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musickatia [10]

3 years ago

10

The height of a ball above the ground in feet is defined by the function h(t) = -16t Squared + 80t +3, where t is the number of

seconds after the ball is thrown. What is the value of h(t), two seconds after the ball is thrown?

1 answer:

jasenka [17]3 years ago

6 0

Plug in 2 for t.
=> -16*(2^2) +80*2 +3
=> -16*4 +160 +3
=> -64+163
=> 99

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A line is parallel to y=3x-2 and intersects the point (3,5)

Digiron [165]

-2 is the answer
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Suppose a simple random sample of size 50 is selected from a population with σ=10σ=10. Find the value of the standard error of t

bogdanovich [222]

Answer:

a) $\sigma_{\bar x} = 1.414$

b) $\sigma_{\bar x} = 1.414$

c) $\sigma_{\bar x} = 1.414$

d) $\sigma _{\bar x} = 1.343$

Step-by-step explanation:

Given that:

The random sample is of size 50 i.e the population standard deviation =10

Size of the sample n = 50

a) The population size is infinite;

The standard error is determined as:

$\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}$

$\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}$

$\sigma_{\bar x} = 1.414$

b) When the population size N= 50000

n/N = 50/50000 = 0.001 < 0.05

Thus ; the finite population of the standard error is not applicable in this scenario;

Therefore;

The standard error is determined as:

$\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}$

$\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}$

$\sigma_{\bar x} = 1.414$

c) When the population size N= 5000

n/N = 50/5000 = 0.01 < 0.05

Thus ; the finite population of the standard error is not applicable in this scenario;

Therefore;

The standard error is determined as:

$\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}$

$\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}$

$\sigma_{\bar x} = 1.414$

d) When the population size N= 500

n/N = 50/500 = 0.1 > 0.05

So; the finite population of the standard error is applicable in this scenario;

Therefore;

The standard error is determined as:

$\sigma _{\bar x} = \sqrt{\dfrac{N-n}{N-1} }\dfrac{\sigma}{\sqrt{n} } }$

$\sigma _{\bar x} = \sqrt{\dfrac{500-50}{500-1} }\dfrac{10}{\sqrt{50} } }$

$\sigma _{\bar x} = 1.343$

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3 years ago

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19X+2

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Step-by-step explanation:

6 0

2 years ago

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sukhopar [10]

Answer:

The average rate of change is equal to $\frac{3}{4}\frac{homeruns}{game}$

Step-by-step explanation:

we have

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we know that

The rate of change of a linear variation is a constant

The rate of change of a linear variation is equal to the slope of the line

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therefore

The average rate of change is equal to $\frac{3}{4}\frac{homeruns}{game}$

<u>Verify</u>

the average rate of change is equal to

$\frac{n(b)-n(a)}{b-a}$

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$a=12\ games$

$b=60\ games$

Substitute

$\frac{45-9}{60-12}=\frac{3}{4}\frac{homeruns}{game}$

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3 years ago

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Answer:

Option (2). None

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A quadrilateral ABCD has been given with a property,

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Therefore, by the given property we can not deduce AB║DC.

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