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Nitella [24]
3 years ago
6

Because sample means were being tested the ____ is used to calculate the z-score

Mathematics
1 answer:
SSSSS [86.1K]3 years ago
7 0

Answer: standard deviation

Step-by-step explanation:

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If the area of the region bounded by the curve y^2 =4ax and the line x= 4a is 256/3 Sq units, using integration find the value o
almond37 [142]

If the area of the region bounded by the curve y^2 =4ax  and the line x= 4a  is \frac{256}{3} Sq units, then the value of  a will be 2 .

<h3>What is area of the region bounded by the curve ?</h3>

An area bounded by two curves is the area under the smaller curve subtracted from the area under the larger curve. This will get you the difference, or the area between the two curves.

Area bounded by the curve  =\int\limits^a_b {x} \, dx

We have,

y^2 =4ax  

⇒ y=\sqrt{4ax}

x= 4a,

Area of the region  =\frac{256}{3}  Sq units

Now comparing both given equation to get the intersection between points;

y^2=16a^2

y=4a

So,

Area bounded by the curve  =   \[ \int_{0}^{4a} y \,dx \] ​

 \frac{256}{3} =\[  \int_{0}^{4a} \sqrt{4ax}  \,dx \]

\frac{256}{3}=   \[\sqrt{4a}  \int_{0}^{4a} \sqrt{x}  \,dx \]

 \frac{256}{3}=   \[2\sqrt{a}  \int_{0}^{4a} x^{\frac{1}{2} }   \,dx \]                                            

\frac{256}{3}= 2\sqrt{a}  \left[\begin{array}{ccc}\frac{(x)^{\frac{1}{2}+1 } }{\frac{1}{2}+1 }\end{array}\right] _{0}^{4a}

\frac{256}{3}= 2\sqrt{a}  \left[\begin{array}{ccc}\frac{(x)^{\frac{3}{2} } }{\frac{3}{2} }\end{array}\right] _{0}^{4a}

\frac{256}{3}= 2\sqrt{a} *\frac{2}{3}  \left[\begin{array}{ccc}(x)^{\frac{3}{2}\end{array}\right] _{0}^{4a}

On applying the limits we get;

\frac{256}{3}= \frac{4}{3} \sqrt{a}   \left[\begin{array}{ccc}(4a)^{\frac{3}{2}  \end{array}\right]

\frac{256}{3}= \frac{4}{3} \sqrt{a} *\sqrt{(4a)^{3} }

\frac{256}{3}= \frac{4}{3} \sqrt{a} *  8 *a^{2}   \sqrt{a}

\frac{256}{3}= \frac{4}{3} *  8 *a^{3}

⇒ a^{3} =8

a=2

Hence, we can say that if the area of the region bounded by the curve y^2 =4ax  and the line x= 4a  is \frac{256}{3} Sq units, then the value of  a will be 2 .

To know more about Area bounded by the curve click here

brainly.com/question/13252576

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8 0
1 year ago
Read 2 more answers
Instruction: use the ratio of a 30-60-90 triangle to solve for the variables. Leave your answers as radicals in simplest form.u=
almond37 [142]

Let's start by identify a 30-60-90 triangle

If we compare those two traingles we can stablish the following relation:

\begin{gathered} a\sqrt[]{3}=6\sqrt[]{3} \\  \end{gathered}

Therefore, a=6

u is the opposite side to the 90° angle, therefore

\begin{gathered} u=2a \\ u=2\cdot6 \\ u=12 \end{gathered}

on the other hand, v is the opposite side to the 30° angles, so

\begin{gathered} v=a \\ v=6 \end{gathered}

6 0
1 year ago
Please help me solve this I will give u brainlst :)
NARA [144]
Equation: 2(.25) + p(.75) = 2.75
Answer: 3 pencils
7 0
2 years ago
Please help me with this!
ahrayia [7]

Answer:

I'm pretty sure its A

Step-by-step explanation:

..................

4 0
3 years ago
Read 2 more answers
C = k - 273.15 (a) Write two other related equations . (b) If the temperature is 23 in degrees Celsius , what is the temperature
Alona [7]

Answer:

(a) k = C + 273.15, F = 1.8(k - 273.15) + 32

(b) k = 296.15

(c) 25 celcius is greater than 300 kelvis.

Step-by-step explanation:

(a) one equation would be to solve for k instead of c, and another related equation could be from kelvin to Farenheit.

C = k - 273.15

C + 273.15 = k

k = C + 273.15

F = 1.8(k - 273.15) + 32

(b) for this one you have to substitute in the previous equation and solve for k.

23 = k - 273.15

k = 23 + 273.15

k = 296.15

(c) you have to convert either kelvins to celcius or celcius to kelvins.

C = k - 273.15

C = 300 - 273.15

C = 26.85

So 25 celcius is greater than 300 kelvis.

5 0
2 years ago
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