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3 years ago

Late work 4 physics

Mathematics

1 answer:

Oliga [24]3 years ago

7 0

Answer:

Step-by-step explanation:

problem 4

Velocity is +2.0m/s and constant acceleration is -0.5m/s^2

For every second the velocity drop by 0.5 m/s, so after 2 seconds the velocity is 1 m/s.

problem 5

Accelerate from rest to 28m/s means go from 0 to 28m/s so with

a constant acceleration of 5.5 m/s^2 it will take 5.09 second

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An oil company is considering 2 sites on which to drill, described as follows:

Olegator [25]

A)site A has a larger probablility of finding oil
B) the difference between amount of profitmade is 30 million.

7 0

3 years ago

Need help Choices 4 3 2 1

-Dominant- [34]

Answer:

Step-by-step explanation:

The amplitude is the maximum distance away from the middle of the wave. Here we can see that the middle of the wave is the x axis and the farthest point (largest difference between the y coordinate of the x-axis, or the line y = 0, and the wave) is 1 unit away.

4 0

3 years ago

Read 2 more answers

Select the correct answer from each drop-down menu.

serious [3.7K]

Answer:

Center: (4,8)

Radius: 2.5

Equation: $(x-4)^2+(y-8)^2=6.25$

Step-by-step explanation:

It was given that; the endpoints of the longest chord on a circle are (4, 5.5) and (4, 10.5).

Note that the longest chord is the diameter;

The midpoint of the ends of the diameter gives us the center;

Use the midpoint formula;

$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )$

The center is at; $(\frac{4+4)}{2} ,\frac{5.5+10.5}{2}=(4,8)$

To find the radius, use the distance formula to find the distance from the center to one of the endpoints.

The distance formula is;

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$r=\sqrt{(4-4)^2+(10.5-8)^2}$

$r=\sqrt{0^2+(2.5)^2}$

$r=\sqrt{0^2+(2.5)^2}=2.5$

The equation of the circle in standard form is given by;

$(x-h)^2+(y-k)^2=r^2$

We substitute the center and the radius into the formula to get;

$(x-4)^2+(y-8)^2=2.5^2$

$(x-4)^2+(y-8)^2=6.25$

6 0

3 years ago

A Pew Internet poll asked cell phone owners about how they used their cell phones. One question asked whether or not during the

EastWind [94]

Answer:

a) $\hat p=\frac{471}{1024}=0.460$

The standard error is given by:

$SE= \sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0156$

And the margin of error is given by:

$ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}=1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0305$

b) The 99% confidence interval would be given by (0.429;0.491)

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Data given and notation

n=1024 represent the random sample taken

X=471 represent the people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

$\hat p=\frac{471}{1024}=0.460$ estimated proportion of people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

p= population proportion of people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

Part a

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.