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Semenov [28]
3 years ago
7

ANSWER THIS QUICKLY PLEASE!!!!!!!!!

Mathematics
2 answers:
Vika [28.1K]3 years ago
8 0
It's the second answer.
Tems11 [23]3 years ago
5 0
C. 2 sq rt 2 + sq rt 5 + sq rt5 + sq rt 2
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I need help pls thank y
otez555 [7]
What do you need help with
5 0
3 years ago
how do you solve this problem. there are approximately 2.1 million deaths per year in country A. Express this quantity as deaths
slamgirl [31]
Well... we know that... there are 365 days in a year, unless is a leap-year, but we'll use 365 anyway

and each day has 24hrs, each hr has 60 minutes

so. let us use those ratios

\bf 2.1million\iff 2.1E6\iff 2,100,000\qquad thus
\\\\\\
\cfrac{2100000\ deaths}{1\ year}\cdot \cfrac{1\ year}{365\ day}\cdot \cfrac{1\ day}{24\ hr}\cdot \cfrac{1\ hr}{60\ minute}

so.. multiply and simplify, cancelling out any like-units atop and bottom

notice, all we do, is use the ratios, in a way, that if we need one unit to be changed, we flip the ratio

for example, to toss away "year" unit, since in the first fraction is at the bottom, then we put it on the top on the ratio, year/year  = 1, effectively cancelling the unit
5 0
3 years ago
Read 2 more answers
.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
I'll give brainlliest to whoever anwsers
postnew [5]

Answer:

Step-by-step explanation:

booom

8 0
3 years ago
Is anyone able to see if there is a counterexample for the following argument in logic?
Colt1911 [192]

if A=B

B=/=C

No A=C

All C are not A

6 0
3 years ago
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