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Kipish [7]
3 years ago
14

Explain how you would solve 16 / C equals 8 then solve the equation

Mathematics
1 answer:
Fantom [35]3 years ago
3 0

Answer: 2

Step-by-step explanation: you divide 16 by 8 then substitute the number you get for C. 16/8=2 Which means C=2.

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What are the square roots of 1?
Marta_Voda [28]

Answer:

1

Step-by-step explanation:

1^2=1

3 0
2 years ago
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Given a = 4 and b = -2, evaluate -|b - a|. Please help!
Sergeeva-Olga [200]
The answer is:  - 6 .
_____________________________________________

Evaluate the value INSIDE the "bars". 
  b - a  =  -2  -  4  = -6 .  The absolute value of the value inside inside the bars is the "positive" number of that value.  The value is negative 6; so, the absolute value, or "magnitude", of "-6", is "6";  Then, then "-" (6) = -6.  So the answer is: -6 .
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6 0
3 years ago
X+3=15 what’s x because I keep on getting 18 and it says I’m wrong on Pearson
icang [17]

Answer:

x = 12

Step-by-step explanation:

Subtract 3 from both sides

p.s. it would be 18 if it was x-3 = 15

8 0
3 years ago
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A data scientist studying a manufacturing process classifies each item produced by the process as either Poor, Acceptable, or Ex
Genrish500 [490]

Answer:

0.214 ; 0.064 ; 0.3859 ; 0.9966 ; 0.135

Step-by-step explanation:

Given the distribution :

Poor ___ Acceptable ____ Excellent

0.25 _____ 0.6 _________ 0.15

The data scientist receives three items produced by the manufacturing process.

Find the chance that:

a. all of items are Acceptable

0.6 * 0.6 * 0.6 = 0.6^3 = 0.214

b. none of the items is Acceptable

P(not acceptable) = 1 - p(acceptable) = 1 - 0.6 = 0.4

0.4 * 0.4 * 0.4 = 0.4^3 = 0.064

c. at least one of the items is Excellent

P(atleast one) = 1 - P(none)

P(not excellent) = 1 - 0.15 = 0.85

P(none is excellent) = 0.85^3 = 0.614125

Hence,

P(atleast one is excellent) = 1 - 0.614125 = 0.3859

d. not all of the items are Excellent

P(not all) = 1 - p(All)

P(all excellent) = 0.15^3 = 0.003375

P(not all are excellent) = 1 - 0.003375 = 0.9966

e. there is one item of each class:

P(poor) * p(acceptable) * p(excellent) * (number of classes)!

0.25 * 0.6 * 0.15 * 3!

= 0.135

3 0
3 years ago
A distribution with µ = 55 and σ = 6 is being standardized so that the new mean and standard deviation will be µ = 50 and σ = 10
Alekssandra [29.7K]

Solution:

\hbox{Let the distribution standarized is z}\\ \hbox{given:-}\mu =50\hbox{ and} \sigma =6\hbox{will become } \mu=50 \hbox{ and} \sigma =10\\\hbox{distribution standarized with value x=52 }\\\rm{so}\\\hbox{the value of score is:}\\z=\frac{x-\mu }{\sigma}\\\rm{if}\\p(x< 52)=p(z

3 0
3 years ago
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