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lora16 [44]
4 years ago
13

A small animal can eat an amount of food that is equivalent to 2/5 it's body weight in 1/2 hour at this rate what fraction of it

s body weight in food can the animal eat in 1 hour
Mathematics
2 answers:
Kruka [31]4 years ago
7 0

Answer:

Step-by-step explanation:

Think:  2/5 its body weight in 1/2 hour immediately becomes 4/5 its body weight in 1 hour.  There's a proportional relationship here.

andre [41]4 years ago
5 0

Answer:The fraction of food that the small animal can eat in one hour is 4/5

Step-by-step explanation:

Let x represent the body weight of the small animal. The small animal can eat an amount of food that is equivalent to 2/5 it's body weight in 1/2 hour. This means that the quantity of food that the small animal can eat in 0.5 hours is 2x/5. Therefore, at the same rate, the amount of food that the animal can eat in one hour would be

2 × 2x/5 = 4x/5

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For this case the statistic calculated is t= 1.37

p_v =0.17

So the p value is a very high value and using the significance level  given\alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and there is enough evidence to conclude that we have a difference between the two means at 10% of significance.

Step-by-step explanation:

Data given and notation

\bar X_{A} represent the mean for the sample for A

\bar X_{B} represent the mean for the sample B

s_{A} represent the sample standard deviation for the sample A

s_{B}= represent the sample standard deviation for the sample B

n_{A}=20 sample size for the group poisoned

n_{B}=20 sample size for the group unpoisoned

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the two means are equal or not , the system of hypothesis would be:

Null hypothesis:\mu_{A} - \mu_{B} =0

Alternative hypothesis:\mu_{A} - \mu_{B} \neq 0

For this case is better apply a t test to compare means, and the statistic is given by:

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

For this case the statistic calculated is t= 1.37

Statistical decision

The significance level is 0.1 \alpha=0.1, but we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case:

df=n_{A}+n_{B}-2

The p value on this case is given

p_v =0.17

So the p value is a very high value and using the significance level  given\alpha=0,1 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and there is enough evidence to conclude that we have a difference between the two means at 10% of significance.

5 0
3 years ago
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