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Simora [160]
2 years ago
11

Find the Diffrence. 4 2/5 - 1 7/10 = ?

Mathematics
2 answers:
Kisachek [45]2 years ago
8 0

Answer:

1. 9/10

2. 7/3

3. 49/3

Step-by-step explanation:

Alika [10]2 years ago
7 0

Answer:

assuming all the / mean fractions (sorry i'm not sure)

<u>first question:</u>

4 and 2/5 is 22/5 (which is 44/10) and 1 and 7/10  is 17/10 so 44/10 - 17/10 = 27/10 (mixed number: 2 7/10)

<u>second question:</u>

5 and 2/3 is 17/3 (34/6) and 2 and 1/2 is 5/2 (15/6) so 34/6 - 15/6 = 19/6 (mixed number: 3 1/6)

<u>third question:</u>

17 (51/3) and 2 1/3 is 7/3 so 51/3 - 7/3 = 44/3 (mixed number: 14 2/3)

i think these are the answers, hope these help :D

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Answer:

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Step-by-step explanation:

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finlep [7]

Answer:

t=\frac{0.475-0.5}{\frac{0.183}{\sqrt{8}}}=-0.386    

p_v =2*P(t_{(7)}  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is different from 0.5 at 5% of signficance.  So then the claim makes sense

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

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In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=0.475

The sample deviation calculated s=0.183

We need to conduct a hypothesis in order to check if the true mean is 0.5 or no, the system of hypothesis would be:  

Null hypothesis:\mu = 0.5  

Alternative hypothesis:\mu \neq 0.5  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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

t-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:  

t=\frac{0.475-0.5}{\frac{0.183}{\sqrt{8}}}=-0.386    

P-value

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

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Since is a two sided test the p value would be:  

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