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Yakvenalex [24]
3 years ago
7

Evaluate 8r -rs when r=6 and s=s

Mathematics
1 answer:
tankabanditka [31]3 years ago
3 0
Well since 8(6) = 48 then 48 - 6s = x, I suppose correct me if i'm wrong. I do hope this helps :)
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Step-by-step explanation:

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Chang wants to attend a two-year university with an annual tuition of $3,250. He has $2,000 in savings and earned a work-study f
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A small company's net income for the first six months of the year was 76,500 for the last six months it was 100,000 what is the
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The ratio of net income in the first 6 months, to the last six months is $76,500 / $100,000. This simplifies intuitively as follows:

76500/100000

Dividing by 100: 765/1000

Dividing by 5: 153/200

The denominator 200 is only divisible by the prime numbers 2 and 5, and since the numerator 153 is not divisible by either 2 or 5, this means that this is in simplest form, and the final answer is 153/200.

Step-by-step explanation:

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3 years ago
A journalist reported that the average amount of time that a French person spends eating lunch at a restaurant is 22 minutes. Pe
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Answer:

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

Where t follows a t distribution with n_1+n_2 -2 degrees of freedom and the pooled variance S^2_p is given by this formula:

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

t=\frac{19 -22)-(0)}{4.095\sqrt{\frac{1}{8}+\frac{1}{7}}}=-1.416

Step-by-step explanation:

Data given

American: 21,17,17,20,25,16,20,16 (Sample 1)

French: 24,18,20,28,18,29,17 (Sample 2)

When we have two independent samples from two normal distributions with equal variances we are assuming that  

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

And the statistic is given by this formula:

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

Where t follows a t distribution with n_1+n_2 -2 degrees of freedom and the pooled variance S^2_p is given by this formula:

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

This last one is an unbiased estimator of the common variance \sigma^2

The system of hypothesis on this case are:

Null hypothesis: \mu_1 = \mu_2

Alternative hypothesis: \mu_1 \neq \mu_2

Or equivalently:

Null hypothesis: \mu_1 - \mu_2 = 0

Alternative hypothesis: \mu_1 -\mu_2 \neq 0

Our notation on this case :

n_1 =8 represent the sample size for group 1

n_2 =7 represent the sample size for group 2

\bar X_1 =19 represent the sample mean for the group 1

\bar X_2 =22 represent the sample mean for the group 2

s_1=3.117 represent the sample standard deviation for group 1

s_2=5.0 represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

S^2_p =\frac{(8-1)(3.117)^2 +(7 -1)(5.0)^2}{8 +7 -2}=16.770

And the deviation would be just the square root of the variance:

S_p=4.095

And now we can calculate the statistic:

t=\frac{19 -22)-(0)}{4.095\sqrt{\frac{1}{8}+\frac{1}{7}}}=-1.416

Now we can calculate the degrees of freedom given by:

df=8+7-2=13

And now we can calculate the p value using the altenative hypothesis:

p_v =2*P(t_{13}

So with the p value obtained and using the significance level assumed \alpha=0.1 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 10% of significance we don't have significant differences between the two means.  

3 0
3 years ago
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