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Natasha_Volkova [10]
2 years ago
11

An object is dropped from a height of 1700 ft above the ground. The function h=-16t^2+1700 gives the object’s height h in feet d

uring free fall at t seconds.
a. When will the object be at 1000ft above the ground?
b. When will the object be 940ft above the ground?
c. What are a reasonable domain and range for the function h?
Mathematics
1 answer:
Softa [21]2 years ago
8 0
The answer is
<span>a) 1000=-16t^2+1700, implies t² = -700 /-16, and t= 6.61s
b) </span><span>970= -16t^2+1700, </span><span>implies t² = -730 /-16, and t=6.75s

c)
reasonable domain of h
h is polynomial function, so its domain is R, (all real number)
its range
the inverse of h is h^-1 = sqrt (1700- t / 16), and its domain is </span>
<span><span><span>1700- t / 16>=0, so t <1700,
the range of h is I= ]-infinity, 1700]</span> </span> </span>




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The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.

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There is  significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units

Step-by-step explanation:

<u>Step:-(1)</u>

Given data the samples of material 1 gave an average (coded) wear of 85 units with a sample standard deviation of 4

Mean of the first sample x₁⁻ =85

standard deviation of the first sample S₁ = 4

Given data the samples of material 2 gave an average of 81 and a sample standard deviation of 5.

Mean of the first sample x₂⁻ =81

standard deviation of the first sample S₂ = 5

<u>Step :-2</u>

<u>Null hypothesis: H₀:</u> there is no significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

<u>Alternative hypothesis :H₁: </u>there is  significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

Assume the populations to be approximately normal with equal variances.σ₁² =σ₂²

The test statistic

                     Z= \frac{x_{1} -x_{2} }{\sqrt{\frac{S^2_{1} }{n_{1} } +\frac{S^2_{2} }{n_{2} }  } }

Given  n₁=n₂=60.

                    Z= \frac{85-81 }{\sqrt{\frac{(4)^2 }{60 } +\frac{5^2 }{60}  } }

On calculation, we get

                   Z =   \frac{4}{\sqrt{0.6833} }

                   z = 4.8389

The tabulated value Z =1.96 at 0.05 level of significance.

The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.

The null hypothesis is rejected.

Conclusion:-

there is  significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units.

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