3) Draw a possibility tree that represents a coin that is tossed 3 times

We have to show that

[tex]\frac{\partial ^{2}u}{\partial t^{2}}=a^{2}\frac{\partial ^{2}u}{\partial x^{2}}[/tex]

for [tex]\frac{\partial ^{2}u}{\partial t^{2}}[/tex] we have

[tex]\frac{\partial ^{2}u}{\partial t^{2}}=a^{2}\frac{\partial ^{2}u}{\partial x^{2}}[/tex]

[tex]\frac{\partial ^{2}u}{\partial t^{2}}=\frac{\partial ^{2}[f(x-at)+g(x+at)]}{\partial t^{2}}[/tex]

[tex]=\frac{\partial }{\partial t}[\frac{\partial[f(x-at)+g(x+at)] }{\partial t}][/tex]

[tex]\frac{\partial }{\partial t}[-a\cdot f'(x-at)+a\cdot g'(x+at)][/tex]

[tex]=a^{2}f''(x-at)+a^{2}g''(x+at)[/tex]

[tex]=a^{2}[f''(x-at)+g''(x+at)].............(i)[/tex]

similarly,

[tex]\frac{\partial ^{2}u}{\partial x^{2}}=\frac{\partial ^{2}[f(x-at)+g(x+at)]}{\partial x^{2}}[/tex]

[tex]=\frac{\partial }{\partial x}[\frac{\partial[f(x-at)+g(x+at)] }{\partial x}][/tex]

[tex]=\frac{\partial }{\partial x}[f'(x-at)+g'(x+at)][/tex]

[tex]=f''(x-at)+g''(x+at).......(ii)[/tex]  

Comparing i and ii we get  

[tex]a^{2}\frac{\partial ^{2}u}{\partial x^{2}}=\frac{\partial ^{2}u}{\partial t^{2}}[/tex]

Hence proved

Answer:

Odd numbers that are multiple of 5 and are in between 22 and 66 are-

25, 35, 45, 55, 65

Let this set be represented by A

A= {25, 35, 45, 55, 65}

the above form represents the set in its roster form

The set notation for the odd numbers between 22 and 66 that are multiples of 5 is { x ∈ N | x is odd, 22 < x < 66, x ≡ 0 (mod 5) }.

The set notation for the odd numbers between 22 and 66 that are multiples of 5 can be written as:

{ x ∈ N | x is odd, 22 < x < 66, x ≡ 0 (mod 5) }

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Answer:

if x= -1 then its is NOT in the domain of h.

Step-by-step explanation:

Domain is the set of values for which the function is defined.

we are given the function

[tex]h(x) =\frac{x+1}{x^2 + 2x + 1}[/tex]

Solving the denominator by factorization

[tex]h(x) =\frac{x+1}{x^2 + x+x + 1}\\h(x) =\frac{x+1}{x(x+1)+1(x + 1)}\\h(x) =\frac{x+1}{(x+1)(x+1)}\\h(x) =\frac{1}{(x+1)}[/tex]

So, if x = -1 then its is NOT in the domain of h.

Answer:

Yes

Step-by-step explanation:

We are given that a function [tex]\phi(x)=x^2-x^{-1}[/tex]

We have to find that given function is an explicit solution to the linear equation

[tex]\frac{d^2y}{dx^2}-\frac{2}{x^2}y=0[/tex]

If given function is an explicit solution of given linear equation then it satisfied the given differential equation

Differentiate w.r.t x

Then we get [tex]\phi'(x)=2x+x^{-2}[/tex]

Again differentiate w.r.t x

Then we get

[tex]\phi''(x)=2-\frac{2}{x^3}[/tex]

Substitute all values in the given differential equation

[tex]2-\frac{2}{x^3}-\frac{2}{x^2}(x^2-x^{-1})[/tex]

=[tex]2-\frac{2}{x^3}-2+\frac{2}{x^3}=0[/tex]

Hence, given function is an explicit solution of given differential equation.

Therefore, answer is yes.

Answer:

$5.80 Option B.

Step-by-step explanation:

It is given that a 20% tip on a meal that costs $29.17.

The cost of the meal = $29.17

Tip on a meal = 20%

Therefore, 20% of $29.17

= [tex]\frac{20}{100}[/tex] × 29.17

= 0.20 × 29.17

= 5.834

= $5.80

The correct estimate would be $5.80 Option B.

Answer:

CF+PI=[tex]c_1e^{2x}+c_2e^{x}[/tex]+[tex]2e^{3t}[/tex]

Step-by-step explanation:

we have given y"-3y'=2y=[tex]4e^{3t}[/tex]

this differential equation solution have two part that CF and PI

CALCULATION OF CF :

[tex]m^2-3m+2=0[/tex]

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

[tex](m-1)(m-2)=0[/tex]

m=1 and m=2

so CF=[tex]c_1e^{2x}+c_2e^{x}[/tex]

CALCULATION OF PI :

PI =   [tex]\frac{4e^{3t}}{(m-1)(m-2)}[/tex]

at m= 3 in PI

[tex]PI=\frac{4e^{3t}}{2}=2e^{3t}[/tex]

so the complete solution is

CF+PI=[tex]c_1e^{2x}+c_2e^{x}[/tex]+[tex]2e^{3t}[/tex]

Answer:

The probability is 0.2356.

Step-by-step explanation:

Let X is the event of using the smartphone in meetings or classes,

Given,

The probability of using the smartphone in meetings or classes, p = 51 % = 0.51,

So, the probability of not using smartphone in meetings or classes, q = 1 - p = 1 - 0.51 = 0.49,

Thus, the probability that fewer than 5 of them use their smartphones in meetings or classes.

P(X<5) = P(X=0) + P(X=1) + P(X=2) + P(X=3)+P(X=4)

Since, binomial distribution formula is,

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

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

Here, n = 11,

Hence,  the probability that fewer than 5 of them use their smartphones in meetings or classes

[tex]=^{11}C_0 (0.5)^0 0.49^{11}+^{11}C_1 (0.5)^1 0.49^{10}+^{11}C_2 (0.5)^2 0.49^{9}+^{11}C_3 (0.5)^3 0.49^{8}+^{11}C_4 (0.5)^4 0.49^{7} [/tex]

[tex]=(0.5)^0 0.49^{11}+11(0.5)0.49^{10} + 55(0.5)^2 0.49^{9}+165 (0.5)^3 0.49^{8} +330(0.5)^4 0.49^{7} [/tex]

[tex]=0.235596671797[/tex]

[tex]\approx 0.2356[/tex]

Answer:

  a) at 16%: $10,936.47

  b) at 8%: $124,177.02

Step-by-step explanation:

At annual rate of return "r", the multiplier of Sarah's initial investment will be ...

  k = (1+r)^34

For r = 0.16, k ≈ 155.433166, and Sarah's investment needs to be ...

  $1.7·10^6/k ≈ $10,936.47

__

For r= 0.08, k ≈ 13.6901336, and Sarah's investment needs to be ...

  $1.7·10^6/k ≈ $124,177.02

Answer:

The only bijection is f(x)=2x+1.

I took r to be a constant.

Step-by-step explanation:

Bijections are both onto and one-to-one.

Onto means every element of the codomain gets hit.  Here the codomain is the set of real numbers.  So you want every y to get hit.

One-to-one means you don't want any y to get hit more than once.

f(x)=2x+1 is a linear function.  It is diagonal line so every element of the codomain will get hit and hit only once so this function is onto and one-to-one.

f(x)=x^2+1 is a quadratic function.  It is parabola so not every element of our codomain will get not get hit and of those that do get hit they get hit more than once.  So this is neither onto or one-to-one.

f(x)=r^3 is a constant function.  It is a horizontal line so not every y will get hit so it isn't onto.  The same y is being hit multiple times so it isn't one-to-one.

f(x)=(x^2+1)/(r^2+2) is a quadratic. It is a parabola. Quadratic functions are not onto or one-to-one.

A horizontal line is a line where all of the [tex]y[/tex] values are the same. In this case, [tex]\boxed{y=5}[/tex], so that is the equation.

A vertical line is where all of the [tex]x[/tex] values are the same.  Here, [tex]\boxed{x=7}[/tex], so that's the equation.

Answer:

see below

Step-by-step explanation:

A horizontal line has the same y value  and has a constant y value

y=5

A vertical line has the same x value  and has a constant x value

x=7

Answer: 0.3439

Step-by-step explanation:

Total number of digits : 10

The number of digits except zero =9

The probability of selecting non-zero number :-

[tex]\dfrac{9}{10}=0.9[/tex]

Then, the probability that one such phone number , the last four digits do not include any zero:-

[tex]\text{P(no zero )}=(0.9)^4[/tex]

Then , the probability that for one such phone​ number, the last four digits include at least one 0:-

[tex]\text{P(at-least one zero )}=1-(0.9)^4=0.3439[/tex]

Hence, the probability that for one such phone​ number, the last four digits include at least one 0 is 0.3439 .

To find the probability of at least one 0 in a four-digit number, one can use the complement of the probability of no 0s occurring in any of the digits. The probability of at least one 0 is approximately 34.39%.

The student is inquiring about the probability of a certain event, specifically related to randomly generating telephone numbers. To find the probability that at least one digit is a 0 in a randomly generated four-digit number, we can use the complementary probability approach. The probability of not getting a 0 in one digit is 9/10 since there are 9 other possible digits (1-9). Hence, the probability of not getting a 0 in any of the four digits is (9/10)^4. Subtracting this from 1 will give the probability of having at least one 0 in the four-digit sequence:

Probability of at least one 0 = 1 - (Probability of no 0 in any digit)

Probability of at least one 0 = 1 - (9/10)^4

After calculating, we have:

Probability of at least one 0 = 1 - 0.6561 = 0.3439

Therefore, the probability of having at least one 0 in a randomly chosen telephone number with four digits is approximately 34.39%.

Answer:

Step-by-step explanation:

The Inverse of the matrix doesn't exist because the determinant is equal to 0.

Answer:

The inverse of given matrix is not exist, since determinant is 0.

Step-by-step explanation:

The inverse of a square matrix [tex]A[/tex] is [tex]A^{-1}[/tex] such that

[tex]A A^{-1}=I[/tex] where I is the identity matrix.

Consider, [tex]A = \left[\begin{array}{ccc}3&3&-1\\-12&-12&4\\2&6&0\end{array}\right][/tex]

[tex]\mathrm{Matrix\:can\:only\:be\:inverted\:if\:it\:is\:non-singular,\:that\:is:}[/tex]

[tex]\det \begin{pmatrix}3&3&-1\\ -12&-12&4\\ 2&6&0\end{pmatrix}\ne 0[/tex]

[tex]\det \begin{pmatrix}3&3&-1\\ -12&-12&4\\ 2&6&0\end{pmatrix}[/tex]

[tex]\mathrm{Find\:the\:matrix\:determinant\:according\:to\:formula}:\quad \:[/tex]

[tex]\det \begin{pmatrix}a&b&c\\ d&e&f\\ g&h&i\end{pmatrix}=a\cdot \det \begin{pmatrix}e&f\\ h&i\end{pmatrix}-b\cdot \det \begin{pmatrix}d&f\\ g&i\end{pmatrix}+c\cdot \det \begin{pmatrix}d&e\\ g&h\end{pmatrix}[/tex]

[tex]=3\cdot \det \begin{pmatrix}-12&4\\ 6&0\end{pmatrix}-3\cdot \det \begin{pmatrix}-12&4\\ 2&0\end{pmatrix}-1\cdot \det \begin{pmatrix}-12&-12\\ 2&6\end{pmatrix}[/tex]

[tex]=3\left(-24\right)-3\left(-8\right)-1\cdot \left(-48\right)[/tex]

[tex]3\left(-24\right)-3\left(-8\right)-1\cdot \left(-48\right)=0[/tex]

Therefore, the inverse of given matrix is not exist, since determinant is 0.

Answer:

It's actually C

Step-by-step explanation:

don't forget about the probability of 0 too

you have to add the two probability formulas

The probability of observing, at most, one patient who will be cured of breast Cancer will be P = (20 / q²⁰) + 20 / (0.20 x 0.80¹⁹). Then the correct option is C.

Its basic premise is that something will almost certainly happen. The percentage of favorable events to the total number of occurrences.

A new drug on the market is known to cure 20% of patients with breast cancer.

p = 0.20

q = 1 – 0.20

q = 0.80

If a group of 20 patients is randomly selected.

The probability of observing, at most, one patient who will be cured of breast Cancer will be

P = (20 / q²⁰) + 20 / (0.20 x 0.80¹⁹)

Then the correct option is C.

More about the probability link is given below.

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Answer:

[tex]a_{n}=\frac{1}{2n}[/tex] [Where a ≥ 1 ]

Step-by-step explanation:

The pattern of the given sequence is {[tex]\frac{1}{2},\frac{1}{4},\frac{1}{6},\frac{1}{8},\frac{1}{10},......[/tex]

We have to find a formula for the general term [tex]a_{n}[/tex] of the given sequence.

We can rewrite the terms of the sequence as

[tex]\frac{1}{2}=\frac{1}{(2)(1)}[/tex]

[tex]\frac{1}{4}=\frac{1}{(2)(2)}[/tex]

[tex]\frac{1}{6}=\frac{1}{(2)(3)}[/tex]

[tex]\frac{1}{8}=\frac{1}{(2)(4)}[/tex]

[tex]\frac{1}{10}=\frac{1}{(2)(5)}[/tex]

Now we can write the term [tex]a_{n}[/tex] as

[tex]a_{n}=\frac{1}{2n}[/tex]

Where n = 1, 2, 3, 4, 5......

Answer:

233 days.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 200, \sigma = 10[/tex]

To be 99% sure that we will not be late in completing the project, we should request a completion time of ...

This is the value of X when Z has a pvalue of 0.99. So X when Z = 2.325.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]2.325 = \frac{X - 200}{10}[/tex]

[tex]X - 200 = 10*2.325[/tex]

[tex]X = 232.5[/tex]

So the correct answer is 233 days.

[tex]1.\rightarrow \frac{dy}{dx}-y=e^x y^2\\\\\rightarrow \frac{1}{y^2}\frac{dy}{dx}-\frac{1}{y}=e^x\\\\ \text{put},\frac{-1}{y}=z\\\\ \frac{dy}{y^2} =d z\\\\ \frac{dy}{dx} \times \frac{1}{y^2}=\frac{dz}{dx}\\\\\frac{dz}{dx} +z=e^x\\\\ \text{Integrating factor}=e^{\int {1} \, dx}\\\\=e^x \\\\ \text{Multiplying both sides by }e^x\\\\e^x(\frac{dz}{dx} +z)=e^{2x}\\\\ \text{Integrating both sides}\\\\z e^x=\frac{e^{2x}}{2}+C\\\\ \frac{-e^x}{y}=\frac{e^{2x}}{2}+C[/tex]

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[tex]\rightarrow \frac{dy}{dx}=x+\frac{y}{x}-y\\\\\rightarrow \frac{dy}{dx}-x=\frac{y}{x}-y\\\\\rightarrow \frac{dy}{dx}+y(1-\frac{1}{x})=x\\\\\text{Integrating factor}=e^{\int{1-\frac{1}{x}}\,dx}\\\\=e^{x-\log x}\\\\ \text{Multiplying both sides by} e^{x-\log x}\\\\e^{x-\log x}\times[\frac{dy}{dx}+y(1-\frac{1}{x})]=x \times e^{x-\log x}\\\\y\times e^{x-\log x} =\int x \times e^{x-\log x} \, dx}\\\\y\times e^{x-\log x}=\int x \times \frac{e^x}{e^{\log x}}\,dx\\\\y\times e^{x-\log x}=\int x \times \frac{e^x}{x} \, dx\\\\y\times e^{x-\log x}=e^x+K[/tex]

[tex]HTT,HTH,HHH,TTT,THT,THH,TTH[/tex]

The magnetic field H(z,t) of an electromagnetic wave is related to the electric field E(z,t) by a factor of the speed of light. Therefore, if E(z,t) = 4cos(6 x 10^8 t - 2z) +3 sin(6 x 10^8 t -2z), the associated magnetic field would be H(z,t) = (4/c) cos(6 x 10^8 t - 2z) +(3/c) sin(6 x 10^8 t -2z), where c is speed of light, approximately 3 x 10^8 m/s.

The question is asking for the associated magnetic field H(z,t) of an Electromagnetic wave given the electric field E(z,t). A crucial fact to know for this question is that the electric and magnetic fields in an electromagnetic wave are perpendicular to each other and the direction of propagation. They also have a constant ratio of magnitudes in free space or air, which is the speed of light given by c = 1/√εOMO. Because of these relations, we know that we can find the magnetic field by simply dividing the given electric field by the speed of light in units that match the given Electric field.

So, if E(z,t) = 4cos(6 x 10^8 t - 2z) +3 sin(6 x 10^8 t -2z), then the associated magnetic field would be H(z,t) = (4/c) cos(6 x 10^8 t - 2z) +(3/c) sin(6 x 10^8 t -2z), where c is the speed of light, approximately 3 x 10^8 m/s.

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To find the associated magnetic field H(z, t), you can use Faraday's law of electromagnetic induction. This law states that the rate of change of magnetic flux through a surface is equal to the induced electromotive force (EMF) along the boundary of the surface. By following a step-by-step process, you can find the magnetic field B(z, t) using the given electric field E(z, t).

The associated magnetic field H(z, t) can be found by using Faraday's law of electromagnetic induction. Faraday's law states that the rate of change of magnetic flux through a surface is equal to the electromotive force (EMF) along the boundary of the surface. In this case, the magnetic field is changing due to the time-dependent electric field, so we can use Faraday's law to find the magnetic field.

By following these steps, you can find the associated magnetic field H(z, t) using Faraday's law of electromagnetic induction.

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Answer:

y = -3/5x - 12/5

Step-by-step explanation:

The equation I'm going to give is going to be in slope-intercept form. If you need it in point-slope, I can do so in an edit or the comments.

Slope-intercept form is: y = mx + b where m is the slope, b is the y-intercept.

So let's plug in our given slope:

y = -3/5x + b

Using this, we now plug in our x- and y-coordinates from the given point to solve for b.

-3 = -3/5(1) + b

-3 = -3/5 + b

Add 3/5 to both sides to isolate variable b.

-3 + 3/5 = b

-15/5 + 3/5 = b

-12/5 = b

Plug this new info back into the original equation and your answer is

y = -3/5x - 12/5

Answer: [tex]H_0:\mu\geq33[/tex]

[tex]H_a:\mu<33[/tex]

Step-by-step explanation:

Let [tex]\mu[/tex] be the average number of hours a person drive without adding the product.

Given claim : An infomercial claims that a woman drove 33 hours without​ oil.

i.e. [tex]\mu\geq33[/tex]

It is known that the null hypothesis always contains equal sign and alternative hypothesis is just opposite of the null hypothesis.

Thus the null and alternative hypothesis for the given situation will be :-

[tex]H_0:\mu\geq33[/tex]

[tex]H_a:\mu<33[/tex]

Answer:

7/32

Step-by-step explanation:

Answer:

a > 12

Step-by-step explanation:

Isolate the variable a. Treat the > sign like the = sign. What you do to one side, you do to the other. Add 1 to both sides:

a - 1 (+1) > 11 (+1)

a > 11 + 1

a > 12

a > 12 is your answer.

~

Answer:

[tex]\huge \boxed{a>12}[/tex]

Step-by-step explanation:

Add by 1 from both sides.

[tex]\displaystyle a-1+1>11+1[/tex]

Simplify, to find the answer.

[tex]\displaystyle 11+1=12[/tex]

[tex]\displaystyle a>12[/tex], which is our answer.

Answer:

The cutoff score should be 75.8 marks

Step-by-step explanation:

The cut-off should be set as the value corresponding to an area 15% in the normal distribution diagram.

Using the standard distribution tables we have value of standard normal deviate (Z) corresponding to area of 15% = -1.04

Thus we have

[tex]-1.04=\frac{X-\bar{X}}{\sigma }\\\\X=-1.04\times 5.5+81\\X=75.8[/tex]

Answer:

The future value of the annuity due to the nearest cent is $2956.

Step-by-step explanation:

Consider the provided information:

It is provided that monthly payment is $175, interest is 7% and time is 11 years.

The formula for the future value of the annuity due is:

[tex]FV of Annuity Due = (1+r)\times P[\frac{(1+r)^{n}-1}{r}][/tex]

Now, substitute P = 175, r = 0.07 and t = 11 in above formula.

[tex]FV of Annuity Due = (1+0.07)\times 175[\frac{(1+0.07)^{11}-1}{0.07}][/tex]

[tex]FV of Annuity Due = (1.07)\times 175[\frac{1.10485}{0.07}][/tex]

[tex]FV of Annuity Due = 187.5(15.7835)[/tex]

[tex]FV of Annuity Due = 2955.4789[/tex]

Hence, the future value of the annuity due to the nearest cent is $2956.

Answer:

72 months approx.

Step-by-step explanation:

Monthly deposit = m = $140

r = 3.8% or 0.038

Amount needed in the account = A = $11300

The formula will be :

[tex]11300=140(\frac{(1+0.038/12)-1}{0.038/12} )[/tex]

[tex]11300=140(\frac{(1+0.038/12)-1}{0.003166})[/tex]

[tex]11300=44219.83((1.003166)^{m}-1)[/tex]

[tex]0.2555=(1.003166)^{m}-1[/tex]

[tex]1.2555=(1.003166)^{m}[/tex]

m=log1.2555/log1.003166

m =71.98 ≈ 72

Hence, it will take 72 months approx.

Answer:

flow rate of to infuse the 1000 mL over 5 hours is 33.33 gtt/min

Step-by-step explanation:

Given data

volume = 1000 mL

time =  5 hours

drop factor = 10 gtt/min

to find out

flow rate of to infuse the 1000 mL over 5 hours

Solution

we know flow rate formula  i.e.

flow rate =  ( volume × drop factor )  / time   .................1

here time will be in min so time =  5 hours = 5  × 60 = 300 min

put volume, drop factor and time value in equation 1 and we get flow rate

flow rate =  ( volume × drop factor )  / time

flow rate =  ( 1000 × 10 )  / 300

flow rate = 33.33 gtt/min

flow rate of to infuse the 1000 mL over 5 hours is 33.33 gtt/min

Divide both sides by [tex]x^2-1[/tex] to get a linear ODE,

[tex]\dfrac{\mathrm dy}{\mathrm dx}+\dfrac2{x^2-1}y=\dfrac{x+1}{x-1}[/tex]

In order for this operation to be valid in the first place, we require that [tex]x\neq\pm1[/tex] (since that would make [tex]\dfrac1{x^2-1}[/tex] undefined, which we don't want to happen). Then we are forcing any solution to the ODE to exist on any of the three intervals, [tex](-\infty,-1)[/tex], [tex](-1, 1)[/tex], or [tex](1,\infty)[/tex], and either the first or third of these can be chosen as the largest interval.

In case you also need to solve the ODE: Multiply both sides by [tex]\dfrac{1-x}{1+x}[/tex], so that

[tex]\dfrac{1-x}{1+x}\dfrac{\mathrm dy}{\mathrm dx}-\dfrac2{(1+x)^2}y=-1[/tex]

Then the left side can be condensed as the derivative of a product, since

[tex]\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac{1-x}{1+x}\right]=-\dfrac2{(1+x)^2}[/tex]

and we have

[tex]\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac{1-x}{1+x}y\right]=-1[/tex]

Integrate both sides:

[tex]\displaystyle\int\frac{\mathrm d}{\mathrm dx}\left[\frac{1-x}{1+x}y\right]\,\mathrm dx=-\int\mathrm dx[/tex]

[tex]\dfrac{1-x}{1+x}y=-x+C[/tex]

[tex]\implies\boxed{y=\dfrac{(-x+C)(1+x)}{1-x}}[/tex]

The largest interval over which the general solution is defined for the given differential equation is [-1, ∞).

Here's how:

To find x, find the difference between the outer angle and middle angle, then divide by two.

X = (210 - 72) / 2

x = 138 / 2

x = 69

The answer is D.

Answer

subtract 210 with 72 and then divide that by 2 and you get 138. so x=138.

Answer:

B. 200 doses

Step-by-step explanation:

Given,

1 dose is required for 100 mg,

Since, 1 mg = 0.001 g,

⇒ 100 mg = 0.1 g

⇒ 1 dost is required for 0.1 g,

Thus, the ratio of doses and quantity ( in gram ) is [tex]\frac{1}{0.1}=10[/tex]

Let x be the doses required for 20 grams,

So, the ratio of doses and quantity is [tex]\frac{x}{20}[/tex]

[tex]\implies \frac{x}{20}=10[/tex]

[tex]\implies x=200[/tex]

Hence, 200 doses can be obtained from 20 grams of the drug.

Option 'B' is correct.

To solve this question, we will follow these steps:
1. We need to ensure that we use the same units for both the drug amount and the dose. Since the drug amount is given in grams and the dose in milligrams, we will convert the drug amount from grams to milligrams.
2. We know that 1 gram is equivalent to 1000 milligrams.
3. Now, let's convert the drug amount from grams to milligrams:
  We have 20 grams of the drug, so the conversion to milligrams is:
  \(20 \text{ grams} \times \dfrac{1000 \text{ milligrams}}{1 \text{ gram}} = 20,000 \text{ milligrams}\)
4. Next, we will divide the total milligrams of the drug by the milligram dosage that is recommended per dose to find out how many doses we can get from the drug amount.
5. Given that each dose is 100 mg, we now divide the total drug amount in milligrams by the dose in milligrams:
  \(20,000 \text{ milligrams} \div 100 \text{ milligrams per dose} = 200 \text{ doses}\)
Therefore, from 20 grams of the drug, we can obtain 200 doses.
The correct answer is:
  B. 200 doses

Answer:

12 minutes

Step-by-step explanation:

First we figure out how much of a car each person can wash in 1 minute.

Shea can wash the car by herself in 20 minutes

Therefore, she can wash a car in 1 minute = [tex]\frac{1}{20}[/tex]

Tucker can wash a car in 30 minutes.

Tucker can wash the car in 1 minute = [tex]\frac{1}{30}[/tex]

Thus, in one minute together they wash a car

[tex]\frac{1}{20}[/tex] + [tex]\frac{1}{30}[/tex] = [tex]\frac{50}{600}[/tex] = [tex]\frac{1}{12}[/tex]

In one minute together they can wash  = [tex]\frac{1}{12}[/tex] of a car

Time needed to wash entire car together = 12 minutes.

It takes them 12 minutes to wash the car.

Final answer:

Shea and Tucker can wash their father's car in 12 minutes if they work together.

Explanation:

To find out how long it takes Shea and Tucker to wash the car together, we can use the concept of work rates.

Shea can wash the car by herself in 20 minutes, so her work rate is 1/20 of the car per minute.

Tucker can wash the car by himself in 30 minutes, so his work rate is 1/30 of the car per minute.

When two people work together, their work rates are additive. So, if Shea and Tucker work together, their combined work rate is 1/20 + 1/30 = 1/12 of the car per minute.

Since the work rate is the reciprocal of the time taken, the time it takes them to wash the car together is 12 minutes. Therefore, it takes Shea and Tucker 12 minutes to wash their father's car if they work together.