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ValentinkaMS [17]
2 years ago
10

-10+16 divided by (-2)+7 *Please show work!*

Mathematics
2 answers:
wel2 years ago
8 0

Answer:

= 1 1/5

Step-by-step explanation:

= -10+16/(-2)+7

= 6/5

= 1 1/5

Verdich [7]2 years ago
4 0
Here’s my work to your question!

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Sarah has 273 feet of wire to make bead necklaces. If each necklace requires 1 2/3 feet of wire, how many necklaces can she make
o-na [289]

Oh boy, here we go again

First we must convert  1 2/3 to an improper fraction.  By doing this, we get 5/3  (3/3 + 2/3)

So now we have 273 /  5/3

To divide this easier we can do something that when I learned it was called (keep, change, flip)  which basically means keep the first fraction, change the sign from division to multiplication, and flip the second fraction

This now turns into:  273/1 * 3/5

Combine 273 and 3/5

273⋅3/5

Multiply 273

by 3

819/5

is your answer

5 0
3 years ago
Val has 3/4 gallon of milk. He gives 1/2 of it to a friend. How many gallons of milk does Val have left
Murljashka [212]

Step-by-step explanation:

given, ¾-½= ¼gallon milk.

hope this helps you.

8 0
2 years ago
What is the inverse operation of 2/3x=6
Artemon [7]

2 3 x - 6

Since g(f(x))=x g ( f ( x ) ) = x , f−1(x)=3x2+9 f - 1 ( x ) = 3 x 2 + 9 is the inverse of f(x)=23x−6 f ( x ) = 2 3 x - 6

pls brain

8 0
3 years ago
Solve for n: a= p(1+nr)
Harrizon [31]
N=    1     a          this is the answer hope i helped
     -  _ +  _
         r      pr
8 0
3 years ago
Read 2 more answers
In a recent poll, 778 adults were asked to identify their favorite seat when they fly, and 492 of them chose a window seat. Use
seropon [69]

Answer:

Null hypothesis:p \leq 0.5  

Alternative hypothesis:p > 0.5  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

p_v =P(Z>7.36)=9.19x10^{-14}  

Since the p value obtained was a very low value and using the significance level given \alpha=0.01 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of adults prefer window seats when they fly is significantly higher than 0.5 .  

Step-by-step explanation:

1) Data given and notation

n=778 represent the random sample taken

X=492 represent the people that chose a window seat.

\hat p=\frac{492}{778}=0.632 estimated proportion of people that chose a window seat.

p_o=0.5 is the value that we want to test

\alpha=0.01 represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the majority of adults prefer window seats when they fly:  

Null hypothesis:p \leq 0.5  

Alternative hypothesis:p > 0.5  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.632 -0.5}{\sqrt{\frac{0.5(1-0.5)}{778}}}=7.36  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.01. The next step would be calculate the p value for this test.  

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

p_v =P(Z>7.36)=9.19x10^{-14}  

5) Conclusion

Since the p value obtained was a very low value and using the significance level given \alpha=0.01 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of adults prefer window seats when they fly is significantly higher than 0.5 .  

6 0
3 years ago
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