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slamgirl [31]
1 year ago
15

-

Mathematics
1 answer:
pav-90 [236]1 year ago
3 0

Using quadratic function concepts, it is found that:

a) The equation is: 16t² - 64 = 0.

b) It takes 4 seconds before the weight is caught by the net.

<h3>What is the quadratic equation for a projectile's height?</h3>

It is given by:

h(t) = -16t^2 + v(0)t + h(0).

In which:

  • v(0) is the initial velocity.
  • h(0) is the initial height.

Item a:

In this problem, the object starts from rest at a height of 67 ft, hence v(0) = 0 and h(0) = 67.

Then, the equation for the height is given by:

h(t) = -16t^2 + v(0)t + h(0)

h(t) = -16t^2 + 67

It hits the net when h(t) = 3, hence:

3 = -16t^2 + 67

16t² - 64 = 0.

Item b:

16t² = 64

t = \pm \sqrt{4}

t = 4, as time is positive.

It takes 4 seconds before the weight is caught by the net.

More can be learned about quadratic function concepts at brainly.com/question/24737967

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Answer:

  • $1,094,748.09
  • $8600.28
  • 11 years, 10 months

Step-by-step explanation:

a) The compound interest formula can be used:

  A = P(1 +r/n)^(nt)

where P is the principal invested at annual rate r compounded n times per year for t years.

  A = $750,000(1 +.038/4)^(4·10) ≈ $1,094,748.09

Tamsyn's account will have a balance of $1,094,748.09 when she retires.

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b1) The amortization formula is good for this.

  A = P(r/n)/(1 -(1 +r/n)^(-nt))

where P is the amount earning interest at annual rate r compounded n times per year for t years.

  A = $1,094,748.09(0.049/12)/(1 - (1 +0.049/12)^(-12·15)) ≈ $8600.28

Tamsyn can withdraw $8600.28 per month for 15 years.

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b2) The account balance after n months will be ...

  B = P(1 +r/12)^n -A((1+r/12)^n -1)/(r/12)

Filling in the known values and solving for n, we have ...

  300,000 = 1,094,748.09(1.1.00408333^n) -8600.28(1.00408333^n -1)/.000408333

  300,000 = 1,094,748.09(1.1.00408333^n) -2,106,191.02(1.00408333^n -1)

  -1,806,191.02 = -1,011,442.93(1.00408333^n)

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  n = log(1.785757)/log(1.00408333) = 142.3

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Describe a normal probability distribution. a. bell-shaped.b. mean, median, and mode all equivalent.c. bimodal.d. symmetric arou
garri49 [273]

Answer:

a) bell-shaped.              

b) mean, median, and mode all equivalent.      

d) symmetric around the mean.        

g) most of the data fall within 3 standard deviations from the mean.

Step-by-step explanation:

We have to describe a normal distribution.

a. bell-shaped.

This is true a normal distribution is a bell shaped distribution.

b. mean, median, and mode all equivalent.

This is true for a normal distribution.

Mean = Mode = Median

c. Bimodal

The is not true about the normal distribution. A normal distribution is unimodal and the mode is equal to the mean of the distribution.

d. symmetric around the mean.

This is true. The normal distribution is centered around the mean

e. skewed to the right.

This is not a property of normal distribution.

f. models discrete random variables.

Normal distribution is a continuous distribution.

g. most of the data fall within 3 standard deviations from the mean.

This is true. According to Empirical rule, almost all the data lies within three standard deviation of mean.

h. uniform-shaped

This is not true. A normal distribution is bell shaped.

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