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marissa [1.9K]
3 years ago
5

Please help quickly with this ​

Mathematics
2 answers:
ladessa [460]3 years ago
8 0

Answer:

x=574

Step-by-step explanation:

it says that on my caculater.

Helen [10]3 years ago
7 0
X= 574 is the answer
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A researcher wants to know how undergraduate college students at her university feel about a list of 1196 undergraduate college
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The complete question is shown on the first uploaded image

Answer:

Population

All undergraduate college students at the  researcher's university ,

Sample

The 107 undergraduate college students who returned the questionnaire

Step-by-step explanation:

From the question we are told that

   The list of undergraduate college students obtained from her university is  n= 1196

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      The number of e-mail that was returned is  k = 107

     The proportion that believes that the federal student loan program needs to be completely overhauled is  \^{p} =  0.71

Generally the population for this particular study is

All undergraduate college students at the  researcher's university ,

this is because they are the total pool of data from which the statistical sample are randomly selected from

Generally the sample for this particular study is

The 107 undergraduate college students who returned the questionnaire

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3 years ago
What is the value of x?<br> help PLS
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8 0
3 years ago
The radius of a right circular cone is increasing at a rate of 1.9 in/s while its height is decreasing at a rate of 2.2 in/s. At
musickatia [10]

Answer:

Volume of the cone is increasing at the rate 9916\pi \frac{in^3}{s}.

Step-by-step explanation:

Given: The radius of a right circular cone is increasing at a rate of 1.9 in/s while its height is decreasing at a rate of 2.2 in/s.

To find: The rate at which volume of the cone changing when the radius is 134 in. and the height is 136 in.

Solution:

We have,

\frac{dr}{dt} =1.9 \:\text{in/s}, \frac{dh}{dt}=-2.2\:\text{in/s}, r=134 \:\text{in}, h=136\:\text{in}

Now, let V be the volume of the cone.

So, V=\frac{1}{3}\pi r^{2}h

Differentiate with respect to t.

\frac{dv}{dt} =\frac{1}{3}\pi \left [ r^2\frac{dh}{dt}+h\left ( 2r \right )\frac{dr}{dt} \right ]

Now, on substituting the values, we get

\frac{dv}{dt} =\frac{1}{3}\pi\left [ \left ( 134 \right )^2\left ( -2.2 \right )+\left (  136\right )\left ( 2 \right )\left ( 134 \right )\left ( 1.9 \right ) \right ]

\frac{dv}{dt} =\frac{1}{3}\pi\left [  -39503.2+69251.2 \right ]  

\frac{dv}{dt} =\frac{1}{3}\pi\left [ 29748 \right ]

\frac{dv}{dt} =9916\pi \frac{in^3}{s}

Hence, the volume of the cone is increasing at the rate 9916\pi \frac{in^3}{s}.

6 0
3 years ago
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