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3 years ago

HELPPPPPP HURRYYYY Which best explains why these two figures are similar or not similar?

Mathematics

2 answers:

Aloiza [94]3 years ago

7 0

Answer:

These two figures are similar because 5/3 equals 15/9.

Step-by-step explanation:

If two figures are similar, their corresponding sides are proportional.

This means comparing the sides, we get equivalent ratios.

The ratio of the sides of the smaller rectangle is 5/3. The ratio of the sides of the larger rectangle is 15/9.

5/3(3/3) = 15/9; this means these are equivalent ratios, which means the corresponding sides are proportional and the figures are similar.

elixir [45]3 years ago

4 0

Answer:

The answer is the bottom right answer.

Step-by-step explanation:

The figures are similar because the lengths of corresponding pairs of sides are in the same ratio.

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Step-by-step explanation:

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3 years ago

An educator claims that the average salary of substitute teachers in school districts is less than $60 per day. A random sample

valentina_108 [34]

Answer:

$t=\frac{58.875-60}{\frac{5.083}{\sqrt{8}}}=-0.626$

The degrees of freedom are given by:

$df=n-1=8-1=7$

The p value would be given by:

$p_v =P(t_{(7)}$

Since the p value is higher than 0.1 we have enough evidence to FAIl to reject the null hypothesis and we can't conclude that the true mean is less than 60

Step-by-step explanation:

Information given

60, 56, 60, 55, 70, 55, 60, and 55.

We can calculate the mean and deviation with these formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

Replacing we got:

$\bar X=58.875$ represent the mean

$s=5.083$ represent the sample standard deviation for the sample

$n=8$ sample size

$\mu_o =60$ represent the value that we want to test

$\alpha=0.1$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to test if the true mean is less than 60, the system of hypothesis would be:

Null hypothesis: $\mu \geq 60$

Alternative hypothesis: $\mu < 60$

The statistic would be given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info we got:

$t=\frac{58.875-60}{\frac{5.083}{\sqrt{8}}}=-0.626$

The degrees of freedom are given by:

$df=n-1=8-1=7$

The p value would be given by:

$p_v =P(t_{(7)}$

Since the p value is higher than 0.1 we have enough evidence to FAIl to reject the null hypothesis and we can't conclude that the true mean is less than 60

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3 years ago

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Because everything in the problem is positive, no negative numbers are needed.

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3 years ago

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