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rosijanka [135]
3 years ago
9

Find the radius of a circle in which an inscribed square has a side of 8 inches.

Mathematics
1 answer:
tatiyna3 years ago
4 0

Answer:

r = 4

Step-by-step explanation:

side of square is the same as the diameter of the circle

if diameter is 8 then radius is 4

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A collection of dimes and nickels is worth $8.40 there are 106 coins how many of each are there
kotegsom [21]
There are 62 dimes and 44 nickels.

N + D = 106 (There are 106 coins in all.)
1 nickel = 5 cents.
1 dime = 10 cents.
5*N + 10*D = 840 ($8.40 = 840 cents)
N + D = 106 -------(1)
N = (106 - D)
5*N + 10*D = 840 ---------(2)
5*(106-D) + 10*D = 840
530 - 5*D + 10*D = 840
5*D = 840 - 530
5*D = 310
D = 310/5 = 62
Eq(1) N + D = 106
N + 62 = 106
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A zoologist is studying four very closely related feline species. She wishes to compare their gestation periods. An observationa
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Answer:

p_v= 0.01

Since the significance level is 0.05 we see that pv so we have enough evidence to reject the null hypothesis. And the best conclusion for this case would be:

b. at least some, but not all, of the gestation periods across all four species are the same

Because is only to identify if AT LEAST one mean is different, NOT to conclude that the all the means are different.

Step-by-step explanation:

Previous concepts

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".  

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"  

Solution to the problem

The hypothesis for this case are:

Null hypothesis: \mu_{A}=\mu_{B}=\mu_{C}= \mu_D

Alternative hypothesis: Not all the means are equal \mu_{i}\neq \mu_{j}, i,j=A,B,C,D

In order to find the mean square between treatments (MSTR), we need to find first the sum of squares and the degrees of freedom.

If we assume that we have p=4 groups and on each group from j=1,\dots,p we have n_j individuals on each group we can define the following formulas of variation:  

SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2  

SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2  

SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2  

And we have this property  

SST=SS_{between}+SS_{within}  

And in order to test this hypothesis we need to ue an F statistic and for this case the p value calculated is

p_v= 0.01

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b. at least some, but not all, of the gestation periods across all four species are the same

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