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iris [78.8K]
3 years ago
12

The equation gives the speed at impact, V metres per second, of an object dropped from a height of h metres. SHOW WORK From what

height must an object be dropped to impact the ground at a speed of 18.6 m/s?
Mathematics
1 answer:
devlian [24]3 years ago
4 0

Answer:

h = 17.65 m

Step-by-step explanation:

The given equation gives the speed at impact :

v=\sqrt{2gh}

h is height form where the object is dropped

Put v = 18.6 m/s in the above equation.

h=\dfrac{v^2}{2g}\\\\h=\dfrac{(18.6)^2}{2\times 9.8}\\\\h=17.65\ m

So, the object must be dropped from a height of 17.65 m.

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3 years ago
An automotive manufacturer wants to know the proportion of new car buyers who prefer foreign cars over domestic. Step 2 of 2 : S
ziro4ka [17]

Answer:

The 85% onfidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars is (0.151, 0.205).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

Sample of 421 new car buyers, 75 preferred foreign cars. So n = 421, \pi = \frac{75}{421} = 0.178

85% confidence level

So \alpha = 0.15, z is the value of Z that has a pvalue of 1 - \frac{0.15}{2} = 0.925, so Z = 1.44.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.178 - 1.44\sqrt{\frac{0.178*0.822}{421}} = 0.151

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.178 + 1.44\sqrt{\frac{0.178*0.822}{421}} = 0.205

The 85% onfidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars is (0.151, 0.205).

8 0
2 years ago
Can someone help me on this pls? It’s urgent, so ASAP (it’s geometry)
GarryVolchara [31]

<u>Question 6</u>

1) \overline{AB} \cong \overline{BD}, \overline{CD} \perp \overline{BD}, O is the midpoint of \overline{BD}, \overline{AB} \cong \overline{CD} (given)

2) \angle ABO, \angle ODC are right angles (perpendicular lines form right angles)

3) \triangle ABO, \triangle CDO are right triangles (a triangle with a right angle is a right triangle)

4) \overline{BO} \cong \overline{OD} (a midpoint splits a segment into two congruent parts)

5) \triangle ABO \cong \triangle CDO (LL)

<u>Question 7</u>

1) \angle ADC, \angle BDC are right angles), \overline{AD} \cong \overline{BD}

2) \overline{CD} \cong \overline{CD} (reflexive property)

3) \triangle CDA, \triangle CDB are right triangles (a triangle with a right angle is a right triangle)

4) \triangle ADC \cong \triangle BDC (LL)

5) \overline{AC} \cong \overline{BC} (CPCTC)

<u>Question 8</u>

1) \overline{CD} \perp \overline{AB}, point D bisects \overline{AB} (given)

2) \angle CDA, \angle CDB are right angles (perpendicular lines form right angles)

3) \triangle CDA, \triangle CDB are right triangles (a triangle with a right angle is a right triangle)

4) \overline{AD} \cong \overline{DB} (definition of a bisector)

5) \overline{CD} \cong \overline{CD} (reflexive property)

6)  \triangle ADC \cong \triangle BDC (LL)

7) \angle ACD \cong \angle BCD (CPCTC)

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1 year ago
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dusya [7]

Answer:

10

Step-by-step explanation:

To calculate the distance d use the distance formula

d = √ (x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = (- 5, 2) and (x₂, y₂ ) = (- 5, - 8)

d = \sqrt{(-5+5)^2+(-8-2)^2}

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6 0
3 years ago
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