If you are given the graph of g(x)=log2x, how could you graph f(x)=log2x+5

Answer:

  54

Step-by-step explanation:

x is half the difference of the two arcs:

  x = (136 -28)/2 = 54

The value of x is 54.

Answer:

Let the number be x

The statement

A number is increased by five is written as

x + 5

Then it's squared

So we the final answer as

Hope this helps

Answer:

x = {9, -3}

Step-by-step explanation:

Or it can be shown as:

Answer:8

Step-by-step explanation:

The smallest prime is 2

cube of 2 is equal to 8

2*2*2=8

Answer:

729

Step-by-step explanation:

The second smallest prime number is 3 (preceded by 2). We have (3^2)^3=3^6=729.

Hope this helped! :)

Answer:

Step-by-step explanation:

The formula of a slope:

[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]

We have the points

[tex](3;\ 9)\to x_1=3;\ y_1=9\\(1;\ 1)\to x_2=1;\ y_2=1[/tex]

Substitute:

[tex]m=\dfrac{1-9}{1-3}=\dfrac{-8}{-2}=4[/tex]

Answer:

m=4

Step-by-step explanation:

Slope can be found using the following formula:

[tex]m=\frac{y_{2} -y_{1} }{x_{2} -x_{1} }[/tex]

where [tex](x_{1},y_{1})[/tex] and [tex](x_{2},y_{2})[/tex] are points on the line.

We are given the points (3,9) and (1,1). Therefore,

[tex]x_{1}=3\\y_{1}=9 \\x_{2}=1\\y_{2}=1[/tex]

Substitute each value into the formula.

[tex]m=\frac{1-9}{1-3}[/tex]

Subtract in the numerator first.

[tex]m=\frac{-8}{1-3}[/tex]

Subtract in the denominator.

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

Divide.

[tex]m=4[/tex]

The slope of the line is 4.

Answer:

( 6, pi/6)

Step-by-step explanation:

( 3 sqrt(3), 3)

To get r we use x^2 + y ^2 = r^2

( 3 sqrt(3) )^2 + 3^2 = r^2

9 *3 +9 = r^2

27+9 = r^2

36 = r^2

Taking the square root of each side

sqrt(36) = sqrt(r^2)

6 =r

Now we need to find theta

tan theta = y/x

tan theta = 3 / 3 sqrt(3)

tan theta = 1/ sqrt(3)

Taking the inverse tan of each side

tan ^-1 ( tan theta) = tan ^ -1 ( 1/ sqrt(3))

theta = pi /6

You want to end up with [tex]A\sin(\omega t+\phi)[/tex]. Expand this using the angle sum identity for sine:

[tex]A\sin(\omega t+\phi)=A\sin(\omega t)\cos\phi+A\cos(\omega t)\sin\phi[/tex]

We want this to line up with [tex]2\sin(4\pi t)+5\cos(4\pi t)[/tex]. Right away, we know [tex]\omega=4\pi[/tex].

We also need to have

[tex]\begin{cases}A\cos\phi=2\\A\sin\phi=5\end{cases}[/tex]

Recall that [tex]\sin^2x+\cos^2x=1[/tex] for all [tex]x[/tex]; this means

[tex](A\cos\phi)^2+(A\sin\phi)^2=2^2+5^2\implies A^2=29\implies A=\sqrt{29}[/tex]

Then

[tex]\begin{cases}\cos\phi=\frac2{\sqrt{29}}\\\sin\phi=\frac5{\sqrt{29}}\end{cases}\implies\tan\phi=\dfrac{\sin\phi}{\cos\phi}=\dfrac52\implies\phi=\tan^{-1}\left(\dfrac52\right)[/tex]

So we end up with

[tex]2\sin(4\pi t)+5\cos(4\pi t)=\sqrt{29}\sin\left(4\pi t+\tan^{-1}\left(\dfrac52\right)\right)[/tex]

Answer:

Step-by-step explanation:

In the form ...

  y(t) = Asin(ωt +φ)

you have ...

The translation from ...

  y(t) = 2sin(4πt) +5cos(4πt)

is ...

  A = √(2² +5²) = √29 . . . . the amplitude

  ω = 4π . . . . the angular frequency in radians per second

  φ = arctan(5/2) ≈ 1.1903 . . . . radians phase shift

Then, ...

  y(t) = √29·sin(4πt +1.1903)

_____

Comment on the conversion

You will notice we used "2" and "5" to find the amplitude and phase shift. In the generic case, these are "coefficient of sin( )" and "coefficient of cos( )". When determining phase shift, pay attention to whether your calculator is giving you degrees or radians. (Set the mode to what you want.)

If you have a negative coefficient for sin( ), you will need to add 180° (π radians) to the phase shift value given by the arctan( ) function.

Answer:

The range is found by subtracting the minimum data entry from the maximum data entry.

Step-by-step explanation:

The range is found by subtracting the minimum data entry from the maximum data entry.

It is easy to compute.

It uses only two entries from the data set.

Answer: 680

Step-by-step explanation:

When order doesn't matter,then the number of combinations of choosing r things out of n = [tex]^nC_r=\dfrac{n!}{r!(n-r)!}[/tex]

Given: Total participants = 17

From these, a group of 3 participants is to be tested under a special condition.

Number of groups of 3 participants chosen = [tex]^{17}C_3=\dfrac{17!}{3!(17-3)!}\[/tex]

[tex]^{17}C_3=\dfrac{17!}{3!(17-3)!}\\\\=\dfrac{17\times16\times15\times14!}{3\times2\times14!}\\\\=680[/tex]

Hence, there are 680 groups of 3 participants can  be chosen,.

Answer:

[tex]g(x)=4x^{2} +10[/tex]

Step-by-step explanation:

If we perform a vertical translation of a function, the graph will move from one point to another certain point in the direction of the y-axis, in another words: up or down.

Let:

[tex]a>0,\hspace{10}a\in R[/tex]

For:

If:

[tex]f(x)=4x^{2} +6[/tex]

and is translated vertically upward by 4 units, this means:

[tex]a=4[/tex]

and:

[tex]g(x)=f(x)+a=(4x^{2} +6)+4=4x^{2} +10[/tex]

Therefore:

[tex]g(x)=4x^{2} +10[/tex]

I attached you the graphs, so you can verify the result easily.

Answer:

Step-by-step explanation:

The average salary in this city is $45,600.

Using the formula

z score = x - u /(sd/√n)

Where x is 46,356, u is 45,600 sd is 15,930 and n is 53.

z = 46,356 - 45600 / (15930/√53)

z = 756/(15930/7.2801)

z = 756/(2188.1568)

z = 0.3455

To draw a conclusion, we have to determine the p value, at 0.05 level of significance for a two tailed test, the p value is 0.7297. The p value is higher than the significance level, thus we will fail to reject the null and can conclude that there is not enough statistical evidence to prove that the average is any different for single people.

Answer:

Option B

an increasing exponential graph

Answer:

0.0668 or 6.68%

Step-by-step explanation:

Variance (V) = 10,000

Standard deviation (σ) = √V= 100

Mean score (μ) = 500

The z-score for any test score X is:

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

For X = 650:

[tex]z=\frac{650-500}{100}\\z=1.5[/tex]

A z-score of 1.5 is equivalent to the 93.32nd percentile of a normal distribution. Therefore, the probability that he or she will make a score of 650 or more is:

[tex]P(X\geq 650)=1-P(X\leq 650)\\P(X\geq 650)=1-0.9332\\P(X\geq 650)=0.0668=6.68\%[/tex]

The probability is 0.0668 or 6.68%

The probability that he or she will make a score of 650 or more is 0.0668.

Let X = Scores made on a certain aptitude test by nursing students

X follows normal distribution with mean = 500 and variance of 10,000.

So, standard deviation = [tex]\sqrt{10000}=100[/tex].

z score of 650 is = [tex]\frac{\left(650-500\right)}{100}=1.5[/tex].

The probability that he or she will make a score of 650 or more is:

[tex]P(X\geq 650)\\=P(z\geq 1.5)\\=1-P(z<1.5)\\=1-0.9332\\=0.0668[/tex]

Learn more: https://brainly.com/question/14109853

Answer:

Null - p= 71%

Alternative - p =/ 71%

Step-by-step explanation:

The null hypothesis is always the default statement in an experiment. While the alternative hypothesis is always tested against the null hypothesis.

Null hypothesis: 71% of their readers own a personal computer- p = 71%

Alternative hypothesis: Not 71% of their readers own a personal computer - p =/ 71%

Answer:

7y⁴- 10yz - y²z - 5

Step-by-step explanation:

First collect like terms

3y²+ 4y²- 6yz - 4yz - y²z - 7+2

7y⁴-10yz - y²z - 5

Answer:

Its C

Step-by-step explanation:

Answer:

The probability that no more than 70% would prefer to start their own business is 0.1423.

Step-by-step explanation:

We are given that a Gallup survey indicated that 72% of 18- to 29-year-olds, if given choice, would prefer to start their own business rather than work for someone else.

Let [tex]\hat p[/tex] = sample proportion of people who prefer to start their own business

The z-score probability distribution for the sample proportion is given by;

                               Z  =  [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex]  ~ N(0,1)

where, p = population proportion who would prefer to start their own business = 72%

            n = sample of 18-29 year-olds = 600

Now, the probability that no more than 70% would prefer to start their own business is given by = P( [tex]\hat p[/tex] [tex]\leq[/tex] 70%)

       P( [tex]\hat p[/tex] [tex]\leq[/tex] 70%) = P( [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] [tex]\leq[/tex] [tex]\frac{0.70-0.72}{\sqrt{\frac{0.70(1-0.70)}{600} } }[/tex] ) = P(Z [tex]\leq[/tex] -1.07) = 1 - P(Z < 1.07)

                                                                       = 1 - 0.8577 = 0.1423

The above probability is calculated by looking at the value of x = 1.07 in the z table which has an area of 0.8577.

Answer:

Step by step explanation

Total length of pole = 21.3 m

Length of pole inside the ground = 0.2 m

Let length of pole outside the ground be X,

So, according to the Question,

[tex]x + 0.2 = 21.3[/tex]

Move constant to R.H.S and change its sign

[tex]x = 21.3 - 0.2[/tex]

Calculate the difference

[tex]x = 21.1 \: m[/tex]

Hope this helps...

Good luck on your assignment...

Answer:

1. [tex] P(x) [/tex] ÷ [tex] Q(x) [/tex]---> [tex] \frac{-3x + 2}{3(3x - 1)} [/tex]

2. [tex] P(x) + Q(x) [/tex]---> [tex]\frac{2(6x - 1)}{(3x - 1)(-3x + 2)}[/tex]

3.  [tex] P(x) - Q(x) [/tex]---> [tex] \frac{-2(12x - 5)}{(3x - 1)(-3x + 2)} [/tex]

4. [tex] P(x)*Q(x) [/tex] --> [tex] \frac{12}{(3x - 1)(-3x + 2)} [/tex]

Step-by-step explanation:

Given that:

1. [tex] P(x) = \frac{2}{3x - 1} [/tex]

[tex] Q(x) = \frac{6}{-3x + 2} [/tex]

Thus,

[tex] P(x) [/tex] ÷ [tex] Q(x) [/tex] = [tex] \frac{2}{3x - 1} [/tex] ÷ [tex] \frac{6}{-3x + 2} [/tex]

Flip the 2nd function, Q(x), upside down to change the process to multiplication.

[tex] \frac{2}{3x - 1}*\frac{-3x + 2}{6} [/tex]

[tex] \frac{2(-3x + 2)}{6(3x - 1)} [/tex]

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

2. [tex] P(x) + Q(x) [/tex] = [tex] \frac{2}{3x - 1} + \frac{6}{-3x + 2} [/tex]

Make both expressions as a single fraction by finding, the common denominator, divide the common denominator by each denominator, and then multiply by the numerator. You'd have the following below:

[tex] \frac{2(-3x + 2) + 6(3x - 1)}{(3x - 1)(-3x + 2)} [/tex]

[tex] \frac{-6x + 4 + 18x - 6}{(3x - 1)(-3x + 2)} [/tex]

[tex] \frac{-6x + 18x + 4 - 6}{(3x - 1)(-3x + 2)} [/tex]

[tex] \frac{12x - 2}{(3x - 1)(-3x + 2)} [/tex]

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

3. [tex] P(x) - Q(x) [/tex] = [tex] \frac{2}{3x - 1} - \frac{6}{-3x + 2} [/tex]

[tex] \frac{2(-3x + 2) - 6(3x - 1)}{(3x - 1)(-3x + 2)} [/tex]

[tex] \frac{-6x + 4 - 18x + 6}{(3x - 1)(-3x + 2)} [/tex]

[tex] \frac{-6x - 18x + 4 + 6}{(3x - 1)(-3x + 2)} [/tex]

[tex] \frac{-24x + 10}{(3x - 1)(-3x + 2)} [/tex]

[tex] = \frac{-2(12x - 5}{(3x - 1)(-3x + 2)} [/tex]

4. [tex] P(x)*Q(x) = \frac{2}{3x - 1}* \frac{6}{-3x + 2} [/tex]

[tex] P(x)*Q(x) = \frac{2*6}{(3x - 1)(-3x + 2)} [/tex]

[tex] P(x)*Q(x) = \frac{12}{(3x - 1)(-3x + 2)} [/tex]

Composite functions involve combining multiple functions to form a new function

The functions are given as:

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

[tex]Q(x) = \frac{6}{-3x + 2}[/tex]

[tex]P(x) \div Q(x)[/tex] is calculated as follows:

[tex]P(x) \div Q(x) = \frac{2}{3x - 1} \div \frac{6}{-3x + 2}[/tex]

Express as a product

[tex]P(x) \div Q(x) = \frac{2}{3x - 1} \times \frac{-3x + 2}{6}[/tex]

Divide 2 by 6

[tex]P(x) \div Q(x) = \frac{1}{3x - 1} \times \frac{-3x + 2}{3}[/tex]

Multiply

[tex]P(x) \div Q(x) = \frac{-3x + 2}{3(3x - 1)}[/tex]

Hence, the value of [tex]P(x) \div Q(x)[/tex] is [tex]\frac{-3x + 2}{3(3x - 1)}[/tex]

P(x) + Q(x) is calculated as follows:

[tex]P(x) + Q(x) = \frac{2}{3x - 1} + \frac{6}{-3x + 2}[/tex]

Take LCM

[tex]P(x) + Q(x) = \frac{2(-3x + 2) + 6(3x - 1)}{(3x - 1)(-3x + 2)}[/tex]

Open brackets

[tex]P(x) + Q(x) = \frac{-6x + 4 + 18x - 6}{(3x - 1)(-3x + 2)}[/tex]

Collect like terms

[tex]P(x) + Q(x) = \frac{18x-6x + 4 - 6}{(3x - 1)(-3x + 2)}[/tex]

[tex]P(x) + Q(x) = \frac{12x - 2}{(3x - 1)(-3x + 2)}[/tex]

Factor out 2

[tex]P(x) + Q(x) = \frac{2(6x -1)}{(3x - 1)(-3x + 2)}[/tex]

Hence, the value of P(x) + Q(x) is [tex]\frac{2(6x -1)}{(3x - 1)(-3x + 2)}[/tex]

P(x) - Q(x) is calculated as follows:

[tex]P(x) - Q(x) = \frac{2}{3x - 1} - \frac{6}{-3x + 2}[/tex]

Take LCM

[tex]P(x) - Q(x) = \frac{2(-3x + 2) - 6(3x - 1)}{(3x - 1)(-3x + 2)}[/tex]

Open brackets

[tex]P(x) - Q(x) = \frac{-6x + 4 - 18x +6}{(3x - 1)(-3x + 2)}[/tex]

Collect like terms

[tex]P(x) - Q(x) = \frac{-18x-6x + 4 + 6}{(3x - 1)(-3x + 2)}[/tex]

[tex]P(x) - Q(x) = \frac{-24x +10}{(3x - 1)(-3x + 2)}[/tex]

Factor out -2

[tex]P(x) - Q(x) = \frac{-2(12x -5)}{(3x - 1)(-3x + 2)}[/tex]

Hence, the value of P(x) - Q(x) is [tex]\frac{-2(12x -5)}{(3x - 1)(-3x + 2)}[/tex]

P(x) * Q(x) is calculated as follows:

[tex]P(x) \times Q(x) = \frac{2}{3x - 1} \times \frac{6}{-3x + 2}[/tex]

Multiply

[tex]P(x) \times Q(x) = \frac{12}{(3x - 1)(-3x + 2)}[/tex]

Hence, the value of P(x) * Q(x) is [tex]\frac{12}{(3x - 1)(-3x + 2)}[/tex]

Read more about composite functions at:

https://brainly.com/question/10687170

Answer:

dI/dt = 0.0001(2000 - I)I

I(0) = 20

[tex]I(t)=\frac{2000}{1+99e^{-0.2t}}[/tex]

Step-by-step explanation:

It is given in the question that the rate of spread of the disease is proportional to the product of the non infected and the infected population.

Also given I(t) is the number of the infected individual at a time t.

[tex]\frac{dI}{dt}\propto \textup{ the product of the infected and the non infected populations}[/tex]

Given total population is 2000. So the non infected population = 2000 - I.

[tex]\frac{dI}{dt}\propto (2000-I)I\\\frac{dI}{dt}=k (2000-I)I, \ \textup{ k is proportionality constant.}\\\textup{Since}\ k = 0.0001\\ \therefore \frac{dI}{dt}=0.0001 (2000-I)I[/tex]

Now, I(0) is the number of infected persons at time t = 0.

So, I(0) = 1% of 2000

            = 20

Now, we have dI/dt = 0.0001(2000 - I)I  and  I(0) = 20

[tex]\frac{dI}{dt}=0.0001(2000-I)I\\\frac{dI}{(2000-I)I}=0.0001 dt\\\left ( \frac{1}{2000I}-\frac{1}{2000(I-2000)} \right )dI=0.0001dt\\\frac{dI}{2000I}-\frac{dI}{2000(I-2000)}=0.0001dt\\\textup{Integrating we get},\\\frac{lnI}{2000}-\frac{ln(I-2000)}{2000}=0.0001t+k \ \ \ (k \text{ is constant})\\ln\left ( \frac{I}{I-222} \right )=0.2t+2000k[/tex]

[tex]\frac{I}{I-2000}=Ae^{0.2t}\\\frac{I-2000}{I}=Be^{-0.2t}\\\frac{2000}{I}=1-Be^{-0.2t}\\I(t)=\frac{2000}{1-Be^{-0.2t}}\textup{Now we have}, I(0)=20\\\frac{2000}{1-B}=20\\\frac{100}{1-B}=1\\B=-99\\ \therefore I(t)=\frac{2000}{1+99e^{-0.2t}}[/tex]

The required expressions are presented below:

[tex]\frac{dI}{dt} = 0.0001\cdot I\cdot (2000-I)[/tex] [tex]\blacksquare[/tex]

[tex]I(0) = \frac{1}{100}[/tex] [tex]\blacksquare[/tex]

[tex]I(t) = \frac{20\cdot e^{\frac{t}{5} }}{1+20\cdot e^{\frac{t}{5} }}[/tex] [tex]\blacksquare[/tex]

After reading the statement, we obtain the following differential equation:

[tex]\frac{dI}{dt} = k\cdot I\cdot (n-I)[/tex] (1)

Where:

Then, we solve the expression by variable separation and partial fraction integration:

[tex]\frac{1}{k} \int {\frac{dI}{I\cdot (n-I)} } = \int {dt}[/tex]

[tex]\frac{1}{k\cdot n} \int {\frac{dl}{l} } + \frac{1}{kn}\int {\frac{dI}{n-I} } = \int {dt}[/tex]

[tex]\frac{1}{k\cdot n} \cdot \ln |I| -\frac{1}{k\cdot n}\cdot \ln|n-I| = t + C[/tex]

[tex]\frac{1}{k\cdot n}\cdot \ln \left|\frac{I}{n-I} \right| = C\cdot e^{k\cdot n \cdot t}[/tex]

[tex]I(t) = \frac{n\cdot C\cdot e^{k\cdot n\cdot t}}{1+C\cdot e^{k\cdot n \cdot t}}[/tex], where [tex]C = \frac{I_{o}}{n}[/tex] (2, 3)

Note - Please notice that [tex]I_{o}[/tex] is the initial infected population.

If we know that [tex]n = 2000[/tex], [tex]k = 0.0001[/tex] and [tex]I_{o} = 20[/tex], then we have the following set of expressions:

[tex]\frac{dI}{dt} = 0.0001\cdot I\cdot (2000-I)[/tex] [tex]\blacksquare[/tex]

[tex]I(0) = \frac{1}{100}[/tex] [tex]\blacksquare[/tex]

[tex]I(t) = \frac{20\cdot e^{\frac{t}{5} }}{1+20\cdot e^{\frac{t}{5} }}[/tex] [tex]\blacksquare[/tex]

To learn more on differential equations, we kindly invite to check this verified question: https://brainly.com/question/1164377

Set up two equations and set equal to each other. Let number of years = x:

College A = 14100+750x

College B = 42100-1250x

Set equal:

14100 + 750x = 42100 - 1250x

Subtract 750x from both sides:

14100 = 42100 - 2000x

Subtract 42100 from both sides:

-28000 = -2000x

Divide both sides by -2000:

x = -28000 / -2000

x = 14

It will take 14 years for the schools to have the same enrollment.

Enrollment will be:

14100 + 750(14) = 14100 + 10500 = 24,600

Answer:

(a)2019 (14 years after)

(b)24,600

Step-by-step explanation:

Let the number of years =n

College A

Initial Population in 2005 = 14,100

Increase per year = 750

Therefore, the population after n years = 14,100+750n

College B

Initial Population in 2005 = 42,100

Decline per year = 1250

Therefore, the population after n years = 42,100-1250n

When the enrollments are the same

14,100+750n=42,100-1250n

1250n+750n=42100-14100

2000n=28000

n=14

Therefore, in 2019 (14 years after), the colleges will have the same​ enrollment.

Enrollment in 2019 =42,100-1250(14)

=24,600

Answer:

It is a negative correlation

Step-by-step explanation:

As the x value increases the y value decreases. This causes it to be a negative.

Answer:

33800

Step-by-step explanation:

100 x 338 = 33800

Answer:

33800

Step-by-step explanation:

338x10=3380 then 3380x10=33800

-------------------------------------------------------

Good luck with your assignment...

Answer:

(a) The probability of getting someone who was not sent to prison is 0.55.

(b) If a study subject is randomly selected and it is then found that the subject entered a guilty plea, the probability that this person was not sent to prison is 0.63.

Step-by-step explanation:

We are given that in a study of pleas and prison sentences, it is found that 45% of the subjects studied were sent to prison. Among those sent to prison, 40% chose to plead guilty. Among those not sent to prison, 55% chose to plead guilty.

Let the probability that subjects studied were sent to prison = P(A) = 0.45

Let G = event that subject chose to plead guilty

So, the probability that the subjects chose to plead guilty given that they were sent to prison = P(G/A) = 0.40

and the probability that the subjects chose to plead guilty given that they were not sent to prison = P(G/A') = 0.55

(a) The probability of getting someone who was not sent to prison = 1 - Probability of getting someone who was sent to prison

      P(A') = 1 - P(A)

               = 1 - 0.45 = 0.55

(b) If a study subject is randomly selected and it is then found that the subject entered a guilty plea, the probability that this person was not sent to prison is given by = P(A'/G)

We will use Bayes' Theorem here to calculate the above probability;

    P(A'/G) =  [tex]\frac{P(A') \times P(G/A')}{P(A') \times P(G/A') +P(A) \times P(G/A)}[/tex]      

                 =  [tex]\frac{0.55 \times 0.55}{0.55\times 0.55 +0.45 \times 0.40}[/tex]

                 =  [tex]\frac{0.3025}{0.4825}[/tex]

                 =  0.63

Complete Question

If w'(t) is the rate of growth of a child in pounds per year, what does

[tex]\int\limits^{7}_{4} {w'(t)} \, dt[/tex]  represent?

a) The change in the child's weight (in pounds) between the ages of 4 and 7.

b) The change in the child's age (in years) between the ages of 4 and 7.

c) The child's weight at age 7.

d) The child's weight at age 4. The child's initial weight at birth.

Answer:

The correct option is  option a

Step-by-step explanation:

From the question we are told that

       [tex]w'(t)[/tex] represents the rate of growth of a child in   [tex]\frac{pounds}{year}[/tex]

So      [tex]{w'(t)} \, dt[/tex]  will be in  [tex]pounds[/tex]

Which then mean that this  [tex]\int\limits^{7}_{4} {w'(t)} \, dt[/tex]  the change in the weight of the child between the ages of  [tex]4 \to 7[/tex] years

Answer:

1/5 cup

Step-by-step explanation:

Sugar: water

1             5

We want 1 cup water, so divide each side by 5

1/5 :  5/5

1/5 : 1

There is 1/5 cup sugar to 1 cup water

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La fórmula para calcular el área de cualquier triángulo es:

base multiplicada por la altura y dividida por dos.

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Bh / 2.

Answer:

a. p`± z₀.₀₂₅[tex]\sqrt{ \frac{p`q`}{n}[/tex]  

b.0.6 ±  1.96 [tex]\sqrt \frac{0.6* 0.4}{1000}[/tex]  

c. { -1.96 ≤  p`± z₀.₀₂₅[tex]\sqrt{ \frac{p`q`}{n}[/tex]     ≥ 1.96} = 0.95  

Step-by-step explanation:

Here the total number of trials is n= 1000

The number of successes is p` = 600/1000 = 0.6. The q` is 1 - p`= 1- 0.6 = 0.4

The degree of confidence is 95 %  therefore z₀.₀₂₅ = 1.96 ( α/2 = 0.025)

a.  The formula used will be

p`± z₀.₀₂₅[tex]\sqrt{ \frac{p`q`}{n}[/tex]       ( z with the base alpha by 2 (α/2 = 0.025))

b. Putting the values

0.6 ±  1.96 [tex]\sqrt \frac{0.6* 0.4}{1000}[/tex]  

c. Confidence Interval in Interval Notation.

{ -1.96 ≤  p`± z₀.₀₂₅[tex]\sqrt{ \frac{p`q`}{n}[/tex]     ≥ 1.96} = 0.95  

{ -z( base alpha by 2) ≤  p`± z₀.₀₂₅[tex]\sqrt{ \frac{p`q`}{n}[/tex]     ≥ z( base alpha by 2)  } = 1- α

Answer:  D.) The second difference is constant.

Step-by-step explanation:

The rate of change of a quadratic function is a linear function. The rate of change of that is constant, so second differences of a quadratic number pattern are constant.

Answer:

D.

Step-by-step explanation:

Step-by-step explanation:

81^x² = 27^x

(3^4)^x² = (3^3)^x

3^(4x²) = 3^(3x)

4x² = 3x

4x² − 3x = 0

x (4x − 3) = 0

x = 0 or ¾