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Ulleksa [173]
2 years ago
6

There are two jobs you can apply for. the first job pays $22,000 the first year, with raises of $4,000 each year thereafter. the

second job pays $26,000 the first year with raises of $2,000 each year thereafter. when would you make as much money in the first job as in the second?
Mathematics
2 answers:
jasenka [17]2 years ago
7 0
We let the number of years that the two jobs will have the same payment be denoted as t. Equating the wages of these two jobs after t - 1 years will give us an equation of,
                          22,000 + 4000(t -1) = 26,000 + 2000(t - 1)
The value of t from the generated equation is 3. Therefore, after 3 years the jobs will be paying the same wages.
Margarita [4]2 years ago
5 0

Answer: 3rd year

Step-by-step explanation:

Given : There are two jobs you can apply for.

Let x be the time (in years).

The first job pays $22,000 the first year, with raises of $4,000 each year thereafter.

Then, the amount earned in x years by first job can be written as :-

y=22000+4000(x-1)...................(1)

The second job pays $26,000 the first year with raises of $2,000 each year there after.

Then, the amount earned in x years can be written as :-

y=26000+2000(x-1)...........................(2)

From equation (1) and (2) , we have

22000+4000(x-1)=26000+2000(x-1)\\\\\Rightarrow\ 4000(x-1)-2000(x-1)=26000-22000\\\\\Rightarrow\ 2000(x-1)=4000\\\\\Rightarrow\ x-1=\dfrac{4000}{2000}=2\\\\\Rightarrow\ x=2+1=3

Hence, in 3rd year you would make as much money in the first job as in the second.

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Answer:

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Step-by-step explanation:

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2 years ago
Read 2 more answers
Legalization of marijuana, Part I. The 2010 General Social Survey asked 1,578 US res- idents: \Do you think the use of marijuana
mixer [17]

Answer:

Step-by-step explanation:

a) Sample statistics are used to estimate population value. Since 48% is a sample proportion, therefore, it is a sample statistic.

b) For 95% confidence level, z* = 1.96.

\hat{p}\pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}= 0.61\pm 0.61\sqrt{\frac{0.61(1-0.61)}{1578}}=0.61\pm 0.024 \ or (0.586, 0.634).

We are 95% confident that the true proportion of US residents who think marijuana should be made legal lies between 58.6% and 63.4%.

c)

\\np=1578(0.61)=962.58

\\n(1-p)=1578(1-0.61)=615.42

Since both np and n(1-p), are at least 10, the normal model is a good approximation for these data.

d) As the lower limit of confidence interval is less than 0.5, less than 50% population is also a plausible value of true proportion. This means the statement "Majority of Americans think marijuana should be legalized" is not justified.

4 0
2 years ago
Craig has $1850 dollars in a bank account that he uses to make automatic payments of $400.73 on his car loan. If Craig stops mak
Ivenika [448]

Given:

Amount in the bank account = $1850

Monthly payment of can loan = $400.73

To find:

When would automatic payments make the value of the account zero?

Solution:

Craig stops making deposits to that account. So, amount $1850 in the bank account is used to make monthly payment of can loan.

On dividing the amount by monthly payment, we get

\dfrac{1850}{400.73}=4.61657

It means, the amount is sufficient for 4 payment but for the 5th payment the amount is not sufficient.

Therefore, the 5th automatic payments make the value of the account zero.

6 0
2 years ago
Consider a set of 7500 scores on a national test whose score is known to be distributed normally with a mean of 510 and a standa
german
\mathbb P(X>600)=\mathbb P\left(\dfrac{X-510}{85}>\dfrac{600-510}{85}\right)=\mathbb P(Z>1.059)\approx0.145

So approximately 14.5% of the scores are higher than 600. This means in a sample of 7500, one could expect to see 0.145\times7500\approx10.86 scores above 600.
5 0
2 years ago
It is claimed that 55% of marriages in the state of California end in divorce within the first 15 years. A large study was start
kykrilka [37]

Answer:

0.0045 = 0.45% probability that less than two of them ended in a divorce

Step-by-step explanation:

For each marriage, there are only two possible outcomes. Either it ended in divorce, or it did not. The probability of a marriage ending in divorce is independent of any other marriage. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

55% of marriages in the state of California end in divorce within the first 15 years.

This means that p = 0.55

Suppose 10 marriages are randomly selected.

This means that n = 10

What is the probability that less than two of them ended in a divorce?

This is

P(X < 2) = P(X = 0) + P(X = 1)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{10,0}.(0.55)^{0}.(0.45)^{10} = 0.0003

P(X = 1) = C_{10,1}.(0.55)^{1}.(0.45)^{9} = 0.0042

P(X < 2) = P(X = 0) + P(X = 1) = 0.0003 + 0.0042 = 0.0045

0.0045 = 0.45% probability that less than two of them ended in a divorce

8 0
1 year ago
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