1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Kryger [21]
3 years ago
9

Please help math problem

Mathematics
1 answer:
dedylja [7]3 years ago
8 0

Answer :im not very sureeee

Step-by-step explanation:

You might be interested in
Which graph represents the function below?
KATRIN_1 [288]
The answer to this problem is A
3 0
3 years ago
Read 2 more answers
What is 487557066895 x 589032760679? Best answer explain get Brainliest!!!!!!
Mila [183]
2.8718709e+23
can I be brainliest
8 0
3 years ago
The manager wanted to know the overall quality of a large batch of products. The quality management team selected a sample of 10
ElenaW [278]

Answer:

a) The 95% confidence level two-tail confidence interval for the mean value of the key index of this batch is between 98.22 and 99.78

b) The minimum sample size to achieve this is 246.

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{1-0.95}{2} = 0.025

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

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

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(square root of the variance) and n is the size of the sample. So in this question, \sigma = \sqrt{16} = 4

M = 1.96\frac{4}{\sqrt{100}} = 0.78

The lower end of the interval is the sample mean subtracted by M. So it is 99 - 0.78 = 98.22

The upper end of the interval is the sample mean added to M. So it is 99 + 0.78 = 99.78.

a) The 95% confidence level two-tail confidence interval for the mean value of the key index of this batch is between 98.22 and 99.78

(b) (5 points) If we want the sampling error to be no greater than 0.5, what is the minimum sample size to achieve this based on the same confidence level with part (a)

We need a sample size of n

n is found when M = 0.5

Then

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

0.5 = 1.96*\frac{4}{\sqrt{n}}

0.5\sqrt{n} = 4*1.96

\sqrt{n} = \frac{4*1.96}{0.5}

(\sqrt{n})^{2} = (\frac{4*1.96}{0.5})^{2}

n = 245.86

Rounding up

The minimum sample size to achieve this is 246.

6 0
3 years ago
How to spell 15.062 in word form
Tema [17]
Fifteen and sixty-two thousandths
3 0
3 years ago
Read 2 more answers
Please Help!
seropon [69]

Answer:

46.78 PROBIBILITY

Step-by-step explanation:

MATHS

8 0
3 years ago
Other questions:
  • Aseem and Stana are building model cars. Stana's car is 5 less than 2 times the length of Aseem's car. The sum of the lengths of
    7·1 answer
  • What is the perimeter of a sandbox with the coordinates of (12,14),(12,17),and (16,14)
    7·1 answer
  • What is cosine ? I really need this answer for my test tomorrow. Please help
    14·2 answers
  • The length of a rectangular state park is 2 miles longer than twice the width. The area of the park is 84 square miles. What is
    6·1 answer
  • Rewrite 9 over 6 as a equivalent fraction and percent
    13·1 answer
  • Do any three points not on the same line always, sometimes, or never determine a plane? Explain
    7·1 answer
  • Evaluate tan( – 33pi/4)​
    13·2 answers
  • Help mesothelioma this pleaseeee
    6·1 answer
  • Abe has 150 t-shirts to sell before the playoffs begin. The table shows the linear relationship between the number of t-shirts r
    8·1 answer
  • Find angle A and angle E
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!