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AysviL [449]
3 years ago
5

Jada solved the equation Negative StartFraction 4 over 9 EndFraction = StartFraction x over 108 EndFraction for x using the step

s below. What was Jada’s error?
Negative StartFraction 4 over 9 EndFraction = StartFraction x over 108 EndFraction
Negative StartFraction 4 over 9 EndFraction (Negative StartFraction 9 over 4 EndFraction) = StartFraction x over 108 EndFraction (Negative StartFraction 9 over 4 EndFraction)
x = negative StartFraction 1 over 48 EndFraction

Jada should have multiplied both sides of the equation by 108.
Jada should have multiplied both sides of the equation by Negative StartFraction 4 over 9 EndFraction.
The product of Negative StartFraction 9 over 4 EndFraction and StartFraction 1 over 108 EndFraction is not equal to Negative StartFraction 1 over 48 EndFraction.
The product of Negative StartFraction 9 over 4 EndFraction and 108 should have been the value of x.
Mathematics
2 answers:
netineya [11]3 years ago
8 0

Answer:

A

Step-by-step explanation:

navik [9.2K]3 years ago
5 0

Answer:

A

Step-by-step explanation:

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Identify the number that does not belong with the other three. Explain your reasoning. 26.6 repeating 6, square root 48, negativ
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What is opposite negative 7 5 over 12
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17. If13=xy-5 and y = 6, what is the value of x? (A) 0 (B) 3 (C) 6 (D) 8​
ludmilkaskok [199]

Answer:

Option B: 3

Step-by-step explanation:

<u>#1: Substitute in 6 for </u>y<u> and 3 for </u>x<u>:</u> 13=(3)(6)-5

<u>#2: Multiply 3 and 6:</u> 13=18-5

<u>#3: Evaluate the equation:</u> 13=18-5 → 13=13

<u>#4: Check if the solution is equal:</u> 13=13 ✓

x is equal to 3 (Option B).

7 0
2 years ago
Given that the population proportion is 0.82, the Normal distribution will provide a good approximation to the true sampling dis
Drupady [299]

Answer:

Then solution as long as n ≥ 56

Step-by-step explanation:

The Normal distribution will provide a good approximation to evaluate proportion tests when the following conditions are met:

n*p₀ > 10

n*q₀ > 10

As  p₀ = 0,82      q₀ = 0,18       n  = 13      30     56    and  100

We test:

A) n = 13        n*q₀  =  13*0,18 = 2,34       2,34 < 10  

B) n = 30       n*q₀  =  30*0,18 = 5,4         5,4 < 10

C) n = 56       n*q₀  =  56 * 0,18 = 10,08

                      n*p₀  = 56 * 0,82 = 45,92

D) n = 100      n*q₀  = 18

                      n*p₀  = 82

If we test 55  we find that  55*0,18  =  9,9

Then solution as long as n ≥ 56

5 0
3 years ago
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