Simplify the following expression. √49^8

Consider a value to be significantly low if its z score less than or equal to minus−2 or consider a value to be significantly hi

katrin2010 [14]

Answer:

Test scores of 10.2 or lower are significantly low.

Test scores of 31.4 or higher are significantly high.

Step-by-step explanation:

Problems of normally distributed samples can be 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.

In this problem, we have that:

$\mu = 20.8, \sigma = 5.3$

Identify the test scores that are significantly low or significantly high.

Significantly low

Z = -2 and lower.

So the significantly low scores are thoses values that are lower or equal than X when Z = -2. So

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

$-2 = \frac{X - 20.8}{5.3}$

$X - 20.8 = -2*5.3$

$X = 10.2$

Test scores of 10.2 or lower are significantly low.

Significantly high

Z = 2 and higher.

So the significantly high scores are thoses values that are higherr or equal than X when Z = 2. So

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

$2 = \frac{X - 20.8}{5.3}$

$X - 20.8 = 2*5.3$

$X = 31.4$

Test scores of 31.4 or higher are significantly high.

3 0

3 years ago

Brendon sells tires at the local auto parts store. He told a customer that the tire width for a new tire was 21.5 centimeters. W

miss Akunina [59]

Answer:

Choice D

Step-by-step explanation:

When you convert a unit to a smaller unit it will always get bigger so in this case you will move the decimal point to the right so 21.5 will become 215.

8 0

2 years ago

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What is the reactance of a 25 mF capacitor when the applied frequency is 400 Hz?

LuckyWell [14K]

The Reactance of a Capacitor is the resistance it has when alternating current (A/C) with a specific frequency (f) passes through it. This is measured in ohms (Ω) and its formula is:

$X_{C}=\frac{1}{2\pi fC}$ (1)

Where:

$X_{C}$ is the Reactance in ohms (Ω)

$f$ is the frequency of the alternating current in Hertz ( $Hz$ )

$C$ is the Capacitance of the Capacitor in Farad ( $F$ )

In this case we have a 25 mF capacitor with an applied frequency of 400 Hz.

Note $1mF=1*10^{-3}F=0.001F$

Well, with the given data we have to solve (1):

$X_{C}=\frac{1}{2\pi(400Hz)(25*10^{-3}F)}$

Then:

XC=0.0159 Ω This is the Reactance of the Capacitor

4 0

3 years ago

If (inserted image), then f(2)= A)1/3 B)1 C)7

Lubov Fominskaja [6]

Given : f(x) = 3x + 1

We get f(2) by substituting x = 2 in the given Function

⇒ f(2) = 3(2) + 1

⇒ f(2) = 6 + 1 = 7

Option C is the Answer

6 0

3 years ago

Dr. Park is interested in how the functioning of the brain changes with age and how this, in turn, affects older individuals’ me

a_sh-v [17]

Answer:

Option C. No, because the two populations from which the samples are selected do not appear to have equal variances.

Step-by-step explanation:

The variance measures how far a set of (random) numbers are spread out from their average value.

The fact that the younger adults show diversity in their brain activity while the older adults produce similar activities show that there are no equal variances in the two populations from which the samples are selected.

Therefore it would not be valid for Dr. Park to use the independent-measures t-test to test whether the brain activity of younger adults is different from that of older adults during a visual recognition task

8 0

3 years ago

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