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
mario62 [17]
3 years ago
8

Solve the following: NO BOTS OR PEOPLE WANTING POINTS!

Mathematics
1 answer:
Ksenya-84 [330]3 years ago
8 0
Memes memes memes memes

You might be interested in
A study was conducted to determine whether magnets were effective in treating pain. The values represent measurements of pain us
amm1812

Answer:

Given:

Sham: n= 20,     x=0.44,   s=1.24,

Magnet:n= 20,  x =0.49,   s= 0.95

For Sham:

Sample size, n = 20

Sample mean = 0.44

Standard deviation = 1.24

For Magnet:

Sample size = 20

Sample mean = 0.49

Standard deviation = 0.95

The null and alternative hypotheses:

H0: s1²=s2²

H1: s1² ≠ s2²

a) To find the test statistics, use the formula:

\frac{s1^2}{s2^2}

\frac{1.24^2}{0.95^2} = \frac{1.5376}{0.9025} = 1.7037

Test statistics = 1.7037

b) P-value:

Sham: degrees of freedom = n - 1 = 20 - 1 = 19

Magnet: degrees of freedom = n - 1 = 20 - 1 = 19

The critical values:

[Za/2, df1, df2)], [(1 - Za/2), df1, df2]

f[0.05/2, 19, 19], f[(1 - 0.05/2), 19, 19]

f[0.025, 19, 19], f[0.975, 19, 19]

(2.526, 0.3958)

The rejection region:

Reject H0, if  F < 0.3958 or if F > 2.526

c) Conclusion:

Since the critical values of test statistic is between (0.3958 < 1.7037 < 2.526), we fail to reject null hypothesis H0.

There is insufficient evidence to to support the claim that those given a sham treatment have reductions that vary more than those treated with magnets

3 0
3 years ago
Is 16 divided by 3 a repeating decimal
Nady [450]

Answer:

Yes

Step-by-step explanation:

16 ÷ 3 = 5.333333...

3 0
3 years ago
Read 2 more answers
Offering 40 points and brainliest for answer
Nata [24]
In fact, this problem belongs to the chemistry section.  Recall that many other sciences require mathematical calculations.  The problem will belong to Mathematics only if no knowledge of other sciences are required to solve the problem.

Solubility for the given substances is measured in grams per 100 g of water at a particular temperature (20 deg.C).
This means that the mass (assumed to be the solute) will not change the solubility, just the minimum quantity of solvent (water) will.
Thus the solubility of sodium chloride will remain L=36 g/100g H2O for any quantity of solute.  Similarly, the solubility of lead nitrate will remain as K=54 g/100 g H2O.

The reason that they remain constant is because the quantity of solvent (water) is fixed at 100 g.  Varying amount of solute will affect the quantity of solvent required, but not the solubility.
I'll leave it to you to calculate the difference between K & L.
8 0
3 years ago
Question 20 need an answerw
max2010maxim [7]
Which one is question 20??
8 0
3 years ago
Tessa has $42 to spend at the county fair. Her admission
Semenov [28]

Answer

g is less than or equal to 34.50

Step-by-step explanation:

$42-$7.50=$34.50

7 0
3 years ago
Other questions:
  • Write an expression to find the difference in the maximum area and minimum area of nys high school soccer field. Then evaluate y
    6·1 answer
  • Quick question what’s 2ab+3b+7b+3a•3a•b+4b-8ab•8ab+5a•5a•b-12ab?
    5·1 answer
  • 14. 7 points and brainliest answer if you get it right
    5·1 answer
  • How to find the volume of a rectangular prism
    15·2 answers
  • The difference of two numbers is 5, the numbers are in the ratio 3:4. So the numbers are 15 and 20 True or False
    12·2 answers
  • 2(4x + 1) &lt; 3(2x - 3)
    6·1 answer
  • If x2-y2 = 10 and x+y = 5, what is the value of x-y? 2-5?
    10·1 answer
  • What are the answers?
    9·1 answer
  • The figures above are examples of
    12·2 answers
  • Please Answer <br><br> This is due by tomorrow night and I need help
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!