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1 year ago

7

Suppose you win $40,000 in the lottery (after taxes). If you deposit this in a money market account earning 6% interest, compoun

ded semi-annually, how much total money would you have after 2 years?

1 answer:

Alenkinab [10]1 year ago

8 0

Answer:

I'm totally gonna have $49.600, in two years earn $9.600.

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a triangle has area 49.5 cm^2. if the base of the triangle is 9 cm, what is the height of the triangle

Anika [276]

Area of a triangle = (height * base ) / 2

height = (Area * 2) / base
height = (49.5 * 2) / 9
height = 99 / 9
height = 10 cm²

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3 years ago

Ede4ka [16]

Answer:

7.5in

Step-by-step explanation:

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3 years ago

What is the equation for the line of reflection that maps the trapezoid onto itself?

kvv77 [185]

The line that maps a figure onto itself is a line of symmetry of the figure.

From the given trapezoid, the line of symmetry of the trapezoid is x = -2.

Therefore, the <span>equation for the line of reflection that maps the trapezoid onto itself</span> is x = -2.

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3 years ago

Read 2 more answers

A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$

5 0

3 years ago

Okay google how many pennies are in $25,000

GalinKa [24]

There are 2,500,000 pennies in $25,000

4 0

3 years ago