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3 years ago

What term best describes the point line or curve defined by the intersection of a cone and a plane?

Mathematics

1 answer:

Sindrei [870]3 years ago

6 0

Conic section from apex

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The effectiveness of a blood-pressure drug is being investigated. An experimenter finds that, on average, the reduction in systo

PIT_PIT [208]

Answer:

The estimate is

$52.02 < \mu < 55.78$

Step-by-step explanation:

From the question we are told that

The sample mean is $\ = x = 53.9$

The sample size is n = 24

The standard deviation is $\sigma = 5.6$

Given that the confidence level is 90% the level of significance is mathematically represented as

$\alpha = 100 - 90$

$\alpha = 10 \%$

$\alpha = 0.10$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table.The value is

$Z_{\frac{\alpha }{2} } =Z_{\frac{0.10 }{2} } = 1.645$

The reason we are obtaining critical value of $\frac{\alpha }{2}$ instead of $\alpha$ is because $\alpha$

represents the area under the normal curve where the confidence level interval ( $1 - \alpha$ ) did not cover which include both the left and right tail while

$\frac{\alpha }{2}$ is just the area of one tail which what we required to calculate the margin of error

NOTE: We can also obtain the value using critical value calculator (math dot armstrong dot edu)

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{ \sqrt{n} }$

substituting values

$E = 1.645 * \frac{5.6 }{ \sqrt{24} }$

$E = 1.880$

The estimate of how much the drug will lower a typical patient's systolic blood pressure(using a 90% confidence level) is mathematically represented as

$\= x - E < \mu < \= x + E$

substituting values

$53.9 - 1.880 < \mu < 53.9 + 1.880$

$52.02 < \mu < 55.78$

5 0

3 years ago

For decent amount of points! Hey can someone tell me the awnser?

igor_vitrenko [27]

I highly believe that the answer is 35

4 0

3 years ago

Can you translate the measurements to (cups, teaspoons, tablespoons etc.)

Alex777 [14]

Answer:

Flour: 18.76 Tablespoons

56.28 Teaspoons

1.17 U.S. Cups

Baking Powder: 0.35 Tablespoons

1.04 Teaspoons

0.02 U.S. Cups

Sugar: 0.80 Tablespoons

2.40 Teaspoons

0.05 U.S. Cups

Butter: 2.82 Tablespoons

8.46 Teaspoons

0.18 U.S. Cups

Maple Syrup:

0.127 Cups

2.029 Tablespoons

6.087 Teaspoons

Milk:

0.169 Cups

2.705 Tablespoons

8.115 Teaspoons

Hope this helps!!

Have A Blessed Day <333

4 0

2 years ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

tresset_1 [31]

Because I've gone ahead with trying to parameterize $S$ directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over $S$ straight away, let's close off the hemisphere with the disk $D$ of radius 9 centered at the origin and coincident with the plane $y=0$ . Then by the divergence theorem, since the region $S\cup D$ is closed, we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV$

where $R$ is the interior of $S\cup D$ . $\vec F$ has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(xz)}{\partial x}+\dfrac{\partial(x)}{\partial y}+\dfrac{\partial(y)}{\partial z}=z$

so the flux over the closed region is

$\displaystyle\iiint_Rz\,\mathrm dV=\int_0^\pi\int_0^\pi\int_0^9\rho^3\cos\varphi\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=0$

The total flux over the closed surface is equal to the flux over its component surfaces, so we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iint_S\vec F\cdot\mathrm d\vec S+\iint_D\vec F\cdot\mathrm d\vec S=0$

$\implies\boxed{\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=-\iint_D\vec F\cdot\mathrm d\vec S}$

Parameterize $D$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec k$

with $0\le u\le9$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

$\vec s_u\times\vec s_v=-u\,\vec\jmath$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^9(u^2\cos v\sin v\,\vec\imath+u\cos v\,\vec\jmath)\cdot(-u\,\vec\jmath)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{2\pi}\int_0^9u^2\cos v\,\mathrm du\,\mathrm dv=0$

$\implies\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\boxed{0}$

8 0

3 years ago

Is this right and plz help

FrozenT [24]

Number 6 and 7 are incorrect and I can't read 1/2/3

for 6
7/10 would be 70%
for 7
3 2/5 would be 3.40 not 3.25

8 0

3 years ago