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2 years ago

Which of the following estimates at a 95% confidence level most likely comes form a small sample?

1 answer:

6 0

OptionAStep-by-step explanation:95% confidence interval is obtained using the following formulaCOnfidence interval lower bound = mean - critical value * sigma/sq rt nand upper bound = mean + critical value*sigma/sq rt nThus margin of error = critical value*sigma/rt nvaries indirectly as n provided others are remaining the same.Hence lower sample indicates higher margin of errorOut of 4 options given we find that 1 option has maximum margin of error as 21%Hence option. A is most likely coming from a small sample.

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Find the value of x. (4x) (3x) (x) (2x) Ox=14

stellarik [79]

Answer:

Ox=14

Step-by-step explanation:

7 0

3 years ago

Use cylindrical coordinates to evaluate the triple integral ∭ where E is the solid bounded by the circular paraboloid z = 9 - 16

4vir4ik [10]

Answer:

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

Step-by-step explanation:

The Cylindrical coordinates are:

x = rcosθ, y = rsinθ and z = z

From the question, on the xy-plane;

$9 -16 (x^2 + y^2) = 0 \\ \\ 16 (x^2 + y^2) = 9 \\ \\ x^2+y^2 = \dfrac{9}{16}$

$x^2+y^2 = (\dfrac{3}{4})^2$

where:

0 ≤ r ≤ $\dfrac{3}{4}$ and 0 ≤ θ ≤ 2π

∴

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} \int ^{9-16r^2}_{0} \ r \times rdzdrd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 z|^{z= 9-16r^2}_{z=0} \ \ \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 ( 9-16r^2}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} ( 9r^2-16r^4}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( \dfrac{9r^3}{3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3r^3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) d \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) \theta |^{2 \pi}_{0}$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{64}}-\dfrac{243}{320}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{160}})2 \pi$

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

4 0

3 years ago

Solve the following equation 6x-1=3 (2x+4)-1

DIA [1.3K]

$6x - 1 = 3(2x + 4) - 1$
Distribute
$6x - 1 = 6x + 12 - 1$
Combine like terms
$6x - 1 = 6x + 11$
Add 1 and subtract 6x from each side
0 = 12
The equation has no real solution

4 0

3 years ago

Read 2 more answers

There are 8 sophomores on the academic team. At the last competition, they each took the math test. Their scores were 82%, 92%,

castortr0y [4]

Median would be 80.75 or 79

8 0

3 years ago

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let y =2t+5 be a linear function representing the distance from home for an ant t minutes after starting out from a location nea

Akimi4 [234]

Answer: T=1/2 Y + -5/2