Values for relation g are given in the table. Which ordered pair would be found in the inverse of g? X Y 2 2 3 5 4 9 5 13 A: (4,9) B:(-3.-5) C:(13,5) D:(-2,-2)

Answer: Price-earnings ratio= 22.0

Step-by-step explanation:

Given: A company had a market price of $38.50 per share, earnings per share of $1.75, and dividends per share of $0.90

To find: price-earnings ratio

Required formula: [tex]\text{price-earnings ratio }=\dfrac{\text{ Market Price per Share}}{\text{Earnings Per Share}}[/tex]

Then, Price-earnings ratio = [tex]\dfrac{\$38.50}{\$1.75}[/tex]

⇒Price-earnings ratio = [tex]\dfrac{22}{1}[/tex]

Hence, the price-earnings ratio= 22.0

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:

i think (d) one i think it will help you

Answer: The length of the line B'C" is 1 unit.

Step-by-step explanation:

Given: Triangle ABC is dilated by a scale factor of 0.5 with the origin as the center of dilation , resulting in the image Triangle A'B'C'.

If A (2,2), B= (4,3) and C=(6,3).

Distance between (a,b) and (c,d): [tex]D=\sqrt{(d-b)^2+(c-b)^2}[/tex]

Then, BC [tex]=\sqrt{(3-3)^2+(6-4)^2}[/tex]

[tex]\\\\=\sqrt{0+2^2}\\\\=\sqrt{4}\\\\=2\text{ units}[/tex]

Length of image = scale factor x length in original figure

B'C' = 0.5 × BC

= 0.5 × 2

= 1 unit

Hence, the length of the line B'C" is 1 unit.

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:

( 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

¡Hola! ¡Ojalá esto ayude!

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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:

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]

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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:

It is unlikely.

Step-by-step explanation:

Certain = 100%

Impossible = 0%

Likely = more than 50%

Unlikely = less than 50%

It is less than 50%, so it is unlikely.

Answer:

(A) it is likely

Step-by-step explanation:

i took the test on edge

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:

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: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:

B. The fact that two variables are strongly correlated does not in itself imply a​ cause-and-effect relationship between the variables.

Step-by-step explanation:

The term "correlation does not imply causation", simply means that because we can deduce a link between two factors or sets of data, it does not necessarily prove that there is a cause-and-effect relationship between the two variables. In some cases, there could indeed be a cause-and-effect relationship but it cannot be said for certain that this would always be the case.

While correlation shows the linear relationship between two things, causation implies that an event occurs because of another event. So the phrase is actually saying that because two factors are related, it does not mean that it is as a result of a causal factor. It could simply be a coincidence. This occurs because of our effort to seek an explanation for the occurrence of certain events.

Answer: B. The fact that two variables are strongly correlated does not in itself imply a​ cause-and-effect relationship between the variables.

Step-by-step explanation:

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]

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Answer:

Probability that it'll not rain during the next three days = 0.064

Step-by-step explanation:

Given

Let:

P(R) represent the probability that it'll rain each day

P(R') represent the probability that it'll not

[tex]P(R) = 60\%[/tex]

Required

Probability that it'll not rain during the next three days

From concept of probability;

[tex]P(R) + P(R') = 1[/tex]

Substitute 60% for P(R)

[tex]60\% + P(R') = 1[/tex]

Subtract 60% from both sides

[tex]60\% - 60\% + P(R') = 1 - 60\%[/tex]

[tex]P(R') = 1 - 60\%[/tex]

Convert % to decimal

[tex]P(R') = 1 - 0.6[/tex]

[tex]P(R') = 0.4[/tex]

The probability that it'll not rain during the next 3 days is:

[tex]P(R') * P(R') * P(R')[/tex]

[tex]P(R') * P(R') * P(R') =0.4 * 0.4 * 0.4[/tex]

[tex]P(R') * P(R') * P(R') = 0.064[/tex]

Answer:

33800

Step-by-step explanation:

100 x 338 = 33800

Answer:

33800

Step-by-step explanation:

338x10=3380 then 3380x10=33800

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Good luck with your assignment...

Answer:

-12,-10,-3,-1.4,-1,0.25,2,5.2,12

Step-by-step explanation:

The following number −1.4; 2; −3 1 2 ; −1; − 1 2 ; 0.25; −10; 5.2 in increasing order

-12,-10,-3,-1.4,-1,0.25,2,5.2,12

It's arranged this way starting from the negative sign because positive it's greater than negative and if the negative gets to approach zero it's get smaller

Answer:

-10 ; -3 1/2 ; -1.4 ; -1 ; -1/2 ; 0.25 ; 2 ; 5.2

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.

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:

Option B

an increasing exponential graph

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:

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

7km/ 4km per hour = 1 3/4 hours

3/4 hour = 45 minutes

Total time = 1 hour and 45 minutes.

Answer:

(0,-2)

Step-by-step explanation:

The y-intercept is simply when the function touches or crosses the y-axis.

We're told that the graph crosses the y-axis at -2. In other words, the y-intercept is at -2.

The ordered pair would be (0,-2)

Answer: 0.388 m

Step-by-step explanation:

Ok, if the ball is dropped from 1.8 meters, then the height after the first bounce will be 3/5 times 1.8 meters:

h1 = (3/5)*1.8m = 1.08m

now we can think that the ball is dropped from a height of 1.08 meters, then the height after the second rebound will be:

h2 = (3/5)*1.08m = 0.648m

Now, using the same method as before, the height after the third bounce will be:

h3 = (3/5)*0.648m = 0.388 m

Notice that we can write this relation as:

h(n) = 1.8m*(3/5)^n

where n is the number of bounces.

if n = 0 we have the initial height, and if n = 3 we are on the third bounce, then:

h(3) = 1.8m*(3/5)^3 = 0.388 m

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.

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