Who knows the answer to the rest of this?

Finger [1]

A13x2+390x here is the answer

6 0

3 years ago

Shelia's measured glucose level one hour after a sugary drink varies according to the normal distribution with μ = 117 mg/dl and

TiliK225 [7]

Answer:

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

Step-by-step explanation:

To solve this problem, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 117, \sigma = 10.6, n = 6, s = \frac{10.6}{\sqrt{6}} = 4.33$

What is the level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L ?

This is the value of X when Z has a pvalue of 1-0.01 = 0.99. So X when Z = 2.33.

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$2.33 = \frac{X - 117}{4.33}$

$X - 117 = 2.33*4.33$

$X = 127.1$

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

6 0

3 years ago

G find the area of the surface over the given region. use a computer algebra system to verify your results. the sphere r(u,v) =

Svetach [21]

Presumably you should be doing this using calculus methods, namely computing the surface integral along $\mathbf r(u,v)$ .

But since $\mathbf r(u,v)$ describes a sphere, we can simply recall that the surface area of a sphere of radius $a$ is $4\pi a^2$ .

In calculus terms, we would first find an expression for the surface element, which is given by

$\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k$
$\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j$
$\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k$
$\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u$

So the area of the surface is

$\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u$
$=-2\pi a^2(\cos\pi-\cos 0)$
$=-2\pi a^2(-1-1)$
$=4\pi a^2$

as expected.

6 0

2 years ago

How many solutions can be found for the equation 4z + 2(z - 4) = 3z + 11

Gnom [1K]

4z+2(z-4)=3z+11 perform indicated multiplication on left side

4z+2z-8=3z+11 combine like terms on left side

6z/-8=3z+11 add 8 to both sides

6z=3z+19 subtract 3z from both sides

3z=19 divide both sides by 3

z=19/3

z=6 1/3

So there is only one solution when z=6 1/3

4 0

3 years ago

Read 2 more answers

A ball is thrown from a height of 70 meters with an initial downward velocity of 10/ms. The ball's height h (in meters) after t

dmitriy555 [2]

Answer:h=−5t2−10t+70

Step-by-step explanation:

<!>BRAINLIEST APPERICATED<!> :D

3 0

3 years ago