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goldfiish [28.3K]
2 years ago
7

HELP PLEASE (Geometric Sequences) 15 points

Mathematics
1 answer:
lora16 [44]2 years ago
4 0
1. 35, 5, 5/7
2. a=4/9
3. Both C and D are correct. a5=6. And a7=96.
4. Yes it is geometric. r=4.

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Assume that you have a sample of n 1 equals 6​, with the sample mean Upper X overbar 1 equals 50​, and a sample standard deviati
tigry1 [53]

Answer:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

df=6+5-2=9

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

n_1 =6 represent the sample size for group 1

n_2 =5 represent the sample size for group 2

\bar X_1 =50 represent the sample mean for the group 1

\bar X_2 =38 represent the sample mean for the group 2

s_1=7 represent the sample standard deviation for group 1

s_2=8 represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 > \mu_2

We are assuming that the population variances for each group are the same

\sigma^2_1 =\sigma^2_2 =\sigma^2

The statistic for this case is given by:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

The pooled variance is:

S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

We can find the pooled variance:

S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67

And the pooled deviation is:

S_p=7.46

The statistic is given by:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

The degrees of freedom are given by:

df=6+5-2=9

The p value is given by:

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

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3 years ago
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STatiana [176]

Answer:

275 miles

Step-by-step explanation:

8 0
2 years ago
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What is lcm of 54 and 64
ankoles [38]
The lcm is 8

7×8=56

6×6=64


7 0
2 years ago
What is -1 3/5 ÷ -2/3?​
elena55 [62]

Answer:

0.26666666666

Step-by-step explanation:

5 0
3 years ago
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The following table summarises the data from a survey on the ownership of iPods among families with different levels of income.
AURORKA [14]

The two nominal variables are related.

The following table summarises the data from a survey on the ownership of iPods among families with different levels of income.  

Ownership           C1                  C2                  C3

No                         40                 32                   48

Yes                        30                 48                   52

The first thing to do in order to determine if they are related or not is to state our null and alternative hypothesis

Null hypothesis

\mathbf{H_o: Two \nomial  \ variables  \ are \  related}

Alternative hypothesis

\mathbf{H_a: Two \nomial  \ variables  \ are \  not \  related}

Using the Chi-square test statistics which can be expressed by using the formula  

X ^2 = \sum \dfrac{(O-E)^2}{E}

Ownership           C1                  C2                  C3                  Total

No                         40                 32                   48                 120

Yes                        30                 48                   52                 130

Total                      70                 80                   100                250

The expected values are calculated as:

\mathsf{E_{a,b} = \dfrac{(row \ total \times column \ total )}{grand \ total }}

\mathsf{E_{1,1} = \dfrac{(70 \times120 )}{250 }}

\mathsf{E_{1,1} = 33.6}

\mathsf{E_{1,2} = \dfrac{(70 \times130 )}{250 }}

\mathsf{E_{1,2} = 36.4}

\mathsf{E_{2,1} = \dfrac{(80 \times120 )}{250 }}

\mathsf{E_{2,1} = 38.4}

\mathsf{E_{2,2} = \dfrac{(80 \times 130 )}{250 }}

\mathsf{E_{2,2} = 41.6}

\mathsf{E_{3,1} = \dfrac{(100 \times120 )}{250 }}

\mathsf{E_{3,1} = 48}

\mathsf{E_{3,2} = \dfrac{(70 \times130 )}{250 }}

\mathsf{E_{3,2} = 52}

∴ Using the Chi-square test statistics, we have:

X ^2 = \sum \dfrac{(O-E)^2}{E}

X ^2 = \Bigg(  \dfrac{(40-33.6)^2}{33.6}+   \dfrac{(30-36.4)^2}{36.4}+   \dfrac{(32-38.4)^2}{38.4}+  \dfrac{(48-41.6)^2}{41.6} +   \dfrac{(48-48)^2}{48}+   \dfrac{(52-52)^2}{52} \Bigg)

X ^2 = \Bigg(  \dfrac{40.96}{33.6}+   \dfrac{40.96}{36.4}+   \dfrac{40.96}{38.4}+  \dfrac{40.96}{41.6} +   \dfrac{0}{48}+   \dfrac{0}{52} \Bigg)

X ^2 = \Bigg(  1.2190+   1.1253+  1.0667+ 0.9846+0+ 0 \Bigg)

\mathbf{X ^2 =4.3956}

The degree of freedom df = ((r - 1) × (c - 1))

= (3 - 1) (2 -1 )

= 2 × 1

= 2

∴

Assuming the level of significance = 5%

The p-value of the Chi-square test statistics at df of 2 is:

= \mathbf{P(X^2 > 4.3956) \implies 0.111}

Therefore, we can conclude that since the p-value (0.111) is greater than the level of significance (0.05), we fail to reject the null hypothesis.

Hence, the two nominal variables are related.

Learn more about Chi-square test statistics here:

brainly.com/question/2365682?referrer=searchResults

7 0
2 years ago
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