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sergeinik [125]
3 years ago
12

Out of a class of 150 students, two-fifths opted for German classes and one-third opted for Italian classes. The remaining opted

for French. What fraction opted for French?
Mathematics
1 answer:
Yuliya22 [10]3 years ago
8 0

Fraction of students that opted for French = \frac{4}{15}

The total number of students = 150

Fraction of students that opted for German = \frac{2}{5}

Fraction of students that opted for Italian = \frac{1}{3}

Fraction of students that opted for French = 1 - Fraction of students that opted for German - Fraction of students that opted for Italian

Fraction of students that opted for French = 1-\frac{2}{5} - \frac{1}{3}

Fraction of students that opted for French = \frac{15-6-5}{15}

Fraction of students that opted for French = \frac{4}{15}

Learn more here: brainly.com/question/18819021

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Answer:

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We have to test the hypothesis of the difference between means.

The claim is that Portland has more average yearly rainfall than Seattle.

Being μ1: average rainfall in Portland, μ2: average rainfall in Seattle, the null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2 > 0

The significance level is 0.10.

The sample for Portland, of size n1=45, has a mean of M1=37.50 and standard deviation of s1=1.82.

The sample for Seattle, of size n1=35, has a mean of M1=37.07 and standard deviation of s1=1.68.

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The standard error for the difference between means is:

s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{1.82^2}{45}+\dfrac{1.68^2}{35}}=\sqrt{ 0.0736+0.0688 }=\sqrt{0.1424}\\\\\\s_{M_d}=0.3774

We can calculate the t-statistic as:

t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{0.43-0}{0.3774}=1.1393

The degrees of freedom are:

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Then, the p-value for this one-tailed test with 78 degrees of freedom is:

P-value=P(t>1.1393)=0.1290

As the P-value is bigger than the significance level, the effect is not significant and the null hypothesis failed to be rejected.

There is no enough evidence to to support the claim that Portland has more average yearly rainfall than Seattle.

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