Please help me get this please

In the air, it had an average speed of 16 1616 m/s m/sstart text, m, slash, s, end text. In the water, it had an average speed o

Diano4ka-milaya [45]

Answer:

t_a = 7 s

t_w = 5 s

Step-by-step explanation:

Given:

- The average speed in air v_a = 16 m/s

- The average speed in water v_w = 3 m/s

- The total distance D = 127 m

- The total time T = 12 s

Find:

How long did the stone fall in air and how long did it fall in the water?

Solution:

- Lets time taken by stone to drop through air as t_a = t_a.

- The time taken in water would be t_w = (12 - t_a)

- The total distance covered can be calculated as:

D = v_a*t_a+ v_w*(12 - t_a)

127 = 16*t_a + 3*(12 - t_a)

Simplify and solve for t:

91 = 16*t_a - 3t_a

13*t_a = 91

t_a = 7 s

t_w = 5 s

8 0

3 years ago

Solve for x.<br> Anyone know how to do this?

nasty-shy [4]

Answer:

hope this helps you look it once.

7 0

3 years ago

Which of the x values are solutions to the inequality 4(2 – x) > –2x – 3(4x + 1)? Check all that apply. x = –1.1 x = –2.2 x =

galben [10]

Answer:
x > - 11/10
(I don't know if this is correct or not, if so i'm glad i helped!)

8 0

3 years ago

Read 2 more answers

Give the first ten terms of the following sequences. You can assume that the sequences start with an index of 1. Logs are to bas

Vitek1552 [10]

Answer:

a)

a1 = log(1) = 0 (2⁰ = 1)

a2 = log(2) = 1 (2¹ = 2)

a3 = log(3) = ln(3)/ln(2) = 1.098/0.693 = 1.5849

a4 = log(4) = 2 (2² = 4)

a5 = log(5) = ln(5)/ln(2) = 1.610/0.693 = 2.322

a6 = log(6) = log(3*2) = log(3)+log(2) = 1.5849+1 = 2.5849 (here I use the property log(a*b) = log(a)+log(b)

a7 = log(7) = ln(7)/ln(2) = 1.9459/0.6932 = 2.807

a8 = log(8) = 3 (2³ = 8)

a9 = log(9) = log(3²) = 2*log(3) = 2*1.5849 = 3.1699 (I use the property log(a^k) = k*log(a) )

a10 = log(10) = log(2*5) = log(2)+log(5) = 1+ 2.322= 3.322

b) I can take the results of log n we previously computed above to calculate 2^log(n), however the idea of this exercise is to learn about the definition of log_2:

log(x) is the number L such that 2^L = x. Therefore 2^log(n) = n if we take the log in base 2. This means that

a1 = 1

a2 = 2

a3 = 3

a4 = 4

a5 = 5

a6 = 6

a7 = 7

a8 = 8

a9 = 9

a10 = 10

I hope this works for you!!

3 0

3 years ago

A Packaging Company produces boxes out of cardboard and has a specified weight of 16 oz. A random sample of 36 boxes yielded a s

user100 [1]

Answer:

The margin of error is of 0.73 oz.

The 99% confidence interval for the true mean weight of the boxes is between 14.57 oz and 16.03 oz. This means that we are 99% sure that the true mean weight of all boxed produced by the Packaging Company is between these two values, and that the specified weight is in this interval.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.005 = 0.995$ , so Z = 2.575.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 2.575\frac{1.7}{\sqrt{36}} = 0.73$

The margin of error is of 0.73 oz.

The lower end of the interval is the sample mean subtracted by M. So it is 15.3 - 0.73 = 14.57 oz.

The upper end of the interval is the sample mean added to M. So it is 15.3 + 0.73 = 16.03 oz.

The 99% confidence interval for the true mean weight of the boxes is between 14.57 oz and 16.03 oz. This means that we are 99% sure that the true mean weight of all boxed produced by the Packaging Company is between these two values, and that the specified weight is in this interval.

4 0

2 years ago