Which expression is equivalent to

Which verb form correctly completes the sentence? Jessica __________ one of her longer gowns to wear. A. choosed B. had chose C.

irina1246 [14]

D. chose

Jessica chose one of her longer gowns to wear.

Chose is an irregular verb. It comes form the verb choose, which means to select. The conjugation of choose is chose and chose is a past simple verb.

5 0

3 years ago

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a washer has an outside radius of 6cm and a hole of radius 28cm. calculate the area of the face of the washer ( give your answer

Cerrena [4.2K]

Answer:

the area of face of washer is 22cm sq

4 0

3 years ago

Wires manufactured for use in a computer system are specified to have resistances between 0.11 and 0.13 ohms. The actual measure

Marina CMI [18]

Answer:

a) $P(0.11$

And we can find this probability with this difference and with the normal standard table or excel:

$P(-1.11$

b) $P(0.11 < \bar X < 0.13)$

And we can use the z score defined by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And using the limits we got:

$z = \frac{0.11-0.12}{\frac{0.009}{\sqrt{4}}}= -2.22$

$z = \frac{0.13-0.12}{\frac{0.009}{\sqrt{4}}}= 2.22$

And we want to find this probability:

$P(-2.22< Z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the resitances of a population, and for this case we know the distribution for X is given by:

$X \sim N(0.12,0.009)$

Where $\mu=0.12$ and $\sigma=0.009$

We are interested on this probability

$P(0.11$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(0.11$

And we can find this probability with this difference and with the normal standard table or excel:

$P(-1.11$

Part b

We select a sample size of n =4. And since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by: