How much is 13 part of 6,918?

Answers

Answer 1

Answer: The answer should be 0.19%

Related Questions

I need helppp !!! Pls
Find the product 8x(2x^2+8x-5)
The number line below represents which combined inequality?number line with a closed circle on negative 4 and 3 with shading in betweenx greater than or equal to −4 and x less than or equal to 3x greater than or equal to −4 or x less than or equal to 3x less than or equal to −4 and x greater than or equal to 3x less than or equal to −4 or x greater than or equal to 3
Using the diagram below, what is the measure of ZE?
Y=3x+4. Find an equation of a line that is:parallel to the given liney=3x+4. Find an equation of a line that is:perpendicular to the given linePLZ ANSWER FAST 50 POINTS

Help! - Answer = Brainliest!

Thanks!

Answers

Answer:

Step-by-step explanation:

i copied the person above me :)

sorryyy not sorry lol

#respectfully

5.
X
1 1 2 2 3 3 4
FX) -1 3 4 0 5 1 6
Label:
Explanation:

Answers

Answer:

Not a function

Step-by-step explanation:

Each input can have only have one output.

x(1) can't equal both -1 and 3 at the same time.

x(2) can't equal both 4 and 0 at the same time.

x(3) can't equal both 5 and 1 at the same time.

In other words there can't be multiple X's equalling different things.

Heavy children: Are children heavier now than they were in the past? The National Health and Nutrition Examination Survey (NHANES) taken between 1999 and 2002 reported that the mean weight of six-year-old girls in the United States was 49.3 pounds. Another NHANES survey, published in 2008, reported that a sample of 190 six-year-old girls weighed between 2003 and 2006 had an average weight of 46 pounds. Assume the population standard deviation is =σ17 pounds. Can you conclude that the mean weight of six-year-old girls in 2006 is different from what it was in 2002? Use the =α0.10 level of significance and the critical value method.

Answers

Answer:

Step-by-step explanation:

Hello!

You have two surveys that measure the weight of six-year-old girls in the USA,

1) 1999-2002

μ= 49.3 pounds

(I'll take this mean as the population value since it can be considered "historical data" or point of comparison to make the test.)

2)2003-2006

sample n= 190

sample mean x[bar]= 46 pounds

population standard deviation σ= 17 pounds

Assuming that the study variable X" Weight of six-year-old girls between 2003 - 2006" (pound) has a normal distribution.

If you need to test that the children are heavier now (2003-2006) than in the past (1999-2002) the test hypothesis is:

H₀: μ ≤ 49.3

H₁: μ > 49.3

α: 0.10

The statistic is Z= (x[bar]-μ) ~N(0;1)

(δ/√n)

The critical region is one-tailed to the right.

$Z_(1-\alpha ) = Z_(0.90) = 1.28$

So you'll reject the null hypothesis if the calculated statistic is equal or greater than 1.28.

Z= 46 - 49.3 = -2.67

17/√190

Since the calculated value -2.67 is less than 1.28 you do not reject the null hypothesis. In other words, the six-year-old girls from 2003-2006 are thinner than the girls from 1999-2002.

I hope it helps!

In a study of 788 randomly selected medical malpractice lawsuits, it was found that 494 of them were dropped or dismissed. Use a 0.05 significance level to test the claim that most medical malpractice lawsuits are dropped or dismissed.Which of the following is the hypothesis test to be conducted?
A.
Upper H 0 : p less than 0.5
Upper H 1 : p equals 0.5
B.
Upper H 0 : p greater than 0.5
Upper H 1 : p equals 0.5
C.
Upper H 0 : p equals 0.5
Upper H 1 : p not equals 0.5
D.
Upper H 0 : p equals 0.5
Upper H 1 : p less than 0.5
E.
Upper H 0 : p not equals 0.5
Upper H 1 : p equals 0.5
F.
Upper H 0 : p equals 0.5
Upper H 1 : p greater than 0.5
What is the test statistic?
Z =
(Round to two decimal places as needed.)
What is the conclusion about the null hypothesis?
A. Reject the null hypothesis because the P-value is greater than the significance level, alpha.
B. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, alpha.
C. Fail to reject the null hypothesis because the P-value is greater than the significance level, alpha.
D. Reject the null hypothesis because the P-value is less than or equal to the significance level, alpha.
What is the final conclusion?
A.There is sufficient evidence to support the claim that most medical malpractice lawsuits are dropped or dismissed.
B.There is not sufficient evidence to support the claim that most medical malpractice lawsuits are dropped or dismissed.
C.There is sufficient evidence to warrant rejection of the claim that most medical malpractice lawsuits are dropped or dismissed.
D.There is not sufficient evidence to warrant rejection of the claim that most medical malpractice lawsuits are dropped or dismissed.

Answers

Answer:

Step-by-step explanation:

A. Upper H 0 : p equals 0.5

Upper H 1 : p not equals 0.5

B. Using the tests promotion formula, we have (p - P) / √P(1-P)

Where p (sample promotion) = 494/788 = 0.6269, P (population proportion) = 0.5,

(0.6269 - 0.5) / (√0.5(1-0.5))

0.1269/ √(0.5 (0.5))

0.1269/ √0.25

0.1269/0.5

Test statistics is equal to 0.2538

C. We will use the p value to determine our result, thus the p value at 0.05 level of significance is 0.79965, thus we fail to reject the null hypothesis because the P-value is greater than the significance level, alpha.

Then we conclude that There is not sufficient evidence to support the claim that most medical malpractice lawsuits are dropped or dismissed.

sarah can complete a project in 90 minutes and her sister betty can complete it in 120 minutes if they both work on the project at the same time how long will it take them to complete the project

Answers

Answer:

It will take them approximately 51.43 minutes to complete the project together

Step-by-step explanation:

This is what is called a "shared job" problem.

The best way to work on them is to start by finding the "portion" of the job done by each of the people in the unit of time.

So, for example, Sarah completes the project in 90 minutes, so in the unit of time (that is 1 minute) she completed 1/90 of the total project

Betty completes the project in 120 minutes, so in the unit of time (1 minute) she completes 1/120 of the total project.

We don't know how long it would take for them to complete the project when working together, so we call that time "x" (our unknown).

Now, when they work together completing the entire job in x minutes, in the unit of time they would have done 1/x of the total project.

In the unite of time, the fraction of the job done together (1/x) should equal the fraction of the job done by Sarah (1/90) plus the fraction of the job done by Betty. This in mathematical form becomes:

$(1)/(x) =(1)/(90) +(1)/(120)\n(1)/(x) =(4)/(360) +(3)/(360)\n(1)/(x) =(7)/(360) \nx=(360)/(7) \nx=51.43\,\,min$

So it will take them approximately 51.43 minutes to complete the project together.

The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airports. The maximum possible rating is 10. Suppose a simple random sample of 50 business travelers is selected and each traveler is asked to provide a rating for the Miami International Airport. The ratings obtained from the sample of 50 business travelers follow. Click on the datafile logo to reference the data.
6 4 6 8 7 7 6 3 3 8 10 4 8
7 8 7 5 9 5 8 4 3 8 5 5 4
4 4 8 4 5 6 2 5 9 9 8 4 8
9 9 5 9 7 8 3 10 8 9 6
Develop a 95% confidence interval estimate of the population mean rating for Miami. If required, round your answers to two decimal places. Do not round intermediate calculations.

Answers

Answer:

The 95% confidence interval estimate of the population mean rating for Miami is (5.7, 7.0).

Step-by-step explanation:

The (1 - α)% confidence interval for the population mean, when the population standard deviation is not provided is:

$CI=\bar x\pm t_(\alpha/2, (n-1))\cdot\ (s)/(√(n))$

The sample selected is of size, n = 50.

The critical value of t for 95% confidence level and (n - 1) = 49 degrees of freedom is:

$t_(\alpha/2, (n-1))=t_(0.05/2, 49)=2.000$

*Use a t-table.

Compute the sample mean and sample standard deviation as follows:

$\bar x=(1)/(n)\sum {x}=(1)/(50)* [6+4+6+...+9+6]=6.34\n\ns=\sqrt{(1)/(n-1)\sum (x-\bar x)^(2)}=\sqrt{(1)/(50-1)* 229.22}=2.163$

Compute the 95% confidence interval estimate of the population mean rating for Miami as follows:

$CI=\bar x\pm t_(\alpha/2, (n-1))\cdot\ (s)/(√(n))$

$=6.34\pm 2.00*(2.163)/(√(50))\n\n=6.34\pm 0.612\n\n=(5.728, 6.952)\n\n\approx(5.7, 7.0)$

Thus, the 95% confidence interval estimate of the population mean rating for Miami is (5.7, 7.0).

Other Questions

One sample has a mean of and a second sample has a mean of . The two samples are combined into a single set of scores. What is the mean for the combined set if both of the original samples have scores

Solve the inequality for x. –8x + 16 > –72

A bag contains 10 w h i t e marbles, 5 g r e e n marbles, 6 y e l l o w marbles. If two different marbles are drawn from the bag , what is the probability of drawing a w h i t e marble then a y e l l o w marble?

In a study of factors affecting whether soldiers decide to reenlist, 320 subjects were measured for an index of satisfaction. The sample mean is 28.8 and the sample standard deviation is 7.3. Use the given sample data to construct the 98 percent confidence interval for the population mean.