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alex41 [277]
2 years ago
8

I need help someone solve

Mathematics
2 answers:
ohaa [14]2 years ago
8 0

Answer: what is your question?

Step-by-step explanation:

lozanna [386]2 years ago
5 0

Answer:

What is ur question I'll help

Step-by-step explanation:

But can u pls mark me brainliest

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-10-3+6= -7

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What's the difference between 1 and 1 3/5
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Solve the system:<br> x + 5y = -15<br> -2x -10y = 30*
Ksju [112]

Answer:

x= -5/3 y-5

Step-by-step explanation:

6 0
2 years ago
Please help 20 points!!!
sleet_krkn [62]
The red one is (4.1) and the green one is (-2.4) because is positive to negative so the positive side is that you got and the negative part is you owe someone for example: -20 < 30 so if it says Emily goes to the store and the milk costs $30.99 and she only has $10.00 that means that she owes $30.00 so is -10 <30
3 0
3 years ago
A manufacturer uses production method to produce steel rods. A random sample of 17 steel rods resulted in lengths with a standar
Arisa [49]

Answer:

\chi^2 =\frac{17-1}{12.15} 22.09 =15.87

Traditional method

We can find a critical value in the chi square distribution with df =16 who accumulates \alpha/2 =0.05 of the area on each tail and we got:

\chi^2_{crit}= 7.962

\chi^2_{crit}=26.296

Since the calculated value is between the two critical values we don't have enough evidence to conclude that the true deviation is NOT significantly different from 3.5 cm

Confidence level method

We can find the p value like this

p_v =2*P(\chi^2 >15.87)=0.92

Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is NOT significantly different from 3.5 cm

Step-by-step explanation:

Data provided

n=17 represent the sample selected

\alpha=0.1 represent the significance

s^2 =4.7^2 =22.09 represent the sample variance

\sigma^2_0 =3.5^2 =12.15 represent the value to check

Null and alternative hypothesis

We want to verify if the new production method has lengths with a standard deviation different from 3.5 cm, so the system of hypothesis would be:

Null Hypothesis: \sigma^2 = 12.15

Alternative hypothesis: \sigma^2 \neq 12.15

The statistic for this case is given by:

\chi^2 =\frac{n-1}{\sigma^2_0} s^2

The degrees of freedom are:

df =n-1= 17-1=16

\chi^2 =\frac{17-1}{12.15} 22.09 =15.87

Traditional method

We can find a critical value in the chi square distribution with df =16 who accumulates \alpha/2 =0.05 of the area on each tail and we got:

\chi^2_{crit}= 7.962

\chi^2_{crit}=26.296

Since the calculated value is between the two critical values we don't have enough evidence to conclude that the true deviation is significantly different from 3.5 cm

Confidence level method

We can find the p value like this

p_v =2*P(\chi^2 >15.87)=0.92

Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is significantly different from 3.5 cm

7 0
3 years ago
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