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Oduvanchick [21]
3 years ago
5

Alana will use a 3/4 - gallon pitcher to fill an empty 2/3 -gallon bucket. How much water will she need to completely fill the b

ucket?
Mathematics
1 answer:
aleksklad [387]3 years ago
8 0

Alana will use 8/9 gallons of the pitcher to fill the bucket

Step-by-step explanation:

The question is on division and capacity

Here you divide the capacity required by the bucket with that provided by the pitcher.

This is: 2/3 ÷ 3/4

          =  2/3 * 4/3

          =8/9

Alana will use 8/9 gallons of the pitcher to fill the bucket

Learn More

To perform division:brainly.com/question/6754359

Keywords : gallon, pitcher, bucket, fill, completely

#LearnwithBrainly

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Answer:

Case n =5

z=\frac{4.8-5}{\frac{0.5}{\sqrt{5}}}=-0.894  

Case n =15

z=\frac{4.8-5}{\frac{0.5}{\sqrt{15}}}=-1.549  

Case n = 40

z=\frac{4.8-5}{\frac{0.5}{\sqrt{40}}}=-2.530  

P value

Case n =5

p_v =P(z  

Case n =15

p_v =P(z  

Case n =40

p_v =P(z  

Step-by-step explanation:

Data given and notation  

\bar X=4.8 represent the sample mean

\sigma=0.5 represent the population standard deviation

n sample size  

\mu_o =5 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

z 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 mean is lower than 5, the system of hypothesis would be:  

Null hypothesis:\mu \geq 5  

Alternative hypothesis:\mu < 5  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

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

Case n =5

z=\frac{4.8-5}{\frac{0.5}{\sqrt{5}}}=-0.894  

Case n =15

z=\frac{4.8-5}{\frac{0.5}{\sqrt{15}}}=-1.549  

Case n = 40

z=\frac{4.8-5}{\frac{0.5}{\sqrt{40}}}=-2.530  

P-value  

Since is a left tailed test the p value would be:  

Case n =5

p_v =P(z  

Case n =15

p_v =P(z  

Case n =40

p_v =P(z  

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