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3 years ago

The temperature at a ski resort was 12°F at 5:00 A.M.

Mathematics

1 answer:

zimovet [89]3 years ago

4 0

22.5 i Hope I’m right

Sorry if this is wrong

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At CFIS Intermediate school, 40% of the 500 students ride their bikes or walk to school. How many students ride their bikes or w

ra1l [238]

40% of 500 = 200
10% of 500 = 50 so just times it by 4

7 0

3 years ago

A survey organization has used the methods of our class to construct an approximate 95% confidence interval for the mean annual

lawyer [7]

Answer:

$\bar X = \frac{66000+70000}{2}= 68000$

We can estimate the margin of error with this formula:

$ME= \frac{Upper -Lower}{2}= \frac{70000-66000}{2}= 2000$

And the margin of error is given by:

$ME = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

And we can rewrite the margin of error like this:

$ME =z_{\alpha/2}*SE$

Where $SE= \frac{\sigma}{\sqrt{n}}$

For 95% of confidence the critical value is $z_{\alpha/2}= \pm 1.96$

The Standard error would be:

$SE= \frac{ME}{z_{\alpha/2}}= \frac{2000}{1.96}= 1020.408$

For 99% of confidence the critical value is $z_{\alpha/2}= \pm 2.58$

And the new margin of error would be:

$ME = 2.58* 1020.408 = 2632.653$

And then the interval would be given by:

$Lower = 68000- 2632.653 = 65367.347$

$Upper = 68000+ 2632.653 = 70632.653$

Step-by-step explanation:

Previous concepts

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".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The 95% confidence interval is given by (66000 , 70000)

We can estimate the mean with this formula:

$\bar X = \frac{66000+70000}{2}= 68000$

We can estimate the margin of error with this formula:

$ME= \frac{Upper -Lower}{2}= \frac{70000-66000}{2}= 2000$

And the margin of error is given by:

$ME = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

And we can rewrite the margin of error like this:

$ME =z_{\alpha/2}*SE$

Where $SE= \frac{\sigma}{\sqrt{n}}$

For 95% of confidence the critical value is $z_{\alpha/2}= \pm 1.96$

The Standard error would be:

$SE= \frac{ME}{z_{\alpha/2}}= \frac{2000}{1.96}= 1020.408$

For 99% of confidence the critical value is $z_{\alpha/2}= \pm 2.58$

And the new margin of error would be:

$ME = 2.58* 1020.408 = 2632.653$

And then the interval would be given by:

$Lower = 68000- 2632.653 = 65367.347$

$Upper = 68000+ 2632.653 = 70632.653$

7 0

3 years ago

The following table shows the speeds at which Jeff does 4 activities. For example, biking has a rate of 19 miles in 2 hours. Whi

ExtremeBDS [4]

The type of activity that Jeff does at a speed of 4.5 miles per hour is Swimming

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem !

<em>This problem is about Kinematics.</em>

We will solve it in the following way

$\texttt{Speed of Biking} = \texttt{Distance of Biking} \div \texttt{Time Taken}$

$\texttt{Speed of Biking} = \texttt{19} \div \texttt{2}$

$\texttt{Speed of Biking} = 9.5 \texttt{ miles per hour}$

$\texttt{ }$

$\texttt{Speed of Walking} = \texttt{Distance of Walking} \div \texttt{Time Taken}$

$\texttt{Speed of Walking} = \texttt{14} \div \texttt{4}$

$\texttt{Speed of Walking} = 3.5 \texttt{ miles per hour}$

$\texttt{ }$

$\texttt{Speed of Running} = \texttt{Distance of Running} \div \texttt{Time Taken}$

$\texttt{Speed of Running} = \texttt{15} \div \texttt{3}$

$\texttt{Speed of Running} = 5 \texttt{ miles per hour}$

$\texttt{ }$

$\texttt{Speed of Swimming} = \texttt{Distance of Swimming} \div \texttt{Time Taken}$

$\texttt{Speed of Swimming} = \texttt{18} \div \texttt{4}$

$\texttt{Speed of Swimming} = \boxed{4.5 \texttt{ miles per hour}}$

$\texttt{ }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate

3 0

3 years ago

Read 2 more answers

Please help! need to know how to solve this Urgent!

PilotLPTM [1.2K]

Step-by-step explanation:

I dont know if this helps but that what I came up with

5 0

3 years ago

Read 2 more answers

Simply the problem.

Marat540 [252]

Answer:

Step-by-step explanation:

$\sqrt{3}\cdot\sqrt{21}= \\\\\sqrt{3\cdot 21}= \\\\\sqrt{63}= \\\\3\sqrt{7}$

Therefore, the correct answer is choice C. Hope this helps!

6 0

3 years ago

Read 2 more answers