If angle 1 has a measure of 56° and angle 2 has a measure of 124°, the two angles are complementary.

Question 1 options:
True
False

Notice that

[tex](n+1)^4-n^4=4n^3+6n^2+4n+1[/tex]

so that

[tex]\displaystyle\sum_{i=1}^n((n+1)^4-n^4)=\sum_{i=1}^n(4i^3+6i^2+4i+1)[/tex]

We have

[tex]\displaystyle\sum_{i=1}^n((i+1)^4-i^4)=(2^4-1^4)+(3^4-2^4)+(4^4-3^4)+\cdots+((n+1)^4-n^4)[/tex]

[tex]\implies\displaystyle\sum_{i=1}^n((i+1)^4-i^4)=(n+1)^4-1[/tex]

so that

[tex]\displaystyle(n+1)^4-1=\sum_{i=1}^n(4i^3+6i^2+4i+1)[/tex]

You might already know that

[tex]\displaystyle\sum_{i=1}^n1=n[/tex]

[tex]\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2[/tex]

[tex]\displaystyle\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6[/tex]

so from these formulas we get

[tex]\displaystyle(n+1)^4-1=4\sum_{i=1}^ni^3+n(n+1)(2n+1)+2n(n+1)+n[/tex]

[tex]\implies\displaystyle\sum_{i=1}^ni^3=\frac{(n+1)^4-1-n(n+1)(2n+1)-2n(n+1)-n}4[/tex]

[tex]\implies\boxed{\displaystyle\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4}[/tex]

If you don't know the formulas mentioned above:

Answer:

1 member took literature but not mathematics.

Step-by-step explanation:

We can draw a Venn diagram for the given question.

In a fraternity total number of members = 32

Number of members who took mathematics M = 18

Number of members who took both mathematics and literature (M∩L) = 5

And number of members who took neither mathematics nor literature = 8

Therefore, number of members who took literature but not mathematics

= 32 - [(18 + 5) + 8]

= 32 - [23 + 8]

= 32 - 31

= 1

Therefore, 1 member took literature but not mathematics.

Answer:

1 member took literature but not mathematics.

Step-by-step explanation:

Answer:

$175 = Cost of single dress

Step-by-step explanation:

It is provided that cost of dress is twice that of a skirt.

Let say, cost of skirt = x

Total things bought 3 dresses and 2 skirts.

Cost of single dress = 2x

Cost of all items = 3 [tex]\times[/tex] 2x + 2 [tex]\times[/tex] x = $1,000 - $300 = $700

6x + 2x = $700

x = 700/8 = $87.50

Cost of single dress = 2x = $87.50 [tex]\times[/tex] 2 = $175

Answer:

450 mL/hr

Step-by-step explanation:

Given:

Order for vasopressin = 18 units/hour

Available vasopressin = 200 units in 5000 mL

Now,

Volume of vasopressin per unit =  [tex]\frac{\textup{Volume of vasopressin}}{\textup{Number of units}}[/tex]

or

Volume of vasopressin per unit =  [tex]\frac{\textup{5000}}{\textup{200}}[/tex]

or

Volume of vasopressin per unit = 25 mL/unit

Thus,

Drip rate in mL/hr  

= volume of vasopressin per unit × order for vassopressin

or

Drip rate in mL/hr  = 25 × 18 = 450 mL/hr

The drip rate for an order of Vasopressin 18 units/hr, given a solution concentration of 200 units in 5000 mL, is calculated to be 450 mL/hr.

To find the drip rate in mL/hr, we start by determining the concentration of the vasopressin solution. It is 200 units in 5000 mL D5W, so the concentration is 0.04 units/mL (200 units/5000 mL).

Next, we know the doctor prescribed 18 units/hr of vasopressin. To find out how many mL this corresponds to, we divide the order of 18 units/hr by the concentration in units/mL,  which gives us 450 mL/hr (18 units/hr / 0.04 units/mL).

Therefore, the drip rate for the Vasopressin order is 450 mL/hr.

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Answer:

a=2

Step-by-step explanation:

u=3b-2(2) or u=3b-4

Final answer:

To solve the equation for a, rearrange the terms and isolate 'a' by dividing by the coefficient.

Explanation:

Step 1: Start with the equation u + 3b - 2a + 2 = 0.

Step 2: Rearrange the terms to isolate 'a', which gives  -2a = -u - 3b - 2.

Step 3: Divide by -2 to solve for 'a', resulting in a = (u + 3b + 2) / 2.

Answer:

12,500 units

Step-by-step explanation:

Given:

Selling cost of the pump = $10

Variable cost of producing each unit = 60% of selling cost

or

Variable cost of producing each unit  = 0.60 × $10 = $6

Monthly fixed cost = $50,000

Now,

The profit per unit = Selling cost - Variable cost

or

The profit per unit = $10 - $6 = $4

Therefore,

The breakeven point = [tex]\frac{\textup{Fixed cost}}{\textup{Profit per unit}}[/tex]

or

the breakeven point = [tex]\frac{\textup{50,000}}{\textup{4}}[/tex]

or

the breakeven point = 12,500 units

By considering the given information, we have

Null hypothesis : [tex]H_0: \mu=35.0[/tex]

Alternative hypothesis : [tex]H_1: \mu\neq35.0[/tex]

Since the alternative hypothesis is two-tailed , so the test is a two-tailed test.

Given : Sample size : n= 20, since sample size is less than 30 so the test applied is a t-test.

[tex]\overline{x}=41.0[/tex]    ;    [tex]\sigma= 3.7[/tex]

Test statistic : [tex]t=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]

i.e. [tex]t=\dfrac{41.0-35.0}{\dfrac{3.7}{\sqrt{20}}}=7.252112359\approx7.25[/tex]

Degree of freedom : n-1 = 20-1=19

Significance level = 0.01

For two tailed,  Significance level [tex]=\dfrac{0.01}{2}=0.005[/tex]

By using the t-distribution table, the critical value of t =[tex]t_{19, 0.005}=2.861[/tex]

Since , the observed t-value (7.25) is greater than the critical value (2.861) .

So we reject the null hypothesis, it means we have enough evidence to support the alternative hypothesis.

We conclude that there is some significance difference between the mean score for sober women and 35.0.

Answer: i guess the problem is with P(x) => "x = [tex]x^{2}[/tex]", then P(x) is true if that equality is true, and is false if the equality is false.

so lets see case for case.

a) x = 0, and [tex]0^{2}[/tex] = 0. So p(0) is true.

b) x = 1 and [tex]1^{2}[/tex] = 1, so P(1) is true.

c) x = 2, and [tex]2^{2}[/tex] = 4, and 2 ≠ 4, then P(2) is false.

d) x= -1 and [tex]1^{2}[/tex] = 1, and 1 ≠ -1, so P(-1) is false.

The truth value of P(0) and P(1) is true while the truth value of P(2) and P(-1) is false

The statement is given as:

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

For P(0), we have:

[tex]0 = 0^2[/tex]

[tex]0 = 0[/tex] --- this is true

For P(1), we have:

[tex]1 = 1^2[/tex]

[tex]1 = 1[/tex] --this is true

For P(2), we have:

[tex]2 = 2^2[/tex]

[tex]2= 4[/tex] -- this is false

For P(-1), we have:

[tex](-1) = (-1)^2[/tex]

[tex](-1) = 1[/tex] --- this is false

Hence, the truth value of P(0) and P(1) is true while the truth value of P(2) and P(-1) is false

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Answer:

The correct option is A) [tex]3d+C[/tex].

Step-by-step explanation:

Consider the provided information.

Paco bought 3 CDs that cost d dollars each.

Let d is the cost of each CDs.

The cost of 3 CDs will be 3 times of d.

This can be written as:

[tex]3d[/tex]

He bought a pack of gum for C cent. Thus, we can say that the cost of pack of gum is C.

Now add the cost of gum in above expression.

[tex]3d+C[/tex]

Hence, the required expression is [tex]3d+C[/tex].

Thus, the correct option is A) [tex]3d+C[/tex].

Answer:

The initial population was approximatedly 3535 inhabitants.

Step-by-step explanation:

The population of the city can be given by the following differential equation.

[tex]\frac{dP}{dt} = Pr[/tex],

In which r is the rate of growth of the population.

We can solve this diffential equation by the variable separation method.

[tex]\frac{dP}{dt} = Pr[/tex]

[tex]\frac{dP}{P} = r dt[/tex]

Integrating both sides:

[tex]ln P = rt + c[/tex]

Since ln and the exponential are inverse operations, to write P in function of t, we apply ln to both sides.

[tex]e^{ln P} = e^{rt + C}[/tex]

[tex]P(t) = Ce^{rt}[/tex]

C is the initial population, so:

[tex]P(t) = P(0)e^{rt}[/tex]

Now, we apply the problem's statements to first find the growth rate and then the initial population.

The problem states that:

In two years the population has doubled:

[tex]P(2) = 2P(0)[/tex]

[tex]P(t) = P(0)e^{rt}[/tex]

[tex]2P(0) = P(0)e^{2r}[/tex]

[tex]2 = e^{2r}[/tex]

To isolate r, we apply ln both sides

[tex]e^{2r} = 2[/tex]

[tex]ln e^{2r} = ln 2[/tex]

[tex]2r = 0.69[/tex]

[tex]r = \frac{0.69}{2}[/tex]

[tex]r = 0.3466[/tex]

So

[tex]P(t) = P(0)e^{0.3466t}[/tex]

In two years the population has doubled and a year later there were 10,000 inhabitants.

[tex]P(3) = 10,000[/tex]

[tex]P(t) = P(0)e^{0.3466t}[/tex]

[tex]10,000= P(0)e^{0.3466*3}[/tex]

[tex]P(0) = \frac{10,000}{e^{1.04}}[/tex]

[tex]P(0) = 3534.55[/tex]

The initial population was approximatedly 3535 inhabitants.

Answer:

  (a) even: J, K, M, O; odd: L, N

  (b) L and O are connected to J

  (c) N is of degree 3

Step-by-step explanation:

Count the edge ends that intersect each vertex. You get ...

  J-2, K-2, L-3, M-2, N-3, O-4

These numbers are the degree of the vertex.

a) Vertices with even degree are J, K, M, O, since these have degrees of 2, 2, 2, and 4, respectively--all even numbers.

Vertices with odd degree are L and N, since these both have degree 3, an odd number.

__

b) The vertices that are adjacent to J are the ones at the other ends of the edges that intersect vertex J. One of those two edges connects to vertex L; the other to vertex O. J is adjacent to L and O.

__

c) When we counted edges in part (a), we found vertex N to be of degree 3.

Answer:

(a) Sampling distribution

P(25) = 0,04

P(35) = 0.1 + 0.1 = 0,2

P(42,5) = 0.06 + 0.06 = 0,12

P(45) = 0,25

P(52,5) = 0.15 + 0.15 = 0,3

P(60) = 0,09

(b) E(X) = 45.5 oz

(c) E(X) = μ

Step-by-step explanation:

The variable we want to compute is

[tex]X=(X1+X2)/2[/tex]

For this we need to know all the possible combinations of X1 and X2 and the probability associated with them.

(a) Sampling distribution

Calculating all the 9 combinations (3 repeated, so we end up with 6 unique combinations):

P(25) = P(X1=25) * P(X2=25) = p25*p25 = 0.2 * 0.2 = 0,04

P(35) = p25*p45+p45*p25 = 0.2*0.5 + 0.5*0.2 = 0.1 + 0.1 = 0,2

P(42,5) = p25*p60 + p60*p25 = 0.2*0.3 + 0.3*0.2 = 0.06 + 0.06 = 0,12

P(45) = p45*p45 = 0.5 * 0.5 = 0,25

P(52,5) = p45*p60 + p60*p45 = 0.5*0.3 + 0.3*0.5 = 0.15 + 0.15 = 0,3

P(60) = p60*p60 = 0.3*0.3 = 0,09

(b) Using the sample distribution, E(X) can be expressed as:

[tex]E(X)=\sum_{i=1}^{6}P_{i}*X_{i}\\E(X)=0.04*25+0.2*35+0.12*42.5+0.3*52.5+0.09*60 = 45.5[/tex]

The value of E(X) is 45.5 oz.

(c) The value of μ can be calculated as

[tex]\mu=\sum_{i=1}^{3}P_{i}*X_{i}\\\mu=0.2*25+0.5*45+0.3*60=45.5[/tex]

We can conclude that E(X)=μ

We could have arrived to this conclusion by applying

[tex]E(X)=E((X1+X2)/2)=E(X1)/2+E(X2)/2\\\\\mu = E(X1)=E(X2)\\\\E(X)=\mu /2+ \mu /2 = \mu[/tex]

The equations of the line are:

x = -4,

y = 0.5,

z = 5 + t

A line that is parallel to the z-axis lies in the xy-plane. Since it's parallel to the z-axis, its direction in the xy-plane is determined by the coefficients of x and y in its direction vector.

Let's first find the midpoint between the two given points:

Midpoint = [(x₁ + x₂) / 2, (y₁ + y₂) / 2, (z₁ + z₂) / 2]

Midpoint = [(-8 + 0) / 2, (5 - 4) / 2, (1 + 9) / 2]

Midpoint = [-4, 0.5, 5]

So, the midpoint is (-4, 0.5, 5).

Now, let's create a line that passes through the midpoint and is parallel to the z-axis. The equation of such a line in vector form is:

r(t) = Midpoint + t Direction

Where r(t) is the position vector of a point on the line, t is a scalar parameter, Midpoint is the midpoint we calculated, and Direction is the direction vector.

Since the line is parallel to the z-axis, its direction vector is (0, 0, 1). Thus, the equation of the line is:

r(t) = (-4, 0.5, 5) + t (0, 0, 1)

r(t) = (-4, 0.5, 5 + t)

In component form, the equations of the line are:

x = -4

y = 0.5

z = 5 + t

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The midpoint of the two points is (-4, 0.5, 5). The line parallel to the z-axis passing through this point has the equations: x=-4, y=0.5, z=t.

Firstly, we need to find out the midpoint between the two points (0, -4, 9) and (-8, 5, 1).

The formula to calculate the midpoint of two points in three dimensions is ((x1+x2)/2, (y1+y2)/2, (z1+z2)/2).

Applying this formula to our points, we get the midpoint as ((0-8)/2, (-4+5)/2, (9+1)/2) which is (-4, 0.5, 5).

Lines parallel to the z-axis in three-dimensional space have equations of the form x=a, y=b, z=t, where 'a' and 'b' are constants representing any particular point through which the line passes, and 't' represents a variable that can take any real value.

Since the line we want passes through the point (-4, 0.5, 5), our equations for the desired line become: x=-4, y=0.5, z=t.

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Answer:

  d)  one solution; (4, 1)

Step-by-step explanation:

It often works well to follow problem directions. A graph is attached, showing the one solution to be (4, 1).

_____

You know there will be one solution because the lines have different slopes. Each is in the form ...

  y = mx + b

where m is the slope and b is the y-intercept.

The first line has slope -1 and y-intercept +5; the second line has slope 1 and y-intercept -3. The slope is the number of units of "rise" for each unit of "run", so it can be convenient to graph these by starting at the y-intercept and plotting points with those rise and run from the point you know.

Answer:

Yusuf got 19 toffees

Step-by-step explanation:

The problem tells us that if Yusuf counts his toffees in 4's, he has 3 left. Let's call the number of toffees "t", then we know that [tex]t[/tex]≡[tex]3(mod4)[/tex]

We also know that [tex]t[/tex]≡[tex]4(mod 5)[/tex]

Between 10 and 20, there are only 2 numbers t≡[tex]4(mod 5)[/tex], these numbers are 14 and 19.

However, 14≡[tex]2(mod 4)[/tex] so it doesn't add up.

On the other hand, 19 ≡3[tex](mod4)[/tex] and therefore, this is the answer.

Answer:

19 toffees.

Step-by-step explanation:

It is given that Yusuf has between 10 and 20 toffees.

Let n be the number of toffees.

If he counts his toffees in 4s, he has 3 left over.

[tex]n\equiv 3(mod4)[/tex]      ... (1)

So, the possible values of n between 10 and 20 are 11, 15 and 19.

If he counts his toffees in 5s, he has 4 left over.

[tex]n\equiv 4(mod5)[/tex]       ... (2)

So, the possible values of n between 10 and 20 are 14 and 19.

From (1) and (2) the possible value of n between 10 and 20 is 19.

Therefore, Yusuf has 19 toffees.

Answer:

[tex]z= 5-2i[/tex]

Step-by-step explanation:

Start replacing!

[tex](x+iy) + 4i (x-iy) = -3+18i \\ x+iy +4ix -4i^2 y = -3 + 18i \\\\x + i (4x+y) +4y = -3 + 18i\\x+4y + (4x+y) i = -3 + 18i[/tex]

Now two complex numbers are equal if both the real and imaginary parts are equal, which gives you the system of equation [tex]\left \{x+4y=-3} \atop {4x+y=18} \right.[/tex]

Pick any method to solve it and you'll get [tex] x=5, y= -2 [/tex]

Answer:

  r = -6

Step-by-step explanation:

We presume you want to find the value of r that satisfies the equation.

Subtract 6r+13 from both sides:

  (6r+7) -(6r+13) = (13 +7r) -(6r+13)

  -6 = r . . . . . simplify

_____

More detailed explanation

When we look at the equation we see the only variable is r, and that terms containing it appear on both sides of the equal sign. There is only one "r" term on each side, so we don't need to do any consolidation. We observe that the term with the smallest coefficient is 6r and that it is on the left side.

When we subtract 6r from the equation we will have the remaining "r" term on the right, but we will also have a constant there that we don't want. So, we can subtract that constant as well. That is why we choose to subtract 6r+13 from the equation. Doing so leaves the constants on the left and the "r" terms on the right.

As it happens, the difference between the "r" terms is plain "r", so we're done after we finish the subtraction.

__

When considering the "r" terms, we choose to subtract the term with the smallest coefficient so that the result has a positive coefficient for "r". This helps reduce mistakes in later steps, if there are any later steps.

___

Alternate "steps"

For a linear equation like this one, you can subtract one side from both sides. This might look like ...

  0 = 6+r . . . . . after subtracting 6r+7 (left side) from both sides

Then you can divide by the coefficient of r (which does nothing to this equation), and subtract the resulting constant (on the side with the variable). Here, that would give ...

  -6 = r

These three steps will work to solve any linear equation. Simplification steps may be required depending on the complexity. Again, it might be helpful, though is not essential, to subtract the side with the smallest coefficient of the variable.

___

Final note

The rules of equality say you can do anything you like to an equation, as long as you do the same thing to both sides. We can say "subtract the constant" because we are assured that you know it must be subtracted from both sides of the equation. Beware of any instruction that tells you to do something to one side of an equation and something different to the other side.

Answer:

[tex]\frac{6\cdot 8 \cdot 10 \cdot 12}{13 \cdot 15 \cdot 17\cdot 19}\approx 0.091 [/tex]

Step-by-step explanation:

To compute the probability of not including any matching pairs, we can compute the number of ways in which he could pick 8 gloves with no matching pairs, and divide it by the total number of ways in which he could pick 8 gloves.

The process of choosing 8 gloves with no matching pair can be seen as follows:

He first picks any random gloves out of the 20, in this 1st step he has then 20 available choices. Then he needs to pick some other glove, BUT it cannot be the other glove from the pair he already picked one from. So at this step there aren't 19 choices, but 18 available choices. Now he has 2 gloves from different pairs, and needs to pick another glove. There are 18 gloves left, BUT he cannot pick the remaining glove from any of the 2 pairs he already has chosen one from. Therefore he only has 16 choices left. The process continues like this, until he chooses 8 gloves in total. Hence the total number of ways to choose 8 gloves with no matching pair is

[tex] 20 \cdot 18 \cdot 16 \cdot 14 \cdot 12 \cdot 10 \cdot 8 \cdot 6[/tex]

Now, the total numer of ways in which he could pick any 8 gloves out of 20 can be seen as follows: At the start he has 20 available gloves, he chooses any of them, so having 20 available choices on that first step. He then needs to choose any other glove, so he has 19 choices. He then picks any other glove, so he has 18 choices. And so on, until he has chosen the 8 gloves. Hence the total number of ways to choose 8 gloves is

[tex] 20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13 [/tex]

Therefore the probability of choosing 8 gloves with no matching pair is

[tex]\frac{20 \cdot 18 \cdot 16 \cdot 14 \cdot 12 \cdot10 \cdot 8 \cdot 6}{20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13}=\frac{6\cdot 8 \cdot 10 \cdot 12}{13 \cdot 15 \cdot 17\cdot 19}\approx 0.091 [/tex]

Answer:

the range of the relation is 3

Answer:

0.84

Step-by-step explanation:

Given that Disks of polycarbonate plastic from a supplier are analyzed for scratch and shock resistance. The results from 100 disks are summarized as follows:

P(A) = 0.86, P(B) = 0.79, P(A') = 0.14, P(AUB) = 0.95

We are to find out P(A'UB)

We have

[tex]P(AUB) =P(A)+P(B)-P(A\bigcap B)\\0.95=0.86+0.79-P(A\bigcap B)\\P(A\bigcap B)=0.70[/tex]

[tex]P(A'UB) = P(A')+P(B)-P(A' \bigcap B)\\= 1-P(A) +P(B)-[P(B)-P(A \bigcap B)]\\= 1-0.86+0.79-P(B)+[tex]P(A'UB)=0.14+0.79-0.79+0.70\\=0.84[/tex]P(A \bigcap B)[/tex]

Answer:

1.[tex]\frac{dy}{dt}=ky[/tex]

2.543.6

Step-by-step explanation:

We are given that

y(0)=200

Let y be the number of bacteria at any time

[tex]\frac{dy}{dt}[/tex]=Number of bacteria per unit time

[tex]\frac{dy}{dt}\proportional y[/tex]

[tex]\frac{dy}{dt}=ky[/tex]

Where k=Proportionality constant

2.[tex]\frac{dy}{y}=kdt[/tex],y'(0)=100

Integrating on both sides then, we get

[tex]lny=kt+C[/tex]

We have y(0)=200

Substitute the values then , we get

[tex]ln 200=k(0)+C[/tex]

[tex]C=ln 200[/tex]

Substitute the value of C then we get

[tex]ln y=kt+ln 200[/tex]

[tex]ln y-ln200=kt[/tex]

[tex]ln\frac{y}{200}=kt[/tex]

[tex]\frac{y}{200}=e^{kt}[/tex]

[tex]y=200e^{kt}[/tex]

Differentiate w.r.t

[tex]y'=200ke^{kt}[/tex]

Substitute the given condition then, we get

[tex]100=200ke^{0}=200 \;because \;e^0=1[/tex]

[tex]k=\frac{100}{200}=\frac{1}{2}[/tex]

[tex]y=200e^{\frac{t}{2}}[/tex]

Substitute t=2

Then, we get [tex]y=200e^{\frac{2}{2}}=200e[/tex]

[tex]y=200(2.718)=543.6=543.6[/tex]

e=2.718

Hence, the number of bacteria after 2 hours=543.6

Answer:

164.6 mg

Step-by-step explanation:

Given:

Weight of the patient= 120 lbs

Height of patient = 5'8" = 5 × 12 + 8 = 68 inches

Dose of Taxol to be administered= 100 mg/ m²

Now,

the surface area of the body of patient = [tex]\textup{(Weight in kg)}^{0.425}\times\textup{(Height in cms)}^{0.725}\times0.007184[/tex]

Also,

weight of patient in kg = 120 × 0.454 = 54.48 kg

Height of patient in cm = 68" × 2.54 = 172.72 cm

therefore,

Body surface area = [tex]\textup{(54.48)}^{0.425}\times\textup{(172.72)}^{0.725}\times0.007184[/tex]

or

= 5.47 × 41.89 × 0.007184

or

= 1.646 m²

Hence,  

Dose of Taxol to be received by the patient

= 100 mg/m²  × surface area of the patient

= 100 × 1.646

= 164.6 mg

Answer:

The value of given expression is [tex]\frac{21}{128}[/tex].

Step-by-step explanation:

Given information: n=7, x=2, p=1/2

[tex]q=1-p=1-\frac{1}{2}=\frac{1}{2}[/tex]

The given expression is

[tex]C(n,x)p^xq^{n-x}[/tex]

It can be written as

[tex]^nC_xp^xq^{n-x}[/tex]

Substitute n=7, x=2, p=1/2 and q=1/2 in the above formula.

[tex]^7C_2(\frac{1}{2})^2(\frac{1}{2})^{7-2}[/tex]

[tex]\frac{7!}{2!(7-2)!}(\frac{1}{2})^2(\frac{1}{2})^{5}[/tex]

[tex]\frac{7!}{2!5!}(\frac{1}{2})^{2+5}[/tex]

[tex]\frac{7\times 6\times 5!}{2\times 5!}(\frac{1}{2})^{2+5}[/tex]

[tex]21(\frac{1}{2})^{7}[/tex]

[tex]\frac{21}{128}[/tex]

Therefore the value of given expression is [tex]\frac{21}{128}[/tex].

The number of letters in word "ALGORITHM" = 9

The number of combinations to select r things from n things is given by :-

[tex]C(n,r)=\dfrac{n!}{r!(n-r)!}[/tex]

Now, the number of combinations to select 6 letters from 9 letters will be :-

[tex]C(9,8)=\dfrac{9!}{6!(9-6)!}=\dfrac{9\times8\times7\times6!}{6!\times3!}=84[/tex]

Thus , the number of ways can six of the letters of the word ALGORITHM=84

The number of ways to arrange n things in a row :[tex]n![/tex]

So, the number of ways can the letters of the word ALGORITHM be arranged in a be seated together in the row :-

[tex]9!=362880[/tex]

If GOR comes together, then we consider it as one letter,  then the total number of letters will be = 1+6=7

Number of ways to arrange 9 letters if "GOR" comes together :-

[tex]7!=5040[/tex]

Thus, the number of ways to arrange 9 letters if "GOR" comes together=5040

Answer:

Minimize C =[tex]13x + 3y[/tex]

[tex]12x + 14y \geq  21[/tex]

[tex]15x + 20y \geq 37[/tex]

and x ≥ 0, y ≥ 0.

Plot the the lines on the graph and find the feasible region

[tex]12x + 14y \geq  21[/tex]  -- Blue

[tex]15x + 20y \geq  37[/tex] --- Green

So, the boundary points of feasible region are (-3.267,4.3) , (0,1.85) and (2.467,0)

Substitute the value in Minimize C

Minimize C =[tex]13x + 3y[/tex]

At (-3.267,4.3)

Minimize C =[tex]13(-3.267) + 3(4.3)[/tex]

Minimize C =[tex]-29.571[/tex]

At (0,1.85)

Minimize C =[tex]13(0) + 3(1.85)[/tex]

Minimize C =[tex]5.55[/tex]

At (2.467,0)

Minimize C =[tex]13(2.467) + 3(0)[/tex]

Minimize C =[tex]32.071[/tex]

So, the optimal value of x is -3.267

So, the optimal value of y is 4.3

So, the minimum value of the objective function is -29.571

Answer:

D = 170.6Km

Step-by-step explanation:

First of all, we set the reference (origin) at Grand Bahama.

Nw, from the first displacement of 3h we calculate the distance:

D1 = V1*t = 43.5 * 3 = 130.5 Km

The coordinates of this new location is given by:

r1 = ( -D1*cos(60°); D1*sin(60°)) = (-62.5; 158.835) Km

For the second displacement, the duration was of 1.95 hours (4.95 -3), so the distance traveled was:

D2 = V2*t = 23.5 * 1.95 = 45.825 Km

The coordinates of this new location is given by:

r2 = r1 + ( 0; D2) = (-62.25; 158.835) Km

Now we just need to calculate the magnitude of that vector to know the distance to Grand Bahama:

[tex]Dt = \sqrt{D_{2x}^{2}+D_{2y}^{2}}=170.6 Km[/tex]

Here are some possible answers

Step-by-step explanation:

Consider the provided information.

For the condition statement [tex]p \rightarrow q[/tex] or equivalent "If p then q"

Part (A) If it snows tonight, then I will stay at home.

Here p is If it snows tonight, and q is I will stay at home.

Converse: If I will stay at home then it snows tonight.

[tex]q \rightarrow p[/tex]

Inverse: If it doesn't snows tonight, then I will not stay at home.

[tex]\sim p \rightarrow \sim q[/tex]

Contrapositive: If I will not stay at home then it doesn't snows tonight.

[tex]\sim q \rightarrow \sim p[/tex]

Part (B) I go to the beach whenever it is a sunny summer day.

Here p is I go to the beach, and q is it is a sunny summer day.

Converse: It is a sunny summer day whenever I go to the beach.

[tex]q \rightarrow p[/tex]

Inverse: I don't go to the beach whenever it is not a sunny summer day.

[tex]\sim p \rightarrow \sim q[/tex]

Contrapositive: It is not a sunny summer day whenever I don't go to the beach.

[tex]\sim q \rightarrow \sim p[/tex]

Part (C) When I stay up late, it is necessary that I sleep until noon.

P is I sleep until noon and q is I stay up late.

Converse: If I sleep until noon, then it is necessary that i stay up late.

[tex]q \rightarrow p[/tex]

Inverse: When I don't stay up late, it is necessary that I don't sleep until noon.

[tex]\sim p \rightarrow \sim q[/tex]

Contrapositive: If I don't sleep until noon, then it is not necessary that i stay up late.

[tex]\sim q \rightarrow \sim p[/tex]

The converse, contrapositive, and inverse of each of the specified conditional statements are shown below.

Suppose the conditional statement given is:

[tex]p \rightarrow q[/tex] or 'if p then q'

Then, we get:

For the listed conditional statements, finding their converse, contrapositive, and inverse statements:

This can be taken as: If it is a sunny summar day, then i go to the beach.

Learn more about converse, contrapositive, and inverse statements here:

https://brainly.com/question/13245751?referrer=searchResults

Answer:

The solution of this system is x=9/4, y=5/2, and z=-13/8

Step-by-step explanation:

1. Writing the equations in matrix form

The system of linear equations given can be written in matrix form as

[tex]\left[\begin{array}{ccc}1&0&2\\2&-1&0\\0&3&4\end{array}\right]\left[\begin{array}{c}x&y&z\end{array}\right] = \left[\begin{array}{c}-1&2&1\end{array}\right][/tex]

Writing

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

X = [tex]\left[\begin{array}{c}x&y&z\end{array}\right][/tex]

B = [tex]\left[\begin{array}{c}-1&2&1\end{array}\right][/tex]

we have

AX=B

This is the matrix form of the simultaneous equations.

2. Solving the simultaneous equations

Given

AX=B

we can multiply both sides by the inverse of A

[tex]A^{-1}AX=A^{-1}B[/tex]

We know that [tex]A^{-1}A=I[/tex], the identity matrix, so

[tex]X=A^{-1}B[/tex]

All we need to do is calculate the inverse of the matrix of coefficients, and finally perform matrix multiplication.

3. Calculate the inverse of the matrix of coefficients

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

To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

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

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

[tex]\left[\begin{array}{ccc|ccc}1&0&2&1&0&0\\0&-1&-4&-2&1&0\\0&0&-8&-6&3&1\end{array}\right][/tex]

[tex]\left[\begin{array}{ccc|ccc}1&0&2&1&0&0\\0&1&4&2&-1&0\\0&0&-8&-6&3&1\end{array}\right][/tex]

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

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

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

As can be seen, we have obtained the identity matrix to the left. So, we are done.

[tex]A^{-1} = \left[\begin{array}{ccc}-1/2&3/4&1/4\\-1&1/2&1/2\\3/4&-3/8&-1/8\end{array}\right][/tex]

4. Find the solution [tex]X=A^{-1}B[/tex]

[tex]X= \left[\begin{array}{ccc}-1/2&3/4&1/4\\-1&1/2&1/2\\3/4&-3/8&-1/8\end{array}\right]\cdot \left[\begin{array}{c}-1&2&1\end{array}\right] = \left[\begin{array}{c}9/4&5/2&-13/8\end{array}\right][/tex]

Answer:

  The scale factor is 2/3.

Step-by-step explanation:

When the dilation is about the origin, the ratio of image coordinates to original coordinates is the scale factor:

  4/5 = 6/9 = 2/3

The scale factor is 2/3.