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2 years ago

8

You begin your first day as a police officer earning $30,000/yr. Assuming that you obtain a full 5% increase each year, what wil

l your salary be in three years?

2 answers:

Inessa [10]2 years ago

4 0

Answer: $34728.75

Step-by-step explanation:

Given: The salary of a police officer for a year P= $30,000

The rate of increase r= 5% = 0.05 per year

We know that the compound amount after t years at rate of intersect r is given by :-

$A=P(1+r)^t$

Then, the compound amount after 3 years is given by :-

$A=30,000(1+0.05)^3\\\Rightarrow\ A=30,000(1.05)^3\\\Rightarrow\ A=34728.75$

Hence, the salary of policeman after 3 years will be $34728.75.

spayn [35]2 years ago

3 0

Answer:

The equation for the amount of salary after x number of years is given by:

$y = a(1+r)^x$

where

a is the initial amount

x represents the number of years and

r represents the growth rate( in decimal)

As per the statement:

You begin your first day as a police officer earning $30,000/yr

⇒ a= 30,000

It is also given that you obtain a full 5% increase each year,

⇒ $r = 5\% = 0.05$

then;

$y = 30,000(1+0.05)^x$

$y = 30,000(1.05)^x$ ....[1]

we have to find the salary in three years.

⇒ $x = 3$

Substitute this value in [1] we have

$y = 30,000(1.05)^3$

Simplify:

$y = \$ 34,4728.75$

therefore, your salary be in three years will be $34,4728.75

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Six teachers and 12 students volunteer for a committee to discuss extra-curricular activities. How many committees of 5 people c

Alex17521 [72]

Answer:

a) $X=3300$

b) $Y=7770$

Step-by-step explanation:

From the question we are told that:

Number Teachers $T=6$

Number Student $S=12$

Number in committee $n=5$

a) Generally the equation for exactly 3 students on the committee is mathematically given by

$X=^{S}C_3*^{T}C_3$

$X=^{12}C_3*^{6}C_3$

$X=3300$

b) Generally the equation for at least one teacher and at least one student on the committee is mathematically given by

Total Ways-(no of ways of selection no teacher or student)

Where total Ways

$T=^{(6+12)}C_5$

$T=8568$

Therefore

$Y=8568-^{6}C_0*^{12}C_5+^{12}C_0*^{6}C_5$

$Y=8568-798$

$Y=7770$

8 0

1 year ago

At a college, 72% of courses have final exams and 46% of courses require research papers. Suppose that 31% of courses have a res

beks73 [17]

Answer:

0.25

Step-by-step explanation:

72% of courses have final exams and 46% of courses require research papers which means probability of 0.72 for courses that have final exams and 0.46 for courses that require research papers.

31% of courses have a research paper and a final exam, which means probability of 0.31 for both courses with exams and research papers, using Venn diagram approach, find picture attached to the solution.

P(R or E) = P(R) + P(E) - P(R and E), which gives:

P(R or E) = 0.15 + 0.41 - 0.31

P(R or E) = 0.25.

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2 years ago

"An ordinance requiring that a smoke detector be installed in all previously constructed houses has been in effect in a particul

Galina-37 [17]

Answer:

a) Probability that the claim is rejected when the actual value of p is 0.8 = P(X ≤ 15) = 0.0173

b) Probability of not rejecting the claim when p = 0.7, P(X > 15) = 0.8106

when p = 0.6, P(X > 15) = 0.4246

c) Check Explanation

The error probabilities are evidently lower when 15 is replaced with 14 in the calculations.

Step-by-step explanation:

p is the true proportion of houses with smoke detectors and p = 0.80

The claim that 80% of houses have smoke detectors is rejected if in a sample of 25 houses, not more than 15 houses have smoke detectors.

If X is the number of homes with detectors among the 25 sampled

a) Probability that the claim is rejected when the actual value of p is 0.8 = P(X ≤ 15)

This is a binomial distribution problem

A binomial experiment is one in which the probability of success doesn't change with every run or number of trials (probability that each house has a detector is 0.80)

It usually consists of a number of runs/trials with only two possible outcomes, a success or a failure (we are sampling 25 houses with each of them either having or not having a detector)

The outcome of each trial/run of a binomial experiment is independent of one another.

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 25 houses sampled

x = Number of successes required = less than or equal to 15

p = probability of success = probability that a house has smoke detectors = 0.80

q = probability of failure = probability that a house does NOT have smoke detectors = 1 - p = 1 - 0.80 = 0.20

P(X ≤ 15) = Sum of probabilities from P(X = 0) to P(X = 15) = 0.01733186954 = 0.01733

b) Probability of not rejecting the claim when p= 0.7 when p= 0.6

For us not to reject the claim, we need more than 15 houses with detectors, hence, th is probability = P(X > 15), but p = 0.7 and 0.6 respectively for this question.

n = total number of sample spaces = 25 houses sampled

x = Number of successes required = more than 15

p = probability that a house has smoke detectors = 0.70, then 0.60

q = probability of failure = probability that a house does NOT have smoke detectors = 1 - p = 1 - 0.70 = 0.30

And 1 - 0.60 = 0.40

P(X > 15) = sum of probabilities from P(X = 15) to P(X = 25)

When p = 0.70, P(X > 15) = 0.8105639765 = 0.8106

When p = 0.60, P(X > 15) = 0.42461701767 = 0.4246

c) How do the "error probabilities" of parts (a) and (b) change if the value 15 in the decision rule is replaced by 14.

The error probabilities include the probability of the claim being false.

When X = 15

(Error probability when p = 0.80) = 0.0173

when p = 0.70, error probability = P(X ≤ 15) = 1 - P(X > 15) = 1 - 0.8106 = 0.1894

when p = 0.60, error probability = 1 - 0.4246 = 0.5754

When X = 14

(Error probability when p = 0.80) = P(X ≤ 14) = 0.00555

when p = 0.70, error probability = P(X ≤ 14) = 0.0978

when p = 0.60, error probability = P(X ≤ 14) = 0.4142

The error probabilities are evidently lower when 15 is replaced with 14 in the calculations.

Hope this Helps!!!

6 0

2 years ago

A consensus forecast is the average of a large number of individual analysts' forecasts. Suppose the individual forecasts for a

stealth61 [152]

Answer:

a) 0.954

b) 0.937

c) 0.891

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 6 percent

Standard Deviation, σ = 1.3 percent

We are given that the distribution of particular interest rate is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(At least 3.8 percent.)

$P(x \geq 3.8)$

$P( x \geq 3.8) = P( z \geq \displaystyle\frac{3.8 - 6}{1.3}) = P(z > -1.69)$

$= 1 - P(z < -1.69)$

Calculation the value from standard normal z table, we have,

$P(x \geq 3.8) = 1 - 0.046 = 0.954 = 95.4\%$

b) P(At most 8 percent)

$P(x \leq 8) = P(z \leq \displaystyle\frac{8-6}{1.3}) = P(z \leq 1.53)$

Calculating the value from the standard normal table we have,

$P( x \leq 8) =0.937= 93.7\%$

c) P(Between 3.8 percent and 8 percent. )

$P(3.8 \leq x \leq 8) = P(-1.69 \leq z \leq 1.53)\\\\= P(z \leq 1.53) - P(z < -1.69)\\= 0.937 - 0.046 = 0.891 = 89.1\%$

$P(3.8 \leq x \leq 8) = 89.1\%$

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2 years ago

Which expression is equivalent to 4√6/3√2

VLD [36.1K]

Given

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2 years ago

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