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3 years ago

6

The Free Banking Era, where anyone was free to start their own bank, which resulted in financial chaos and political unrest, is

also known as _____.
wildcat banking

lender of last resort

open market operations

the Bank Bill

2 answers:

zimovet [89]3 years ago

6 0

Answer:

Wildcat banking. For sure.

ratelena [41]3 years ago

4 0

Wildcat banking is right

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You expect to receive $2,600 upon your graduation and will invest your windfall at an interest rate of 0.33 percent per quarter

kupik [55]

Answer:

n= 39.49 years

Explanation:

Giving the following information:

Present value (PV)= $2,600

Future value (FV)= $4,375

Interest rate (i)= 0.33/100= 0.0033

<u>To calculate the number of years, we need to use the following formula:</u>

n= ln(FV/PV) / ln(1+i)

n= ln(4,375/2,600) / ln(1.0033)

n= 157.96/4

n= 39.49 years

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2 years ago

Name and briefly describe the three essential principles of test construction

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Three essential <span>principles of test construction are:
1. consistency. It refers to the consistency </span><span>between the goal of the subject, course, and the methods of teaching goals. Also consistency between the methods and what the test measures. </span><span>
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3. reliability. The <span>test must be constructed in such a way that it will produce reliable, consistent results .</span>

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3 years ago

The College Board reported the following mean scores for the three parts of the Scholastic Aptitude Test (SAT) (The World Almana

storchak [24]

Answer:

a) $P(492$

And we can use excel or the normal standard table to find this probability:

$P(-0.949 < Z< 0.949)= P(Z$

b) $P(505$

And we can use excel or the normal standard table to find this probability:

$P(-0.949 < Z< 1.898)= P(Z$

c) $P(484$

And we can use excel or the normal standard table to find this probability:

$P(-1 < Z< 1)= P(Z$

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 scores for critical reading of a population, and for this case we know the distribution for X is given by:

$X \sim N(502,100)$

Where $\mu=502$ and $\sigma=100$

We select a sample of size n=90, since the distribution for X is normal then the distribution for the sample size is also normal

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}}=\frac{100}{\sqrt{90}}=10.54)$

And for this case we want this probability:

$P(502-10 < \bar X < 502+10)$

And for this case we can use the z score given by:

$z= \frac{\bar X -\mu}{\sigma_{\bar x}}$

And if we use this formula we got:

$P(492$

And we can use excel or the normal standard table to find this probability:

$P(-0.949 < Z< 0.949)= P(Z$

Part b

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

$X \sim N(515,100)$

Where $\mu=515$ and $\sigma=100$

We select a sample of size n=90, since the distribution for X is normal then the distribution for the sample size is also normal

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}}=\frac{100}{\sqrt{90}}=10.54)$

And for this case we want this probability:

$P(515-10 < \bar X < 515+10)$

And for this case we can use the z score given by:

$z= \frac{\bar X -\mu}{\sigma_{\bar x}}$

And if we use this formula we got:

$P(505$

And we can use excel or the normal standard table to find this probability:

$P(-0.949 < Z< 1.898)= P(Z$

Part c

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

$X \sim N(494,100)$

Where $\mu=494$ and $\sigma=100$

We select a sample of size n=100, since the distribution for X is normal then the distribution for the sample size is also normal

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}}=\frac{100}{\sqrt{100}}=10)$

And for this case we want this probability:

$P(494-10 < \bar X < 494+10)$

And for this case we can use the z score given by:

$z= \frac{\bar X -\mu}{\sigma_{\bar x}}$

And if we use this formula we got:

$P(484$

And we can use excel or the normal standard table to find this probability:

$P(-1 < Z< 1)= P(Z$

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ValentinkaMS [17]

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2 years ago

The competitive power of a company resource strength or competitive capability does not hinge on

olga55 [171]

It does not hinge on <span>whether the strength or capability represents a distinctive competence.
A company could be considered to have a strong competitive power without distinctive competence with these following conditions
- The competitors are equally incompetent
- They receive some sort of incentives from the government
- Their product is completely unique compared to any others</span>

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3 years ago