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4 years ago

The value of the cumulative standardized normal distribution at 1.5x is 0.9332. the value of x is

Mathematics

1 answer:

ollegr [7]4 years ago

3 0

Use the table of standardized normal cumulative function.

Look up the corresponding x' for the value z=0.9332: I found x'=0.8232 (approximately).

We know that x'=1.5x, so x = 0.8232/1.5 = 0.5488

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Dafna1 [17]

Answer:

144 cm²

Step-by-step explanation:

The given translation doesn't change the shape or dimensions, hence the area remains same:

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what is the quotient (65y3 15y2 − 25y) ÷ 5y? a. 13y2 3y − 5 b. 13y3 3y2 − 5y c. 13y2 − 3y 5 d.13y2 − 3y − 5

Aneli [31]

Keywords:

Division, quotient, polynomial, monomial

For this case we must solve a division between a polynomial and a monomial and indicate which is the quotient.

By definition, if we have a division of the form: $\frac {a} {b} = c$ , the quotient is given by "c".

We have the following polynomial:

$65y ^ 3 + 15y ^ 2 - 25y$ that must be divided between monomy $5y$ , then:

$C (y)$ represents the quotient of the division:

$C (y) = \frac {65y ^ 3 + 15y ^ 2 - 25y} {5y}$

$C (y) = \frac {65y ^ 3} {5y} + \frac {15y ^ 2} {5y} - \frac {25y} {5y}$

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Thus, the quotient of the division between the polynomial and the monomial is given by:

$C (y) = 13y ^ 2 + 3y-5$

Answer:

The quotient is: $C (y) = 13y ^ 2 + 3y-5$

Option: A

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3 years ago

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A news station would like to conduct an exit poll to determine the likelihood that a highly debated amendment will receive enoug

vladimir1956 [14]

Answer:

The expression is $n = (\frac{1.645*0.5}{0.03})^2$

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

The margin of error is of:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a p-value of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

What expression would give the smallest sample size that will result in a margin of error of no more than 3 percentage points?

We have to find n for which M = 0.03.

We have no prior estimate for the proportion, so we use $\pi = 0.5$ . So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 1.645\sqrt{\frac{0.5*0.5}{n}}$

$0.03\sqrt{n} = 1.645*0.5$

$\sqrt{n} = \frac{1.645*0.5}{0.03}$

$(\sqrt{n})^2 = (\frac{1.645*0.5}{0.03})^2$

$n = (\frac{1.645*0.5}{0.03})^2$

The expression is $n = (\frac{1.645*0.5}{0.03})^2$

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3 years ago

Please give me an approximation of how long it would take to evacuate a population of 91,000 to a destination 21 miles away, usi

RoseWind [281]

Answer:

The time will depend on the number of people who move on each trip from the point of origin to the destination. If done at the maximum speed allowed using 100 vehicles of 50 seats each, the evacuation would be done in 63.68 hours

Step-by-step explanation:

Population = 91,000 ppl

Speed limit = 60 mph

Distance = 21 miles.

1. Assuming that people is evacuated at the max. speed allowed, it means that each trip will take:

T = D/V

D= 21 miles

V = 60 mph:

So;

T = 21 miles / 60mph

T= 0,35 h

2. Asumming that we are going to use an amount of 100 vehicles with 50-seats in each trip for evacuating people, it means that we could evacuate

500 people every 0,35 h ≈ 1,429 ppl/hour (evacuation rate)

To know how long it would take us to evacuate 91,000 people under these conditions, we would have to divide the total amount by the previously calculated evacuation rate

T= 91,000/ 1,429 = 63,68 hours

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4 years ago