30 points! Please help!<br><br> https://brainly.com/question/20274185

A chess tournament starts at 12:30 p.m. and ends at 5:00 p.m. Does the tournament last more than 5 hours? Use the drop‐down menu

pychu [463]

The tournament does not last more than 5 hours because the duration of the tournament is 4.5 hours.

<h3>What is an arithmetic operation?</h3>

It is defined as the operation in which we do the addition of numbers, subtraction, multiplication, and division. It has a basic four operators that is +, -, ×, and ÷.

We have a chess tournament starts at 12:30 p.m. and ends at 5:00 p.m.

The difference between 12:30 p.m. and 5:00 p.m. can be calculated after changing time into 24 hours clock

12:30 p.m. = 12 hours 30 minutes = 12.50

5: 00 p.m. = 17 hours 00 minutes = 17.00

Difference between them:

= 17.00 - 12.50

= 4.5 hours

4.5 < 5

Thus, the tournament does not last more than 5 hours because the duration of the tournament is 4.5 hours.

Learn more about the arithmetic operation here:

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6 0

2 years ago

Bartholomew went hiking over the weekend. He hiked all 4 trails in 3 hours. Which is the best estimate for the number of miles h

Helen [10]

Answer:

a?

Step-by-step explanation:

5 0

4 years ago

Multiply (-3x4y3z)(-4x2y3z).

velikii [3]

Answer:

-3×4ycube z square (-4)×2y3

-12cube z square (-4)×2y3

48ycube z square×2y3

96ycube z square y cube

288y3 (yz)square

I hope you will understand these squares or cubes

3 0

3 years ago

Slope=2/3, y-intercept=-5<br>please helphelp

Shtirlitz [24]

What do u need help w
its y=2/3x - 5

3 0

3 years ago

A union of restaurant and foodservice workers would like to estimate the mean hourly wage, μ, of foodservice workers in the U.S.

pav-90 [236]

Answer:

$n=(\frac{1.960(2.25)}{0.35})^2 =158.76 \approx 159$

So the answer for this case would be n=159 rounded up to the nearest integer

Step-by-step explanation:

Assuming this complete question: A union of restaurant and foodservice workers would like to estimate the mean hourly wage, , of foodservice workers in the U.S. The union will choose a random sample of wages and then estimate using the mean of the sample. What is the minimum sample size needed in order for the union to be 95% confident that its estimate is within $0.35 of ? Suppose that the standard deviation of wages of foodservice workers in the U.S. is about $2.15 .

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)

$\sigma=2.25$ represent the population standard deviation

n represent the sample size

Solution to the problem

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (a)

And on this case we have that ME =0.35 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.960$ , replacing into formula (b) we got:

$n=(\frac{1.960(2.25)}{0.35})^2 =158.76 \approx 159$

So the answer for this case would be n=159 rounded up to the nearest integer

3 0

3 years ago

30 points! Please help! https://brainly.com/question/20274185