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olchik [2.2K]
2 years ago
14

I'll give brainiest who ever helps me fast!)524x51

Mathematics
1 answer:
DaniilM [7]2 years ago
3 0

Answer:

26,724 is the answer

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Can someone tell me how to list these numbers from least to greatest. -4, 6.2, 18 1/2, -5,9, 21, -1/4, and 1.75. Thanks
Fiesta28 [93]
I think it is -5,-4,-1/4, 1.75, 6.2, 9, 18 1/2, 21
4 0
3 years ago
What have I done wrong and where please help me
lys-0071 [83]
12x - 18 - 4x -2
12x - 4x = 8x (yours 12x - 4x = 8)
-18 - 2 = -20 (yours -18 -1 = 17)

answer:

8x - 20
6 0
3 years ago
A linear pair of<br> angles have<br> measures of 3m - 6<br> and 3m + 18. What<br> is the value of m?
timama [110]

Answer:

the value of m = 28°

Step-by-step explanation:

the sum of adjacent angles which form linear pair is 180°

so,

=》3m - 6° + 3m + 18° = 180°

=》6m + 12° = 180°

=》6m = 180° - 12°

=》m = 168° ÷ 6

=》m = 28°

4 0
3 years ago
Can you help me please ​
sammy [17]

Answer:

196

Step-by-step explanation:

3 0
3 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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