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zzz [600]
3 years ago
5

Luke Bryant is a train conductor. He earns $36/hour and time and a half for overtime. Last week he worked 37 hours plus 8 hours

of overtime. Solve for a. straight-time pay, b. over-time pay, and c. total pay ?
Mathematics
1 answer:
balandron [24]3 years ago
3 0

Answer:

Step-by-step explanation:

a Calculate Remas wage in a week where she works 37 hours. ... To calculate the hourly rate earned when working overtime we multiply the normal hourly rate by ... overtime factor, which is 1 2 for time and a half and 2 for double time. ... 3 Gareth works as a train driver and is normally paid $11.48 per hour.

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How do I simplify a fraction? For example if the question is 7 1/5 divided by 2 2/15 what would I do to simplify?
Westkost [7]
7 1/5:
Multiply the denominator and the whole number: 7*5 = 35 and then add the numerator: 35+1= 36/5. Then, do this for the following fraction, the improper fraction value of 2 2/15 is 32/15.
Now we need 36/5 & 32/15 to have common denominators so that we can simplify it a little less complicated.

For Instance,

36/5*3= 108/15

Now we divide, (108/15)/(32/15)
108/32= 3.375 = 3 3/8

Answer is 3 3/8.

Hope this was easy to understand!
3 0
3 years ago
X+x+x+x-20 what expression is equivalent to it?
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X + x + x + x - 20 is equivalent to 4x - 20 because you can add the like terms, or the x's.
5 0
3 years ago
Read 2 more answers
A multiple choice test consists of six questions, each of which has four choices. Each question has exactly one correct answer.
mario62 [17]
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6 0
3 years ago
g A certain financial services company uses surveys of adults age 18 and older to determine if personal financial fitness is cha
leva [86]

Answer:

Null hypothesis:p_{1} = p_{2}    

Alternative hypothesis:p_{1} \neq p_{2}  

z=\frac{0.410-0.35}{\sqrt{0.385(1-0.385)(\frac{1}{1000}+\frac{1}{700})}}=2.502    

p_v =2*P(Z>2.502)= 0.0123    

Comparing the p value with the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to to reject the null hypothesis, and we can say that the proportion analyzed is significantly different between the two groups at 5% of significance.    

Step-by-step explanation:

Data given and notation    

X_{1}=410 represent the number of people indicating that their financial security was more than fair for the recent year

X_{2}=245 represent the number of people indicating that their financial security was more than fair for the year before

n_{1}=1000 sample 1 selected  

n_{2}=700 sample 2 selected  

p_{1}=\frac{410}{1000}=0.410 represent the proportion estimated of indicating that their financial security was more than fair this year

p_{2}=\frac{245}{700}=0.35 represent the proportion estimated of indicating that their financial security was more than fair the year before

\hat p represent the pooled estimate of p

z would represent the statistic (variable of interest)    

p_v represent the value for the test (variable of interest)  

\alpha significance level given  

Concepts and formulas to use    

We need to conduct a hypothesis in order to check if is there is a difference between the two proportions, the system of hypothesis would be:    

Null hypothesis:p_{1} = p_{2}    

Alternative hypothesis:p_{1} \neq p_{2}    

We need to apply a z test to compare proportions, and the statistic is given by:    

z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}   (1)  

Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{410+245}{1000+700}=0.385  

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.    

Calculate the statistic  

Replacing in formula (1) the values obtained we got this:    

z=\frac{0.410-0.35}{\sqrt{0.385(1-0.385)(\frac{1}{1000}+\frac{1}{700})}}=2.502    

Statistical decision  

Since is a two sided test the p value would be:    

p_v =2*P(Z>2.502)= 0.0123    

Comparing the p value with the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to to reject the null hypothesis, and we can say that the proportion analyzed is significantly different between the two groups at 5% of significance.    

7 0
3 years ago
A poll found that 53% of a sample of 1170 teens in a certain large country go online several times a day. (Treat this sample as
posledela

Answer:(a) margin error = 2.4%

(b) The margin error gives the measure in percentage of how the population parameter determined differ from the real population statistics or value.

(c) in 90% of the samples of teens in the country, the percent who go online several times a day will be within 50.6% and 55.4%. of the estimated 100%

Step-by-step explanation:

Using the proportion formulae

Margin error = z √p(1-p)/n

n= 1170, p = 53% = 0.53, 1-p = 0.47

and the z value at 90% C.I = 1.645

M error= 1.645 √0.53×0.47/1170

Margin error = 0.024 = 0.024 ×100

Margin error = 2.4%

53 - 2.4 = 50.6% and 53 + 2.4 = 55.4%

In other words 90% of the time: the number of teens who go online several time a day will be between 50.6 and 55.4%.

8 0
3 years ago
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