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taurus [48]
3 years ago
7

What is the value of |5| - 4(3² - 2)

Mathematics
1 answer:
marta [7]3 years ago
8 0

Answer:

-23

Step-by-step explanation:

Follow pemdas

|5|-4(3²-2)

1.  3²=9

2.(9-2)= 7

3. 4×7=28

4. |5|-28= -23

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The answer is going to be 25 2/5.
Steps i used to get the answer
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Which statement about proportional relationships is false?
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Step-by-step explanation:

I just took the quiz it’s right !

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What is the converse of the statement "If Bob does his homework, then George gets candy"? *
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Which equation would best help solve the following problem? Hideki throws a baseball ball upward with an initial velocity of 12
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8 0
3 years ago
A professor would like to test the hypothesis that the average number of minutes that a student needs to complete a statistics e
professor190 [17]

Answer:

\chi^2 =\frac{15-1}{25} 16 =8.96

The degrees of freedom are given by:

df = n-1 = 15-1=14

The p value for this case would be given by:

p_v =P(\chi^2

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not ignificantly lower than 5 minutes

Step-by-step explanation:

Information given

n=15 represent the sample size

\alpha=0.05 represent the confidence level  

s^2 =16 represent the sample variance

\sigma^2_0 =25 represent the value that we want to  verify

System of hypothesis

We want to test if the true deviation for this case is lesss than 5minutes, so the system of hypothesis would be:

Null Hypothesis: \sigma^2 \geq 25

Alternative hypothesis: \sigma^2

The statistic is given by:

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

And replacing we got:

\chi^2 =\frac{15-1}{25} 16 =8.96

The degrees of freedom are given by:

df = n-1 = 15-1=14

The p value for this case would be given by:

p_v =P(\chi^2

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not ignificantly lower than 5 minutes

6 0
3 years ago
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