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umka2103 [35]
3 years ago
9

Elevator 1 moved up 10 feet from the ground level. Its position is labeled as +10. Elevator 2 moved down 5 feet from the ground

level.
Mathematics
1 answer:
Sindrei [870]3 years ago
7 0
-5.
Idk if this is right
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Jackie’s test scores for the first semester are listed below.
jek_recluse [69]

Answer:

The answer is 91

Step-by-step explanation:

90 x 6 = 540

The numbers need to equal 540 all together to make an average of 90.

97 + 86 + 83 + 95 + 88 = 449

540 - 449 = 91

To check:

97 + 86 + 83 + 95 + 88 + 91 = 540

Therefore, the answer is 91

8 0
3 years ago
Solve −6x+6−7=2x+8 for x. How would I solve this?
MakcuM [25]

Answer:

Step-by-step explanation:

−6x+6−7=2x+8

-1-8=2x+6x

8x=-9

x= -\frac{9}{8}

7 0
3 years ago
What is the simplified answer to 5+3w+3-w​
DaniilM [7]

Answer:

<h2>5 + 3w + 3 - w = 2w + 8</h2>

Step-by-step explanation:

5+3w+3-w\qquad\text{combine like terms}\\\\=(3w-w)+(5+3)\\\\=2w+8

6 0
3 years ago
Are the expressions equivalent? 10(x + 2); 10x + 5
lutik1710 [3]

Answer: yes, indeed it is i forgot how it is but it is

7 0
2 years ago
.A variety of stores offer loyalty programs. Participating shoppers swipe a bar-coded tag at the register when checking out and
Leni [432]

Answer:

a) Null hypothesis:\mu \leq 120  

Alternative hypothesis:\mu > 120  

b) t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236  

The degrees of freedom are given by:

df = n-1 = 80-1=79

The p value for this case taking in count the alternative hypothesis would be:

p_v =P(t_{79}>2.236)=0.0141  

Step-by-step explanation:

Information given

\bar X=130 represent the sample mean for the amount spent each shopper

s=40 represent the sample standard deviation

n=80 sample size  

\mu_o =120 represent the value to verify

t would represent the statistic    

p_v represent the p value f

Part a

We want to verify if the shoppers participating in the loyalty program spent more on average than typical shoppers, the system of hypothesis would be:  

Null hypothesis:\mu \leq 120  

Alternative hypothesis:\mu > 120  

The statistic for this case would be given by:

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

Replacing the info given we got:

t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236  

The degrees of freedom are given by:

df = n-1 = 80-1=79

The p value for this case taking in count the alternative hypothesis would be:

p_v =P(t_{79}>2.236)=0.0141  

5 0
3 years ago
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