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Vika [28.1K]
3 years ago
7

What is the range of the data?

Mathematics
1 answer:
Ghella [55]3 years ago
5 0
O
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4 square root 5 is equivalent to
Lostsunrise [7]

That is equivalent to 8√10


8 0
3 years ago
Cardiovascular disease is a major cause of death and illness worldwide, with high blood pressure and high LDL cholesterol both b
valkas [14]

Answer:

Null and alternative hypotheses:

H0: p1 = p2

H1: p1 ≠ p2

To find the sample proportions, we have the following:

p'1 = \frac{x_1}{n_1} = \frac{113}{3180} = 0.0355

p'2 = \frac{x_2}{n_2} = \frac{157}{3168} = 0.0495

p' = \frac{(x_1 + x_2}{n_1 + n_2} = \frac{113 + 157}{3180 + 3168} = 0.0425

Calculate Z statistics:

z = \frac{(p1 -p2)}{\sqrt{(p'(1-p')*(\frac{1}{n1}+ \frac{1}{n2})}}

= \frac{(0.0355 - 0.0495)}{\sqrt{(0.0425*(1-0.0425) * (\frac{1}{3180} + \frac{1}{3168})}} = -2.764

Z = -2.764

P-value = 0.00285

The pvalue is low.

Since the pvalue is low, reject null hypothesis, H0.

Conclusion:

There is strong evidence that patients experiencing the primary outcome is different for those receiving the treatment compared to those without

7 0
2 years ago
A recipe calls for mixed nuts with 50% peanuts. 1/2 pound of 15% peanuts has already been used. How many pounds of 75% peanuts n
Galina-37 [17]
Now you have
<u />
<u>(.15)0.5 lb peanuts</u>     =     <u>0.075 lb peanuts</u>
0.5 lb  nuts                              0.5 lb nuts

for every pound of 75% mixed nuts you add, you get .75 lb peanuts, how many pounds, x, do you need to add to get the ratio 50 to 100?

<u>0.075 lb peanuts + 0.75x lb peanuts</u><em>       =    <u /></em><u>   50   </u><u>
</u>0.5 lb nuts + x lb nuts                                      100

cross multiply

(0.075 + .75x) 100 = (0.5 + x) 50
7.5 + 75x = 25 + 50x
25x = 17.5
x = 0.7 lb of 75% peanuts

6 0
3 years ago
Lee is thinking of 2 numbers, the product is 18.the quotient is 2. what are yhe two numbers?
jeka94
xy=18\\&#10;\dfrac{x}{y}=2\\\\&#10;xy=18\\&#10;x=2y\\\\&#10;2y\cdot y=18\\&#10;2y^2=18\\&#10;y^2=9\\&#10;y=-3 \vee y=3\\\\&#10;x=2\cdot (-3) \vee x=2\cdot 3\\&#10;x=-6 \vee x=6\\\\&#10;\boxed{(x,y)=\{(-6,-3),(6,3)\}}
4 0
3 years ago
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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