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Elis [28]
3 years ago
9

A 10-item statistics quiz was given to 30 students. The table below gives the scores received along with the corresponding frequ

encies.
A 2-column table with 6 rows. Column 1 is labeled score with entries 5, 6, 7, 8, 9, 10. Column 2 is labeled Frequency with entries 1, 2, 5, 5, 7, 10.

What was the mean score on the quiz?

7.5
8.5
9
10
Mathematics
1 answer:
hichkok12 [17]3 years ago
3 0

Answer:

Should be (B).

8.5

ED2021

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The sum of the angle measures of a polygon with n sides is 18001800. find n.
Contact [7]
Using
(n-2)180 = 18001800
(n-2) = 18001800÷180
n-2 = 100010
n = 100010+2
n= 100012 sides
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2 years ago
Does the equation 6x + 35 = -6x - 35 have no, one, or infinite solutions?
fenix001 [56]

Answer:

The equation has one solution.

Step-by-step explanation:

6x + 35 = -6x - 35

subtract 6x from both sides of the equation

35 = -35 - 12x

add 35 to both sides of the equation

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divide -12x on both sides of the equation

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2 years ago
A forester studying diameter growth of red pine believes that the mean diameter growth will be different from the known mean gro
-Dominant- [34]

Answer:

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

Step-by-step explanation:

Data given and notation  

\bar X=1.6 represent the sample mean

s=0.46 represent the sample deviation

n=32 sample size  

\mu_o =1.35 represent the value that we want to test  

\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.  

We need to conduct a hypothesis in order to check if the true mean is different from 1.35 in/year, the system of hypothesis would be:  

Null hypothesis:\mu =1.35  

Alternative hypothesis:\mu \neq 1.35  

The statistic is given by:

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

Calculate the statistic  

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

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

P-value  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

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Consider the expression 1.1(1.3)^x+4 Which of the following is an equivalent expression? A. 1.1(1.3)^4x B. 1.1(1.3)^x/4 C. 1.1(1
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Answer:

d

Step-by-step explanation:

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3 years ago
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