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2 years ago

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Ainsley works for the owners of a bookstore. Her starting salary was 24,500, and she gets a 3% raise each year. Write an equatio

n in function notation to represent subsidy’s salary as a function of the number of years she has been working at the bookstore.

1 answer:

likoan [24]2 years ago

5 0

Answer: A (t) = 24500 (1+0.03)^t

Step-by-step explanation:

Hi, to answer this question we have to write a growth function:

A = P (1+r) ^t

Where:

A = amount after growth

P = initial amount

r = growth rate (in this case is yearly, since she gets a 3% raise each year)

t = years

Replacing with the values given:

The growth rate must be in decimal form (divided by 100)

A (t) = 24500 (1+3/100)^t

A (t) = 24500 (1+0.03)^t

Feel free to ask for more if needed or if you did not understand something .

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4. Hydrocarbons in the cab of an automobile were measured during trips on the New Jersey Turnpike and trips through the Lincoln

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No these these result do not differ at 95% confidence level

Step-by-step explanation:

From the question we are told that

The first concentrations is $c _1= 30.0 \ g/m^3$

The second concentrations is $c _2 = 52.9 \ g/m^3$

The first sample size is $n_1 = 32$

The second sample size is $n_2 = 32$

The first standard deviation is $\sigma_1 = 30.0$

The first standard deviation is $\sigma_1 = 29.0$

The mean for Turnpike is $\= x _1 = \frac{c_1}{n} = \frac{31.4}{32} = 0.98125$

The mean for Tunnel is $\= x _2 = \frac{c_2}{n} = \frac{52.9}{32} = 1.6531$

The null hypothesis is $H_o : \mu _1 - \mu_2 = 0$

The alternative hypothesis is $H_a : \mu _1 - \mu_2 \ne 0$

Generally the test statistics is mathematically represented as

$t = \frac{\= x_1 - \= x_2}{ \sqrt{\frac{\sigma_1^2}{n_1} +\frac{\sigma_2^2}{n_2} }}$

$t = \frac{0.98125 - 1.6531}{ \sqrt{\frac{30^2}{32} +\frac{29^2}{32} }}$

$t = - 0.0899$

Generally the degree of freedom is mathematically represented as

$df = 32+ 32 - 2$

$df = 62$

The significance $\alpha$ is evaluated as

$\alpha = (C - 100 )\%$

=> $\alpha = (95 - 100 )\%$

=> $\alpha =0.05$

The critical value is evaluated as

$t_c = 2 * t_{0.05 , 62}$

From the student t- distribution table

$t_{0.05, 62} = 1.67$

So

$t_c = 2 * 1.67$

=> $t_c = 3.34$

given that

$t_c > t$ we fail to reject the null hypothesis so this mean that the result do not differ

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