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MAVERICK [17]
2 years ago
13

A train is traveling up slight grade with an angle of inclination of only 2 degrees. After traveling 1 mile, what is the vertica

l change in feet?
Mathematics
1 answer:
patriot [66]2 years ago
7 0
Tan 2° = x / 1 mile,   where x is the vertical change
0.03492 = x / 1 mile
x = 1 mile · 0.03429 = 0.03492 miles
1 mile = 5,280 feet
0.03492 · 5,280 = 184.38 ft
Answer: The vertical change is 184.38 feet 
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Using the information given, select the correct ending balance.
klemol [59]

Step-by-step explanation:

Simple first we will add previous balance and total deposit

498.03 + 604.28

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Then we will subtract it by total withdrawals

1102.31 - 599.49

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The ending balance is 502.82

6 0
2 years ago
Solve the system of linear equations by graphing. Round the solution to the nearest tenth as needed. Y + 2.3 = 0.45x and -2y = 4
HACTEHA [7]

Answer:

(2.4, -1.2)

Step-by-step explanation:

Start by moving the x and the y to the same side and moving the number across the equal sign in both equations. We should now have y-0.45x=-2.3 and 2y+4.2x=7.8. We can use the elimination method by multiplying the first equation by -2 to get -2y+0.9x=4.6 and 2y+4.2x=7.8. From there, add the two equations together, eliminating y (-2+2=0). We now have 5.1x=12.4; divide both sides by 5.1 to get x=2.4. Then, in any of the two equations, let's use y-0.45x=-2.3, substitute x with 2.4. Now we have y-1.08=-2.3. Add 1.08 to both sides to get y=-1.22; round that to the nearest tenth to get -1.2.

4 0
2 years ago
Read 2 more answers
The Acme Candy Company claims that​ 60% of the jawbreakers it produces weigh more than 0.4 ounces. Suppose that 800 jawbreakers
vichka [17]

Answer:

Yes, it would be statistically significant

Step-by-step explanation:

The information given are;

The percentage of jawbreakers it produces that weigh more than 0.4 ounces = 60%

Number of jawbreakers in the sample, n = 800

The mean proportion of jawbreakers that weigh more than 0.4 = 60% = 0.6 = \mu _ {\hat p} =p

The formula for the standard deviation of a proportion is \sigma  _{\hat p} =\sqrt{\dfrac{p(1-p)}{n} }

Solving for the standard deviation gives;

\sigma  _{\hat p} =\sqrt{\dfrac{0.6 \cdot (1-0.6)}{800} } = 0.0173

Given that the mean proportion is 0.6, the expected value of jawbreakers that weigh more than 0.4  in the sample of 800 = 800*0.6 = 480

For statistical significance the difference from the mean = 2×\sigma _{\hat p} = 2*0.0173 = 0.0346 the equivalent number of Jaw breakers = 800*0.0346 = 27.7

The z-score of 494 jawbreakers is given as follows;

Z=\dfrac{x-\mu _{\hat p} }{\sigma _{\hat p}  }

Z=\dfrac{494-480 }{0.0173  } = 230.94

Therefore, the z-score more than 2 ×\sigma _{\hat p} which is significant.

8 0
2 years ago
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gladu [14]

The answer is: 6.25\leq6.3

The explanation is shown below:

1. Cynthia rounds the number, which is identified as x, to one decimal place and the result is 6.3.

2. Based on this, we know that x could have been between 6.25 and 6.35. Therefore, the error interval for x is given by:

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Where \leq indicates that the value 6.25 is included and indicates that the value 6.35 is not included (Because if x had been exactly 6.35, Cynthia would round up to 6.4).

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2 years ago
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Bas_tet [7]
The answer is A

fidnjrjfbdjd
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