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beks73 [17]
3 years ago
10

Now suppose that you have 2 individual and identical oranges. give a generating function for selecting from the oranges.

Mathematics
1 answer:
sweet-ann [11.9K]3 years ago
6 0
12 is the correct answer
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THIS TABLE LINEAR OR NON-LINEAR?
defon

Answer:

yes

Step-by-step explanation:

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2 years ago
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I honestly just don't get this question. I'll give 30 points!!!
Svetllana [295]
I wish I could help but but I can't I'm dumb as a box of rocks
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Subtract:<br> 3 3/5 -1 1/1<br> 1 1/10<br> 2 1/10<br> 2 1/5<br> 2 2/3
VARVARA [1.3K]

Step-by-step explanation:

3 3/5 - 1 1/10

3 6/10 - 1 1/10

2 2/10= 2 1/5 is the answer

5 0
3 years ago
The following prices, in dollars, of 7.5-cubic-foot refrigerators were recorded from a random sample. 314 305 344 283 285 310​ 3
Mariana [72]

Answer:

t=\frac{310.9-300}{\frac{31.09}{\sqrt{10}}}=1.109      

The degrees of freedom are given by:

df=n-1=10-1=9  

And the p value would be given by:

p_v =P(t_{9}>1.109)=0.148  

Since the p value is higher than the significance level of 0.05 we don't have enough evidence to conclude that the true mean is significantly higher than $300 because we FAIL to reject the null hypothesis.

Step-by-step explanation:

Information given

314 305 344 283 285 310​ 383​ 285​ 300​ 300

We can calculate the sample mean and deviation with the following formula:

\bar X =\frac{\sum_{i=1}^n X_i}{n}

s =\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}

\bar X=310.9 represent the sample mean      

s=31.09 represent the standard deviation for the sample      

n=10 sample size      

\mu_o =300 represent the value to test

\alpha=0.05 represent the significance level

t would represent the statistic

p_v represent the p value

Hypothesis to test

We want to verify if the true mean is greater than 300, the system of hypothesis would be:      

Null hypothesis:\mu \leq 300      

Alternative hypothesis:\mu > 300      

The statistic is given by:

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

Replacing the info given we got:

t=\frac{310.9-300}{\frac{31.09}{\sqrt{10}}}=1.109      

The degrees of freedom are given by:

df=n-1=10-1=9  

And the p value would be given by:

p_v =P(t_{9}>1.109)=0.148  

Since the p value is higher than the significance level of 0.05 we don't have enough evidence to conclude that the true mean is significantly higher than $300 because we FAIL to reject the null hypothesis.

6 0
3 years ago
suppose a population consists of 5000 being surveyed could result in sample statistic but not parameter
Talja [164]
Huh? What is the question here???
4 0
2 years ago
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