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Alexus [3.1K]
3 years ago
10

Water safety standards in southern California suggest that if a concentration of adenoviruses in the water is more than 1,000 pe

r liter, further testing would have to be done to determine if the water was hazardous. A random sample of 45 water sources around San Diego showed an average of 1,023 adenoviruses per liter, with a standard deviation of 138.9. Which hypothesis test should be used to determine if the water samples contain more than the allowed 1,000 adenoviruses per liter
Mathematics
1 answer:
Alborosie3 years ago
7 0

Answer:

I should use a t-distribution test.

Step-by-step explanation:

A t-distribution test is used in lieu of a standard normal distribution test if the population standard deviation is unknown.

In this case, the population standard deviation is unknown but the sample standard deviation is known, therefore, a t-distribution hypothesis test is appropriate to use.

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Can anyone pls help me I really need help pls help me I will mark them as brainiest pls!!! Can u do it in a paper and then send
Tom [10]

Answer:

see them bellow

Step-by-step explanation:

6 0
4 years ago
If two marbles are randomly selected from the bag without replacement, what is the probability that the first marble is blue?
joja [24]

Answer:1,0.5, or 0 it depends

Explanation:

The probability that one event occurred depends on the total of random event, being 1 the most of the probabilities

If the two marbles are blue the probability to select a marble blue is 1

If one of the marbles is blue and the other marble is of any color the probability to select the blue marbles is:

P=1 - 1/2=1/2

Being 1/2 or 0.5 the answer

If any of the marbles are blue then the probabilities are 0

8 0
4 years ago
Christine performed the following experiment 20 times using a book that contains 462 pages.
wel

Using the probability concept, it is found that there is a 0.3 = 30% experimental probability that Christine will randomly choose a page that is divisible by 3.

<h3>What is a probability?</h3>

A probability is given by the <u>number of desired outcomes divided by the number of total outcomes</u>.

In an experimental probability, these number of outcomes are taken from previous trials.

In this problem:

  • The experiment was made 20 times.
  • Of those, in 6 experiments(81, 105, 171, 159, 297, 387) she choose a page that was divisible by 3.

Then:

p = \frac{6}{20} = 0.3

0.3 = 30% experimental probability that Christine will randomly choose a page that is divisible by 3.

You can learn more about the probability concept at brainly.com/question/15536019

7 0
2 years ago
3. Decide whether each equation is true or false, and explain how you know.
BARSIC [14]

Answer:

a.false

b.true

c.false

d.I really can't get about it.

e.false

Step-by-step explanation:

So,here;

a.9.9.3=35 is given

or,243=35 (which is false)

b.7+7+7=3+3+3+3+3+3+3

or,21=21 (which is true)

c.2/3/2=2/3

or,2*2/3=2/3

or,4/3=2/3(which is false)

e.6+6+6=63

or,18=63 (which is false)

5 0
3 years ago
For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4
Elina [12.6K]

Answer:

<em>C.</em> (5-\frac{1}{2})^6

Step-by-step explanation:

Given

15(5)^2(-\frac{1}{2})^4

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

Sum = 2 + 4

Sum = 6

Each term of a binomial expansion are always of the form:

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

Where n = the sum above

n = 6

Compare 15(5)^2(-\frac{1}{2})^4 to the above general form of binomial expansion

(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......

Substitute 6 for n

(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......

[Next is to solve for a and b]

<em>From the above expression, the power of (5) is 2</em>

<em>Express 2 as 6 - 4</em>

(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......

By direct comparison of

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

and

(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......

We have;

^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4

Further comparison gives

^nC_r = 15

a^{n-r} =(5)^{6-4}

b^r= (-\frac{1}{2})^4

[Solving for a]

By direct comparison of a^{n-r} =(5)^{6-4}

a = 5

n = 6

r = 4

[Solving for b]

By direct comparison of b^r= (-\frac{1}{2})^4

r = 4

b = \frac{-1}{2}

Substitute values for a, b, n and r in

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......

Solve for ^6C_4

(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......

<em>Check the list of options for the expression on the left hand side</em>

<em>The correct answer is </em>(5-\frac{1}{2})^6<em />

3 0
3 years ago
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