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g100num [7]
3 years ago
5

What is m∠ BAC???????

Mathematics
2 answers:
Korvikt [17]3 years ago
4 0

Answer:

36 degrees

Step-by-step explanation:

I’m pretty sure that’s it

faltersainse [42]3 years ago
4 0
36. That’s the answer mate
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A zoologist is studying four very closely related feline species. She wishes to compare their gestation periods. An observationa
diamong [38]

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

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.

8 0
3 years ago
What is the measure of angle CAE??
raketka [301]
M <CAE = 360 - 240  = 120 degrees answer
4 0
3 years ago
A survey was taken of students in math classes to find out how many hours per day students spend
Hoochie [10]

Answer:

1. Mean, because there are no outliers that affect the center

Step-by-step explanation:

Second period: 3,2,3,1,3, 4, 2, 4, 3, 1, 0, 2, 3, 1, 2

Sorted values : 0, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4

The mean = ΣX / n

n = sample size, n = 15

Mean = 34 / 15 = 2.666

The median = 1/2(n+1)th term.

1/2(16)th term = 8th term.

The 8th term = 2

The best measure of centre is the mean because the values for the second period has no outliers that might have affected the centre of the distribution.

Both interquartile range and standard deviation are measures of spread and not measures of centre.

6 0
3 years ago
Ivan and Tanya share £150 in the ratio 4: 1<br>Work out how much more Ivan gets compared to Tanya.​
Anastasy [175]

Answer:

Ivan got £90 more than Tanya.

Step-by-step explanation:

Ivan and Tanya share the amount £150 in the ratio of 4 : 1

Amount received by Ivan = 150\times \frac{4}{(4+1)}

                                          = 150\times \frac{4}{5}

                                          = £120

Similarly, amount received by Tanya = 150\times \frac{1}{(4+1)}

                                                             = £30

Difference in their amounts = 120 - 30

                                              = £90

Therefore, Ivan got £90 more than Tanya.                                                                                                        

7 0
3 years ago
Express f(x) = |x-2| +|x+2| in the non-modulus form. Hence, sketch the graph of f.
alexgriva [62]
Recall that

|x|=\begin{cases}x&\text{if }x\ge0\\-x&\text{if }x

There are three cases to consider:

(1) When x+2, we have |x+2|=-(x+2) and |x-2|=-(x-2), so

|x-2|+|x+2|=-(x-2)-(x+2)=-2x-4

(2) When x+2\ge0 and x-2, we get |x+2|=x+2 and |x-2|=-(x-2), so

|x-2|+|x+2|=-(x-2)+(x+2)=4

(3) When x-2\ge0, we have |x+2|=x+2 and |x-2|=x-2, so

|x-2|+|x+2|=(x-2)+(x+2)=2x

So

|x-2|+|x+2|=\begin{cases}-2x-4&\text{if }x
4 0
3 years ago
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