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Temka [501]
1 year ago
6

Francine applies the commutative property to the product. (−8/9)⋅(−13)⋅(18) Which expression illustrates the commutative propert

y applied to the product? (−8/9)⋅(18)⋅(−13) [(−8/9)⋅(−13)]⋅(18) (−8/9)⋅[(−13)⋅(18)] (−8/9)⋅(−13)⋅(18)
Mathematics
2 answers:
oksian1 [2.3K]1 year ago
4 0

The commutative property implies that, the order of multiplication is not important.

(-\frac 89) \cdot (18) \cdot (-13) is true

Commutative property of multiplication states that:

If a \times b  = b \times a

The expression is given as:

(-\frac 89)\cdot (-13) \cdot (18)

To get the commutative expression, we simply change the position of each factor in the expression.

So, some possible equivalent expressions are:

(18) \cdot (-13) \cdot (-\frac 89)

(18)\cdot (-\frac 89) \cdot (-13)

(-13) \cdot (18) \cdot (-\frac 89)

There are several other equivalent expressions.

From the list of options;

(-\frac 89) \cdot (18) \cdot (-13) is true

Read more about commutative property at:

brainly.com/question/7037119

Firlakuza [10]1 year ago
4 0

Answer:(−8/9)⋅(18)⋅(−13)

Step-by-step explanation: I took the test ;)

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Granola Crunch cereal is packaged in 1 pound boxes. Susan Torres, a quality control analyst who works for the manufacturer of th
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Answer:

z=\frac{1.02-1}{\frac{0.04}{\sqrt{40}}}=3.162    

p_v =2*P(z>3.162)=0.0016    

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to to reject the null hypothesis, so we can conclude that the true mean is different from 1 pound at 5% of signficance.

Step-by-step explanation:

Data given and notation    

\bar X=1.02 represent the sample mean

\sigma=0.04 represent the population standard deviation    

n=40 sample size    

\mu_o =1 represent the value that we want to test  

\alpha=0.05 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 true mean is equal to 1 pound or no :    

Null hypothesis:\mu =1    

Alternative hypothesis:\mu \neq 1    

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:    

z=\frac{1.02-1}{\frac{0.04}{\sqrt{40}}}=3.162    

P-value    

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

p_v =2*P(z>3.162)=0.0016    

Conclusion    

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to to reject the null hypothesis, so we can conclude that the true mean is different from 1 pound at 5% of signficance.    

5 0
3 years ago
SHORT ANSWER
Zigmanuir [339]
Gradient of the line formed;
m=\frac{change in y}{change in x}
m=\frac{-2-1}{6--5}
m=\frac{-3}{11}
y=mx+c
y=\frac{-3}{11} x+c
Replacing for x and y;
1=-3/11*(-5)+c
1=15/11+c
c=1-15/11
c=-4/11
y=\frac{-3}{11} x- \frac{4}{11}

4 0
3 years ago
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