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marshall27 [118]

2 years ago

8

1. Assume the reserve requirement is 10%. First National Bank receives a deposit of $5,400. If there is no slippage, how much co

uld the money supply expand? Explain and show your work.

1 answer:

Andru [333]2 years ago

3 0

Compose the result function for 400 by replacing the function designators with the actual functions 400

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What two integers does the square root of 429 lie?

Lubov Fominskaja [6]

The answer should be Two and Three ...

7 0

2 years ago

If 40% of a number is 32, what is 25% of that number?

Mekhanik [1.2K]

Well what I would do is split the 40% into 20% so from 40% to 20% that is /2. So 32/2=16 so 20% of a number is 16, we know there is 5, 20% in 100% so multiply 16 and 5 which gives you 80. Now 25% of 80 is the same as 80*.25 or 80/4 which is 20

Your Answer 20

7 0

2 years ago

Read 2 more answers

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = <u><em>the length of the calls, in minutes.</em></u>

So, X ~ Normal( $\mu=4.5,\sigma^{2} =0.70^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time = 4.5 minutes

$\sigma$ = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X $\leq$ 4.50 min)

P(X < 5.30 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{5.30-4.5}{0.7}$ ) = P(Z < 1.14) = 0.8729

P(X $\leq$ 4.50 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.5-4.5}{0.7}$ ) = P(Z $\leq$ 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = <u>0.3729</u>.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

P(X > 5.30 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{5.30-4.5}{0.7}$ ) = P(Z > 1.14) = 1 - P(Z $\leq$ 1.14)

= 1 - 0.8729 = <u>0.1271</u>

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 5.30 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 5.30 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{5.30-4.5}{0.7}$ ) = P(Z $\leq$ 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = <u>0.1109</u>.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 4.00 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 4.00 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.0-4.5}{0.7}$ ) = P(Z $\leq$ -0.71) = 1 - P(Z < 0.71)

= 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = <u>0.745</u>.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

P(X > x) = 0.05 {where x is the required time}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-4.5}{0.7}$ ) = 0.05

P(Z > $\frac{x-4.5}{0.7}$ ) = 0.05

Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;

$\frac{x-4.5}{0.7}=1.645$

${x-4.5}{}=1.645 \times 0.7$

x = 4.5 + 1.15 = 5.65 minutes.

SO, the time is 5.65 minutes.

7 0

2 years ago

Andre and Diego were each trying to solve 2x+6=3x−8. Describe the first step they each make to the equation. The result of Andre

Oksi-84 [34.3K]

Answer:

Andre subtracted 3x from both sides
Diego subtracted 2x from both sides

Step-by-step explanation:

<u>Andre</u>

Comparing the result of Andre's work with the original, we see that the "3x" term on the right is missing, and the x-term on the left is 3x less than it was. It is clear that Andre subtracted 3x from both sides of the equation.

__

<u>Diego</u>

Comparing the result of Diego's work with the original, we see that the "2x" term on the left is missing, and the x-term on the right is 2x less than it was. It is clear that Diego subtracted 2x from both sides of the equation.

_____

<em>Comment on their work</em>

IMO, Diego has the right idea, as his result leaves the x-term with a positive coefficient. He can add 8 and he's finished, having found that x=14.

Andre can subtract 6 to isolate the variable term, and that will give him -x=-14. This requires another step to get to x=14. Sometimes minus signs get lost, so this would not be my preferred sequence of steps.

As a rule, I like to add the opposite of the variable term with the least (most negative) coefficient. This results in the variable having a positive coefficient, making errors easier to avoid.

3 0

1 year ago

41 packages are randomly selected from packages received by a parcel service. the sample has a mean weight of 20.6 pounds and a

lina2011 [118]

Using normal distribution $\alpha = 1-0.95=0.05$
critical probability $p^{*} = 1 - \frac{ \alpha }{2}$
critical value for the critical probability is CV = 1.96
standard error $SE = \frac{3.2}{ \sqrt{41} } = 0.5$
margin of error = CV*SE = 1.96 * 0.5 = 0.98
95% confidence interval is between μ + 0.98 and <span>μ - 0.98, which is (19.62, 21.58)</span>

6 0

2 years ago