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ehidna [41]
3 years ago
6

NEED HELP ASAP! PLEASE HElp!!

Mathematics
1 answer:
mote1985 [20]3 years ago
4 0

the answer should be C

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Difference of 86.42 - 2.1.
densk [106]

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The first one

Step-by-step explanation:

The decimals line up when you subtract them

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2. What number will go in the blank so that the equation will have an
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Answer:

I think option B) -5

I hope help you

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Bhavik bought 333 liters of milk and 555 loaves of bread for a total of \$11$11dollar sign, 11. A month later, he bought 444 lit
AnnZ [28]

Let the price of 1 litre of milk be  x$

Let the price of 1 loaf of bread be y$

According to the problem

333x + 555y =11

444x+444y=10

On subtracting the two,

We get

(333x+555y=11) * 444

(444x+444y= 10) * 333

(-)     (-)         (-)

147852x + 246420y =4884

147852x + 147852y = 3330

(-)            (-)                  (-)

------------------------------                   

 98568y = 1554     

           y = 1554/98568     

              = 0.0157$          

                 =0.02$    

Price of 1 loaf of bread= 2 cents    

On putting the value of y 

333x + 555*0.0157 = 11333x  + 8.7135 = 11333x = 2.2865x = 2.2865/333x = 0.00068$   = 0.01$

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1.5


6 0
3 years ago
What would be the value of x in x+5=-3x-9
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Of interest is to test the hypothesis that the mean length of all face-to-face meetings and the mean length of all Zoom meetings
Goshia [24]

Answer:

Null hypothesis: \mu_1 = \mu_2 = ..... \mu_j , j =1,2,....,n

Alternative hypothesis: \mu_i \neq \mu_j , i,j =1,2,....,n

The alternative hypothesis for this case is that at least one mean is different from the others.

And the best method for this case is an ANOVA test.

Step-by-step explanation:

For this case we wnat to test if all the mean length of all face-to-face meetings and the mean length of all Zoom meetings are the same. So then the system of hypothesis are:

Null hypothesis: \mu_1 = \mu_2 = ..... \mu_j , j =1,2,....,n

Alternative hypothesis: \mu_i \neq \mu_j , i,j =1,2,....,n

The alternative hypothesis for this case is that at least one mean is different from the others.

And the best method for this case is an ANOVA test.

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