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aalyn [17]
3 years ago
11

WHAT ARE THE FOUR MAIN CYCLES THAT OCCUR IN NATURE PLZ HELP

Mathematics
2 answers:
olchik [2.2K]3 years ago
5 0
Water, Carbon Nitrogyn and Rock i believe so yeah that guys right

maw [93]3 years ago
3 0
<span>Rock, Carbon, Water, Nitrogen</span>
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The expression 3(x-9) is equivalent to
Inessa05 [86]
The answer to this question would be D. 3(x)-3(9)
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3 years ago
A cigarette industry spokesperson remarks that current levels of tar are no more than 5 milligrams per cigarette. A reporter doe
d1i1m1o1n [39]

Answer:

t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516  

df=n-1=15-1=14  

Since is a right tailed test the p value would be:  

p_v =P(t_{14}>1.516)=0.076  

If we use a significance level of 0.05 we see that p_v > \alpha and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

Step-by-step explanation:

Data given and notation  

\bar X=5.63 represent the sample mean  

s=1.61 represent the standard deviation for the sample  

n=15 sample size  

\mu_o =15/tex] represent the value that we want to test  &#10;[tex]\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses to be tested  

We need to conduct a hypothesis in order to determine if the average is more than 5.63, the system of hypothesis would be:  

Null hypothesis:\mu \leq 5.63  

Alternative hypothesis:\mu > 5.63  

Compute the test statistic  

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

We can replace in formula (1) the info given like this:  

t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516  

Now we need to find the degrees of freedom for the t distribution given by:

df=n-1=15-1=14  

P value

Since is a right tailed test the p value would be:  

p_v =P(t_{14}>1.516)=0.076  

If we use a significance level of 0.05 we see that p_v > \alpha and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

3 0
3 years ago
Theoretically if a card is chosen 200 times, how many times would you expect to get a card without stripes?
steposvetlana [31]

Answer:

5

Step-by-step explanation:

200/2=100 then after 100 u divide card by 2 so 100/2 and that's 50 and then last divide by 10 because its the circumference so its 5

3 0
2 years ago
an electrician charges $50 to make a house call and $40 for each hour worked. if you have $200 can you afford a repair that take
mariarad [96]
No. you will be short 10$ so it will not work.
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3 years ago
Read 2 more answers
Aileen worked 35.5 hours last week. She earns $7.75 per hour. How much money did she make last week.
andrezito [222]

Answer:

275.125

Step-by-step explanation:

just multiply 35.5 with 7.75 and theres your answer

8 0
3 years ago
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