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3 years ago

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A newspaper found that, on average, 7.6% of people stay overnight in the hospital with a 1.5% margin of error. Construct a 95% c

onfidence interval.

1 answer:

Verizon [17]3 years ago

3 0

Given:

p = 7.6% = 0.076, the percentage of people who stay overnight at the hospital.

E = 1.5% = 0.015, margin of error

95% confidence interval.

The standard error is

Es = \sqrt{ \frac{p(1-p)}{n} }

where

n = the sample size.

The margin of error is

E=z^{*}E_{s}

where

z* = 1.96 at the 95% confidence level.

Because the margin of error is given, there is no need to calculate it.

The 95% confidence interval is

p +/- E = 0.076 +/- 0.015 = (0.061, 0.091) = (6.1%, 9.1%)

Answer:

The 95% confidence interval is between 6.1% and 9.1%.

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Write the following using index notation:<br> C) 4 x 4 x 4 x 5 x 5

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Answer:

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3 years ago

A ball is thrown in the air from a platform that is 96 feet above ground level with an initial vertical velocity of 32 feet per

pishuonlain [190]

Answer:

y = -16 (x - 1)^2 + 112

The object lands on the ground in approximately 3.6s

Explanation:

The equation given is that of a parabola.

Now the maximum (local) point of a parabola is the vertex. Therefore, if we want to rewrite our function in the form that would be used to find the maximum height, then that form must be the vertex form of a parabola.

The vertex form of a parabola is

$y=a(t-h)^2+k$

where (h, k) is the vertex.

The only question is, what is the vertex for our function h(t)?

Remember that if we have an equation of the form

$y=ax^2+bx+c$

then the x-coordinate of the vertex is

$h=-\frac{b}{2a}$

Now in our case b = 32 and a = -16; therefore,

$h=\frac{-32}{2(16)}=1$

We've found the value of the x-coordinate of the vertex. What about the y-coordinate? To get the y-coordinate, we put x = 1 into h(t) and get

$k=-16(1)+32(1)+96=112$

Hence, the y-coordindate is k = 112.

Therefore, the vertex of the parabola is (1, 112).

With the coordinates of the vertex in hand, we now write the equation of the parabola in vertex form.

$h(t)=a(t-1)^2+112$

The only problem is that we don't know what the value of a is. How do we find a?

Note that the point (0, 96) lies on the parabola. In other words,

$h(0)=-16(0)^2+32(0)+96=96$

Therefore, the vertex form of the parabola must also contain the point (0, 96).

Putting in t = 0, h = 96 into the vertex form gives

$96=a(0-1)^2+112$ $96=a+112$

subtracting 112 from both sides gives

$a=-16$

With the value of a in hand, we can finally write the equation of the parabola on vertex form.

$\boxed{h\mleft(t\mright)=-16\left(t-1\right)^2+112.}$

Now when does the object hit the ground? In other words, for what value of t is h(t) = 0? To find out we just have to solve the following for t.

$h(t)=0.$

We could either use h(t) = -16t^2 + 32t + 96 or the h(t) = -16(t - 1)^2 + 112 for the above equation. But it turns out, the vertex form is more convenient.

Thus we solve,

$-16\left(t-1\right)^2+112=0$

Now subtracting 112 from both sides gives

$-16(t-1)^2=-112$

Dividing both sides by -16 gives

$(t-1)^2=\frac{-112}{-16}$ $(t-1)^2=7$

taking the square root of both sides gives

$t-1=\pm\sqrt{7}$

adding 1 to both sides gives

$t=\pm\sqrt{7}+1$

Hence, the two solutions we get are

$t=\sqrt{7}+1=3.6$ $t=-\sqrt{7}+1=-1.6$

Now since time cannot take a negative value, we discard the second solution and say that t = 3.6 is our valid solution.

Therefore, it takes about 3.6 seconds for the object to hit the ground.

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1 year ago

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3 years ago

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