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natulia [17]
3 years ago
10

A plant cell has length 6.6 times 10 Superscript negative 6 m and width 2.2 times 10 Superscript negative 6 m. What is the ratio

of the plant​ cell's length to its​ width? The ratio of the plant​ cell's length to its width is nothing. ​(Type the ratio as a simplified​ fraction.)
Mathematics
1 answer:
Vinvika [58]3 years ago
8 0

The length/width ratio is 3/1 .

<em>l</em> = 6.6 × 10⁻⁶ m; <em>w</em> = 2.2 × 10⁻⁶ m

<em>l</em>/<em>w</em> = 6.6 × 10⁻⁶ m/2.2 × 10⁻⁶ m = 3/1

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To compare two programs for training industrial workers to perform a skilled job, 20 workers are included in an experiment. Of t
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Answer:

t=\frac{19.1-23.3}{\sqrt{\frac{4.818^2}{10}+\frac{5.559^2}{10}}}}=-1.805  

p_v =P(t_{(18)}

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Step-by-step explanation:

Data given and notation

We can calculate the sample mean and deviation with these formulas:

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

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\bar X_{2}=23.3 represent the mean for the sample mean for 2

s_{1}=4.818 represent the sample standard deviation for the sample 1

s_{2}=5.559 represent the sample standard deviation for the sample 2

n_{1}=10 sample size selected 1

n_{2}=10 sample size selected 2

\alpha represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

p_v represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the average time taken when training under method 1 is less than the average time for Method 2, the system of hypothesis would be:

Null hypothesis:\mu_{1} \geq \mu_{2}

Alternative hypothesis:\mu_{1} < \mu_{2}

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}} (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

t=\frac{19.1-23.3}{\sqrt{\frac{4.818^2}{10}+\frac{5.559^2}{10}}}}=-1.805  

P-value

The first step is calculate the degrees of freedom, on this case:

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