correct. It is false that the bus has an engine and the people have money.

Final answer:

Using De Morgan's laws, the equivalent statement for 'The bus does not have an engine or the people do not have money' is 'It is not the case that the bus has an engine and the people have money,' which corresponds to answer choice A.

Explanation:

The question asks to use De Morgan's laws to write an equivalent statement for the sentence: "The bus does not have an engine or the people do not have money." Let's denote 'the bus has an engine' by A and 'the people have money' by B. The original statement can be written in logical form as ¬A ∨ ¬B, which using De Morgan's laws, translates to ¬(A ∧ B). This means that the equivalent statement is: "It is not the case that the bus has an engine and the people have money." Therefore, the correct answer is A.

Answer:

C. If I lower my grades then I work alone

Step-by-step explanation:

In logic, the contrapositive of a conditional statement, [tex]P\to Q[/tex] is [tex]\neg Q\to \neg P[/tex].

In other words, the contrapositive of "If P then Q" is "If not Q, then not P".

The given conditional statement is:

"If I form a study group then I raise my grades."

We can break this conditional statement into the following propositional statements:

P : I form a study group

¬P : I work alone

Q : I raise my grades

¬Q : I lower my grades

The contrapositive of  "If I form a study group then I raise my grades." becomes:

" If I lower my grades then I work alone"

The correct choice is C

To find the contrapositive of the given conditional statement "If I form a study group then I raise my grades," we have to negate both the hypothesis (the "if" or P part) and the conclusion (the "then" or Q part) and also reverse their order.
The original statement is in the form: If P then Q.
P: I form a study group.
Q: I raise my grades.
The contrapositive is: If not Q then not P.
Not Q: I do not raise my grades.
Not P: I do not form a study group.
So, the contrapositive of the original statement would be: "If I do not raise my grades then I do not form a study group."
Now let's match this to one of the choices given:
A. If I work alone then I lower my grades. (This is not the contrapositive; it introduces new concepts that weren't part of the original statement.)
B. If I raise my grades then I form a study group. (This is the converse, not the contrapositive.)
C. If I lower my grades then I work alone. (This is the contrapositive: "lower my grades" is equivalent to "do not raise my grades," and "work alone" is equivalent to "do not form a study group.")
D. If I form a study group then I lower my grades. (This is not the contrapositive; it's the inverse of the original statement.)
E. If I form a study group then I raise my grades. (This is the original statement.)
F. If I raise my grades then I work alone. (This is not the contrapositive; it introduces the concept of working alone.)
The correct choice that represents the contrapositive of the original statement is C. If I lower my grades then I work alone.

Answer:

Step-by-step explanation:

Let us calculate std dev and std error of two samples

Sample     N     Mean    Std dev   Std error

  1              65     52.8      9.9         1.22279

  2             111      54.1       11.09      1.0526

Assuming equal variances

df =111+65-2=174

Pooled std deviation = combined std deviation = 10.6677

Pooled std error = 10.6677/sqrt 174 = 0.808

difference in means =52.8-54.1 = -1.3

Margin of error = 0.5923

( t critical 1.6660)

Confidence interval

=(-4.4923, 1.8923)

Since 0 lies in this interval we can accept null hypothesis that 10 year old boys and girls have the same average height

Answer:

The two lengths are 3.4303 and 9.8606 centimeters

Step-by-step explanation:

The metal bar is divided into two pieces, so we are going to call X the length of the first piece and Y the length of the second piece.

From the phrase: one piece is 3 centimeters longer than twice the length of the other, we can separate and rewrite as:

One piece - is - 3 centimeters - longer than - twice the length of the other

X                   =               3                      +               2         *       Y

So, X=3+2Y is our first equation

From the phrase: the sum of the squares of the two lengths is 109 square centimeters, we can rewrite as:

The sum of the squares of the two lengths - is - 109 square centimeters

               [tex]X^{2} +Y^{2}[/tex]                                            =      109

So, X^2+Y^2=109 is our second equation

Replacing the first equation on the second question we get:

[tex]X^{2} +Y^{2} =109\\(3+2Y)^{2} +Y^{2} =109[/tex]

[tex]9+(2*3*2Y)+4Y^{2} +Y^{2} -109=0[/tex]

[tex]5Y^{2} +12Y-100=0[/tex]

Solving this equation we find two solutions:

Y=3.4303 and Y= -5.8303

Since the question is talking about the length there is no sense use Y=-5.8303, then our first length is 3.4303

So replacing this value on the first equation we get:

X= 3 + 2*Y

X= 3 + 2*3.4303

X= 9.8606

Finally the two length are 3.4303 and 9.8606 centimeters

Answer:

(0.23, 0.30)

Step-by-step explanation:

Number of green peas = 428

Number of yellow peas = 155

Total number of peas = n = 583

Since we have to establish the confidence interval for yellow peas, the sample proportion of yellow peas would be considered as success i.e. p = [tex]\frac{155}{583}[/tex]

q = 1 - p = [tex]\frac{428}{583}[/tex]

Confidence Level = 95%

Z value associated with this confidence level = z = 1.96

Confidence interval for the population proportion is calculated as:

[tex](p-z\sqrt{\frac{pq}{n}} ,p+z\sqrt{\frac{pq}{n}})[/tex]

Using the values, we get:

[tex](\frac{155}{583}-1.96\sqrt{\frac{\frac{155}{583} \times\frac{428}{583}}{583} },\frac{155}{583}+1.96\sqrt{\frac{\frac{155}{583} \times\frac{428}{583}}{583} })\\\\ =(0.23,0.30)[/tex]

Conclusion:

We are 95% confident that true value of population proportion of yellow peas lie between 0.23 and 0.30

Answer:

(x+3)^2 + (y-5)^2 = 4^2

or

(x+3)^2 + (y-5)^2 = 16

Step-by-step explanation:

The equation of a circle can be written in the form

(x-h)^2 + (y-k)^2 = r^2

where (h,k) is the center and r is the radius

(x--3)^2 + (y-5)^2 = 4^2

(x+3)^2 + (y-5)^2 = 4^2

or

(x+3)^2 + (y-5)^2 = 16

Answer with explanation:

Given → A, B ⊆R, are non-empty sets.

Also, A ⊆ B.

It is also given that , B has least upper bound.

To Prove:→A has a least upper bound,

and → lub A ≤ lub B.

Proof:

    A and B are non empty sets.

Also, A ⊆ B

it means there are some elements in B which is not in A and there are some elements in set B which may be greater or smaller than or equal to set A.

The meaning of least upper bound is the the smallest member of the set from which all members are greater.

As, A⊆B

So, there are two possibilities ,either

→ Least upper bound set A = least upper bound set B------(1)

→ Least upper bound set A < least upper bound set B-------(2)

Combining (1) and (2)

→→→l u b(A) ≤ l u b (B)