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zmey [24]
2 years ago
11

Chase tried to solve an equation step by step. \qquad\begin{aligned} \dfrac13(g-3)&=3\\\\ \\ g-3&=9&\green{\text{Ste

p } 1}\\\\ \\ g&=12&\blue{\text{Step } 2}\\\\ \end{aligned} 3 1 ​ (g−3) g−3 g ​ =3 =9 =12 ​ Step 1 Step 2 ​ Find Chase's mistake.
Mathematics
2 answers:
Sonbull [250]2 years ago
6 0

Answer:cant really help with that because the question is messed up sorry

Step-by-step explanation:

AfilCa [17]2 years ago
5 0

Answer:

C

Step-by-step explanation:

Pop in the answer they came up with, 12. The equation will be written like this:

1/3 (12-3) = 3

How to solve:

  1. 12 - 3 = 9
  2. 1/3 (9) = 3
  3. 1/3 of 9 = 3

Therefore, the correct answer is C.

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Please see the attached picture.

5 0
1 year ago
Peter has 96 stamps and Sam has 63. How many stamps should Sam give Peter so that Peter will have twice as many stamps as Sam?
MAXImum [283]
96÷2=48
63-48=15

So Sam should give Peter 15 of his stamps so that Peter will have twice as many stamps as him.
4 0
1 year ago
Read 2 more answers
-6,-2,0,1,3/2 whats the sequence and what do I multiply by
Angelina_Jolie [31]
Its not possible because if you try to multiply back and forth it wont work. 

5 0
1 year ago
5 Adoncia makes a scale drawing of the front of
Molodets [167]

Answer:

Since the length of the drawing is 200 ft. and equivalent to 13.33 in. with a scale of 15 ft to 1 in. and the length of the paper is 11 in., Adoncia's drawing will not fit on the sheet of paper

Step-by-step explanation:

The given parameters are;

The scale of the drawing is 15 ft = 1 in.

The actual dimensions of the monument;

Height = 80 ft.

Length = 200 ft.

Therefore, we have;

The required dimension of the paper height = 80/15 = 16/3 = 5.33 in.

The required dimension of the paper length = 200/15 = 40/3 = 13.33 in.

The given paper dimension by 11 in. which is of a dimension of that of a standard  letter paper size of 8.5 in. by 11 in.

Drawing length, 13.33 in. > Paper length > 11 in.

Adoncia's drawing will not fit on the sheet of paper.

4 0
2 years ago
The number of "destination weddings" has skyrocketed in recent years. For example, many couples are opting to have their wedding
Andru [333]

Answer:

There is no sufficient evidence to support the claim that wedding cost is less than $30000.

Step-by-step explanation:

Values (x) ∑(Xi-X)^2

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

29.1                    0.1702

28.5                  1.0252

28.8                  0.5077

29.4                   0.0127

29.8                  0.0827

29.8                  0.0827

30.1                   0.3452

30.6                   1.1827

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

236.1                 3.4088

Mean = 236.1 / 8 = 29.51

S_{x}=\sqrt{3.4088/(8-1)}=0.6978

Statement of the null hypothesis:

H0: u ≥ 30 the mean wedding cost is not less than $30,000

H1: u < 30 the mean wedding cost is less than $30,000

Test Statistic:

t=\frac{X-u}{S/\sqrt{n}}=\frac{29.51-30}{0.6978/\sqrt{8}}= \frac{-0.49}{0.2467}=-1.9861

Test criteria:

SIgnificance level = 0.05

Degrees of freedom = df = n - 1 = 8 - 1 = 7

Reject null hypothesis (H0) if

t

Finding in the t distribution table α=0.05 with df=7, we have

t_{0.05,7}=2.365

t>-t_{0.05,7} = -1.9861 > -2.365

Result: Fail to reject null hypothesis

Conclusion: Do no reject the null hypothesis

u ≥ 30 the mean wedding cost is not less than $30,000

There is no sufficient evidence to support the claim that wedding cost is less than $30000.

Hope this helps!

5 0
2 years ago
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