An object was launched from a given height. The
equation for the object's height over time is given
as: h= -16t^2 + 19.6t +58.8. What was the
object's initial height?

Answer:

One solution

Step-by-step explanation:

The slopes are different which means, the systems has one solution.

hope this helps and is right p.s. i really need brainliest :)

Answer:

The calculated χ² =  17.0281    falls in the critical region χ² ≥  9.49  so we reject the null hypothesis that  the quality of management and the reputation of the company are independent and conclude quality of management and the reputation of the company are dependent

The p-value is 0 .001909. The result is significant at p < 0.05

Part b:

40 > 8.5

35> 7.5

25> 4

Step-by-step explanation:

1) Let the null and alternative hypothesis as

H0: the quality of management and the reputation of the company are independent

against the claim

Ha: the quality of management and the reputation of the company are dependent

2) The significance level alpha is set at 0.05

3) The test statistic under H0 is

χ²= ∑ (o - e)²/ e where O is the observed and e is the expected frequency

which has an approximate chi square distribution with ( 3-1) (3-1)=  4 d.f

4) Computations:

Under H0 ,

Observed       Expected E              χ²= ∑(O-e)²/e

40                      35.00                          0.71

25                      24.50                         0.01

5                         10.50                         2.88  

35                      40.00                         0.62

35                      28.0                          1.75

10                       12.00                           0.33  

25                      25.00                             0.00

10                        17.50                              3.21

15                       7.50                                 7.50  

∑                                                               17.0281

Column Totals 100 70 30   200  (Grand Total)

5) The critical region is χ² ≥ χ² (0.05)2 = 9.49

6) Conclusion:

The calculated χ² =  17.0281    falls in the critical region χ² ≥  9.49  so we reject the null hypothesis that  the quality of management and the reputation of the company are independent and conclude quality of management and the reputation of the company are dependent.

7) The p-value is 0 .001909. The result is significant at p < 0.05

The p- values tells that the variables are dependent.

Part b:

If we take the excellent row total = 70 and compare it with the excellent column total= 100

If we take the good row total = 70 and compare it with the good column total= 80

If we take the fair row total = 50 and compare it with the fair column total= 30

The two attributes are said to be associated if

Thus we see that ( where (A)(B) are row and columns totals and AB are the cell contents)

AB> (A)(B)/N  

40 > 1700/200

40 > 8.5

35> 1500/200

35> 7.5

25> 800/200

25> 4

and so on.

Hence they are positively associated

Answer:

1 : 6

Step-by-step explanation:

Answer:

92.59 ounces................

Answer:

92.5941501 ounces, 93 ounces if rounded to whole number, and 92.6 if rounded to the nearest tenth.

Step-by-step explanation:

Hope this helps! have a great day

Answer:

Step-by-step explanation:

Answer:

- The population in 2001 is 23.1 million

- It will reach 28.3 million in 2013

Step-by-step explanation:

Given

[tex]A =23.1e^{0.0152t}[/tex]

Solving (a): Population in 2000.

This implies that

[tex]t = 2000 - 2000[/tex]

[tex]t = 0[/tex]

So, we have:

[tex]A =23.1e^{0.0152t}[/tex]

[tex]A =23.1e^{0.0152*0}[/tex]

[tex]A =23.1e^{0}[/tex]

Solve the exponent

[tex]A =23.1*1[/tex]

[tex]A =23.1[/tex]

The population in 2001 is 23.1 million

Solving (b): When will it reach 28.3 million

This implies that:

[tex]A =28.3[/tex]

So, we have:

[tex]A =23.1e^{0.0152t}[/tex]

[tex]28.3 = 23.1e^{0.0152t}[/tex]

Divide both sides by 23.1

[tex]1.225 = e^{0.0152t}[/tex]

Take natural logarithm of both sides

[tex]ln\ 1.225 = ln(e^{0.0152t})[/tex]

Rewrite as:

[tex]ln\ 1.225 = 0.0152t\ ln(e)[/tex]

[tex]ln(e) = 1[/tex]

So:

[tex]ln\ 1.225 = 0.0152t*1[/tex]

[tex]ln\ 1.225 = 0.0152t[/tex]

Make t the subject

[tex]t = \frac{ln\ 1.225}{0.0152}[/tex]

[tex]t = \frac{0.2029}{0.0152}[/tex]

[tex]t \approx 13[/tex]

Add the value of t to 2013 to get the actual year

[tex]Year = 2000 +13[/tex]

[tex]Year = 2013[/tex]

It will reach 28.3 million in 2013

9514 1404 393

Answer:

  9%

Step-by-step explanation:

The interest earned is ...

  I = Prt . . . . . principal P at annual rate r for t years

Solving for r, we find ...

  r = I/(Pt) = $540/($3000×2) = 0.09 = 9%

The interest rate on Beka's loan is 9%.

Answer:

So 4 ÷ 1/5 = 4 × 5. So your answer is 5.

Answer:

I didnt get what the photo shows....

Step-by-step explanation:

Answer:

4.8

Step-by-step explanation:

Answer:

As the pvalue of the test is 0.02097 < 0.1, using a .10 significance level, he can conclude that the average trunk diameter of a pine tree is greater than the average diameter of a spruce tree.

Step-by-step explanation:

Using a .10 significance level, can he conclude that the average trunk diameter of a pine tree is greater than the average diameter of a spruce tree

The null hypothesis is that they are equal(that is, subtraction between the means is 0), while the alternate hypothesis is that they are greater(that is, subtraction between the means is larger than 0). So

[tex]H_0: \mu_1 - \mu_2 = 0[/tex]

[tex]H_a: \mu_1 - \mu_2 > 0[/tex]

In which [tex]\mu_1[/tex] is the mean pine trees height while [tex]\mu_2[/tex] is the mean spruce trees height.

The test statistic is:

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

In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis and s is the standard error.

0 is tested at the null hypothesis:

This means that [tex]\mu = 0[/tex]

Mean trunk diameter (cm) 35 30:

Pine trees - 35

Spruce trees - 30

The sample mean is given by the subtraction of the means. So

[tex]X = 35 - 30 = 5[/tex]

Sample Size 40 80

Population variance 160 160

This means that the standard error for each sample is given by:

[tex]s_1 = \frac{\sqrt{160}}{\sqrt{40}} = 2[/tex]

[tex]s_1 = \frac{\sqrt{160}}{\sqrt{80}} = \sqrt{2}[/tex]

The standard error of the difference is the square root of the sum squared of the standard error of each sample.

[tex]s = \sqrt{2^2 + (\sqrt{2})^2} = \sqrt{6} = 2.4495[/tex]

Test statistic:

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

[tex]z = \frac{5 - 0}{2.4495}[/tex]

[tex]z = 2.04[/tex]

Pvalue of the test:

Probability of a sample mean larger than 5, which is 1 subtracted by the pvalue of Z when X = 5.

Looking at the z-table, z = 2.04 has a pvalue of 0.9793.

1 - 0.9793 = 0.0207

As the pvalue of the test is 0.02097 < 0.1, using a .10 significance level, he can conclude that the average trunk diameter of a pine tree is greater than the average diameter of a spruce tree.

Answer:

Here's a graph to help you solve this answer:

Answer:

12345678910

Step-by-step explanation:

CHARRRRRRRRR joke lang bestie

Answer:

A is 14 and P is 17

Step-by-step explanation:

working is above mentioned

Answer:

all sides are equal so the x is 50.1 degree and nearest 100 will be 100

please mark me as brainliest

Answer:

sampling distribution = 22.4

sampling deviation = 0.68

Step-by-step explanation:

Given that :

Mean , μ = 22.4

Standard deviation, σ = 5.8

n = 73

Therefore, the sampling distribution of the mean and the standard deviation is

[tex]$\mu_{\overline x} = 22.4$[/tex]

So, [tex]$\sigma_{\overline x}=\frac{\sigma}{\sqrt n}$[/tex]

            [tex]$=\frac{5.8}{\sqrt {73}}$[/tex]

          = 0.68

So the sampling distribution of the mean is 22.4

The standard deviation is 0.68

Answer:

∠RQO and ∠STV

∠PQO and ∠UTV

Step-by-step explanation:

Answer:

I believe it is 12

Step-by-step explanation:

but I am not 100 percent sure

Answer:

Using the equation y = abx , substitute both of your given points into that equation.

2 = ab2 and 4 = ab3   Solve each equation for a.

2⁄b2    and 4⁄b3 = a     Therefore, 2⁄b2 = 4⁄b3

Cross multiply: 2b3 = 4b2     Divide both sides by b2

2b = 4      a = 2/4 = 1/2

b = 2

y = 1 (2)x

     2

Step-by-step explanation:

this my answer not sure you like it or not

Answer:

8: 0,-2  9: 7,-3  10: 7,1/2

Step-by-step explanation:

Hello, for question 8 (the top equation), you trying to find y and x so lets find y first. You need y to be by itself so you divide all the numbers by -3 like this:

6x/-3-3y/-3=6/-3

The equation simplifies to -2x+y=-2. You need y to be by itself so you move it to the other side and you get -2 for y. Now you need x. Go to the second equation and take the number that is y and plug it into the equation like this:

6x+8(-2)=-16

Multiply 8 with -2 which is -16. Then, you add -16 with -16 which makes 0.

6x=0

You then divide 6 on both sides and x is 0. You basically do the same process for questions 9 and 10 which are 7,-3 and 7,1/2. And your done!

Answer: #1 is the answer. Alternate exterior angles are the pair of angles that lie on the outer side of the two parallel lines but on either side of the transversal line. Notice how the pairs of alternating exterior angles lie on opposite sides of the transversal but outside the two parallel lines.

Step-by-step explanation: Parallel Lines: Definition: We say that two lines (on the same plane) are parallel to each other if they never intersect each other, ragardless of how far they are extended on either side. Pictorially, parallel lines run along each other like the tracks of a train.

Answer:

y =-10

Step-by-step explanation:

because there is no value of x so the slope of y is-10

Answer:

a) attached below

b) Eigen values : λ1 = -1  and λ2 = - 1/2

  Eigenvector : V1 = - y1  and V2 = ( - 1/2 ) y2

c) since the Oscillator is > 0 hence  it is Overdamped

Step-by-step explanation:

Lets take  : m = 2 , k = 1 and b = 3

a) Second -order differential equation and the first order system

attached below

b) determine the eigenvalues and eigenvectors of the Linear system

Eigen values : λ1 = -1  and λ2 = - 1/2

Eigenvector : V1 = - y1  and V2 = ( - 1/2 ) y2

attached below is the detailed solution

c) The oscillator is classified as OVERDAMPED

To determine if the oscillator is damped, underdamped, or overdamped we will apply the relation below

 b^2 - 4km

 ( 3 )^2 - 4(1)(2)  = 1

since the Oscillator is > 0 hence it is Overdamped

d) Attached below is the phase portrait of the associated linear system

y(0) = 0

v(0) = 0

Answer:

[tex]Factor = 6w^4 v^3[/tex]

Step-by-step explanation:

Given

[tex]30x^8 w^4 v^3[/tex] and [tex]12w^7 v^5[/tex]

Required

Determine the common factor.

To do this, we simply factorize each. SO, we have:

[tex](1)\ 30x^8 w^4 v^3=: 2 * 3 * 5 * x^8 * w^4 * v^3[/tex]

[tex](2)\ 12w^7 v^5:=\ 2 * 2 * 3 * w^4*w^3 * v^3 * v^2[/tex]

Check for similar factors.

[tex]Factor = 2 *3 * w^4 * v^3[/tex]

[tex]Factor = 6w^4 v^3[/tex]

The above represents the similar factor