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3 years ago

The calculation of property tax is based on the

1 answer:

5 0

The assessment rate is a uniform percentage and varies by tax jurisdiction, and could be any percentage below 100%. After getting the assessed value, it is multiplied by the mill levy to determine your taxes due. For example, suppose the assessor determines your property value is $500,000 and the assessment rate is 8%.

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Can somebody anybody please help me with this math problem?

Annette [7]

Answer:

30°

Step-by-step explanation:

Angle C = 55°, Angle D = Angle A = 65°, Angle B = 2x

Sum of angles in a triangle = 180°

65° + 55° + 2x = 180

120 + 2x = 180

2x = 180 - 120

2x = 60

x = 30°

3 0

3 years ago

Read 2 more answers

Find the domain and range with a vertex of (1,-2)

klemol [59]

Answer:

see explanation

Step-by-step explanation:

the domain ( values of x ) that a quadratic can have is all real numbers

domain : - ∞ < x < ∞

the range ( values of y ) are from the vertex upwards , that is

range : y ≥ - 2

3 0

2 years ago

Ysidra paints a room that has 400 square feet of wall space in 2 and a half hours. at this rate, how long will it take her to pa

Pie

1 hour 10 minutes /////

5 0

3 years ago

What are the values of x and y in PQRS? What are PR and SQ?

mamaluj [8]

Step-by-step explanation:

sol;

x+1=y...(1)

3y-7=2x....(2)

or, 3(x+1)-72x [from (1)]

or, 3x+3-7=2x

or, 3x-2x=7-3

x=4

now,

putting the value of x in (1)

y=x+1

=4+1

PR and SQ are the diagonal of PQRS.

4 0

3 years ago

The U.S. government has devoted considerable funding to missile defense research over the past 20 years. The latest development

Bad White [126]

Answer:

a) Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

2) A value of n fixed , for this case is 20 (satisfied)

3) Probability of success p =0.2 fixed (Satisfied)

So then we have all the conditions and we can assume that:

$X \sim Bin(n =20, p=0.8)$

b) $X \sim Bin(n =20, p=0.8)$

c) $P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

d) $P(X \geq 15) = P(X=15)+ .....+P(X=20)$

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

$P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218$

$P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205$

$P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137$

$P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058$

$P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012$

$P(X\geq 15)=0.804208$

e) $E(X) = np = 20*0.8 = 16$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

2) A value of n fixed , for this case is 20 (satisfied)

3) Probability of success p =0.2 fixed (Satisfied)

So then we have all the conditions and we can assume that:

$X \sim Bin(n =20, p=0.8)$

Part b

$X \sim Bin(n =20, p=0.8)$

Part c

For this case we just need to replace into the mass function and we got:

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

Part d

For this case we want this probability: $P(X\geq 15)$

And we can solve this using the complement rule:

$P(X \geq 15) = P(X=15)+ .....+P(X=20)$

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

$P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218$

$P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205$

$P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137$

$P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058$

$P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012$

$P(X\geq 15)=0.804208$

Part e

The expected value is given by:

$E(X) = np = 20*0.8 = 16$

5 0

3 years ago