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qaws [65]
3 years ago
9

The function h = -16t2 + 1700 gives an object’s height h, in feet, at t seconds.

Mathematics
1 answer:
katen-ka-za [31]3 years ago
3 0
What do you want me to solve for?
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How Important Is Regular Exercise? In a recent poll1 of 1000 American adults, the number saying that exercise is an important pa
Elza [17]

Answer:

Step-by-step explanation:

Given that in a recent poll 1 of 1000 American adults, the number saying that exercise is an important part of daily life was 753.

Sample proportion = \frac{753}{1000} =0.753

To find 90% confidence interval we must find std error of proportion first

As per central limit theorem, sample proportion will follow a normal distribution with mean = 0.753

Std error of proportion= \sqrt{0.753*0.247/1000} \\=0.01363

90% Z critical value = 1.645

Margin of error = 1.645*0.01363 = 0.0224

Confidence interval lower bound = 0.753-0.0224=0.731

Upper bound = 0.775

6 0
3 years ago
If the AREA of the rectangle at the right is 22 3/4 square inches, what is the width of the rectangle?
sergejj [24]

Answer:

3.25 or 3 1/4 inches

Step-by-step explanation:

A = lw

22.75 = 7w

w = 3.25 which is 3 1/4

6 0
3 years ago
A construction worker fills a mold with concrete. This solid figure represents the mold.
nadezda [96]

Answer:

volume = 338 x 20 = 6760 cm^3

Step-by-step explanation:

large rectangular top area = 14 x 22 = 308 cm^2

small rectangular top area = 6x5 = 30 cm^2

total top area = 308 + 30 = 338 cm^2

volume = 338 x 20 = 6760 cm^3

4 0
3 years ago
$6 is what percent of $8?
iren2701 [21]
Hello there!
The answer to your question is 75%
If you multiply 8 × 75% it equals 6
So, your answer is 75%
Hope I helped!
3 0
3 years ago
Read 2 more answers
Find the exact length of the curve. 36y2 = (x2 − 4)3, 5 ≤ x ≤ 9, y ≥ 0
IrinaK [193]
We are looking for the length of a curve, also known as the arc length. Before we get to the formula for arc length, it would help if we re-wrote the equation in y = form.

We are given: 36 y^{2} =( x^{2} -4)^3
We divide by 36 and take the root of both sides to obtain: y = \sqrt{ \frac{( x^{2} -4)^3}{36} }

Note that the square root can be written as an exponent of 1/2 and so we can further simplify the above to obtain: y =  \frac{( x^{2} -4)^{3/2}}{6} }=( \frac{1}{6} )(x^{2} -4)^{3/2}}

Let's leave that for the moment and look at the formula for arc length. The formula is L= \int\limits^c_d {ds} where ds is defined differently for equations in rectangular form (which is what we have), polar form or parametric form.

Rectangular form is an equation using x and y where one variable is defined in terms of the other. We have y in terms of x. For this, we define ds as follows: ds= \sqrt{1+( \frac{dy}{dx})^2 } dx

As a note for a function x in terms of y simply switch each dx in the above to dy and vice versa.

As you can see from the formula we need to find dy/dx and square it. Let's do that now.

We can use the chain rule: bring down the 3/2, keep the parenthesis, raise it to the 3/2 - 1 and then take the derivative of what's inside (here x^2-4). More formally, we can let u=x^{2} -4 and then consider the derivative of u^{3/2}du. Either way, we obtain,

\frac{dy}{dx}=( \frac{1}{6})( x^{2} -4)^{1/2}(2x)=( \frac{x}{2})( x^{2} -4)^{1/2}

Looking at the formula for ds you see that dy/dx is squared so let's square the dy/dx we just found.
( \frac{dy}{dx}^2)=( \frac{x^2}{4})( x^{2} -4)= \frac{x^4-4 x^{2} }{4}

This means that in our case:
ds= \sqrt{1+\frac{x^4-4 x^{2} }{4}} dx
ds= \sqrt{\frac{4}{4}+\frac{x^4-4 x^{2} }{4}} dx
ds= \sqrt{\frac{x^4-4 x^{2}+4 }{4}} dx
ds= \sqrt{\frac{( x^{2} -2)^2 }{4}} dx
ds=  \frac{x^2-2}{2}dx =( \frac{1}{2} x^{2} -1)dx

Recall, the formula for arc length: L= \int\limits^c_d {ds}
Here, the limits of integration are given by 5 and 9 from the initial problem (the values of x over which we are computing the length of the curve). Putting it all together we have:

L= \int\limits^9_5 { \frac{1}{2} x^{2} -1 } \, dx = (\frac{1}{2}) ( \frac{x^3}{3}) -x evaluated from 9 to 5 (I cannot seem to get the notation here but usually it is a straight line with the 9 up top and the 5 on the bottom -- just like the integral with the 9 and 5 but a straight line instead). This means we plug 9 into the expression and from that subtract what we get when we plug 5 into the expression.

That is, [(\frac{1}{2}) ( \frac{9^3}{3}) -9]-([(\frac{1}{2}) ( \frac{5^3}{3}) -5]=( \frac{9^3}{6}-9)-( \frac{5^3}{6}-5})=\frac{290}{3}


8 0
3 years ago
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