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Alina [70]
3 years ago
10

27=9•s what does s equal

Mathematics
2 answers:
Andru [333]3 years ago
7 0

Answer:

s=3

Step-by-step explanation:

27=9•s

Divide each side by 9

27/9 = 9s/9

3 =s

ratelena [41]3 years ago
3 0

Answer:3

Step-by-step explanation:

27=9•s

Divide 27/9

S=3

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Lee and Maya are collecting leaves for an art project lee collects 25/100 of the total number leaves needed. Maya collects 4/10
MrRissso [65]

Answer:  The answer is \dfrac{13}{20}.


Step-by-step explanation:  Given that Lee and Maya are collecting leaves for an art project. Let'x' be the total number of leaves needed in the project.

Then, the number of leaves collected by Lee is given by

L_\ell=\dfrac{25}{100}x,

and the fraction of the leaves collected by Maya is given by

M_\ell=\dfrac{4}{10}x.

Therefore, the total number of leaves collected by Lee and Maya altogether is

L=L_\ell+M_\ell=\dfrac{25}{100}x+\dfrac{4}{10}x=\dfrac{25+40}{100}x=\dfrac{65}{100}x=\dfrac{13}{20}x.

Thus, the total fraction of the leaves collected by Lee and Maya is

\dfrac{13}{20}.


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2 years ago
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Allied Corporation is trying to sell its new machines to Ajax. Allied claims that the machine will pay for itself since the time
kvasek [131]

Answer:

z(s) is in the acceptance region. We accept H₀  we did not find a significantly difference in the performance of the two machines therefore we suggest not to buy a new machine

Step-by-step explanation:

We must evaluate the differences of the means of the two machines, to do so, we will assume a CI  of 95%, and as the interest is to find out if the new machine has better performance ( machine has a bigger efficiency or the new machine produces more units per unit of time than the old one) the test will be a one tail-test (to the left).

New machine

Sample mean                  x₁ =    25

Sample variance               s₁  = 27

Sample size                       n₁  = 45

Old machine

Sample mean                    x₂ =  23  

Sample variance               s₂  = 7,56

Sample size                       n₂  = 36

Test Hypothesis:

Null hypothesis                         H₀             x₂  -  x₁  = d = 0

Alternative hypothesis             Hₐ            x₂  -  x₁  <  0

CI = 90 %  ⇒  α =  10 %     α = 0,1      z(c) = - 1,28

To calculate z(s)

z(s)  =  ( x₂  -  x₁ ) / √s₁² / n₁  +  s₂² / n₂

s₁  = 27     ⇒    s₁²  =  729

n₁  = 45    ⇒   s₁² / n₁    = 16,2

s₂  = 7,56   ⇒    s₂²  = 57,15

n₂  = 36     ⇒    s₂² / n₂  =  1,5876

√s₁² / n₁  +  s₂² / n₂  =  √ 16,2  + 1.5876    = 4,2175

z(s) = (23 - 25 )/4,2175

z(s)  =  - 0,4742

Comparing z(s) and  z(c)

|z(s)| < | z(c)|  

z(s) is in the acceptance region. We accept H₀  we did not find a significantly difference in the performance of the two machines therefore we suggest not to buy a new machine

The very hight dispersion of values s₁ = 27 is evidence of frecuent values quite far from the mean

3 0
3 years ago
Tommy has 6 more than 3 times the amount of money in his bank account than Judy. Judy has $7500 more than a number in her accoun
Ad libitum [116K]

Answer:

Amount of money Judy has in her account is \$(7500+n).

Amount of money Tommy has in his account is  \$(22506+3n)

Step-by-step explanation:

To find : Amount of money in Judy account and amount of money in Tommy account:

Solution:

Given:

Judy has $7500 more than a number in her account.

Let the number be 'n'.

So we can say that;

Amount of money in Judy account is equal to 7500 plus number.

framing in equation form we get;

Amount of money in Judy account = \$(7500+n)

Hence Amount of money Judy has in her account is \$(7500+n).

Now Given:

Tommy has 6 more than 3 times the amount of money in his bank account than Judy.

So we can say that;

Amount of money in Tommy account is equal to 3 multiplied by Amount of money in Judy account plus 6.

framing in equation form we get;

Amount of money in Tommy account = 3(7500+n)+6 =22500+3n+6=\$(22506+3n)

Hence Amount of money Tommy has in his account is  \$(22506+3n).

8 0
3 years ago
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