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yan [13]
3 years ago
6

michael is paid $12.25 per hour. he worked 28 hours this week. he put 1/4 of his weekly pay in his savings account. how much did

he have left?
Mathematics
2 answers:
Archy [21]3 years ago
8 0

Answer:

<u><em>257.25</em></u>

Step-by-step explanation:

$12.25 * 28

=  343

343 / 4

= 85.75 (the amount he pays in his savings account)

So 3/4 of the total amount he have left

85.75 * 3

=<u><em>257.25</em></u>

PSYCHO15rus [73]3 years ago
6 0

12.25  ( 28) (3/4)   =  ???

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Twenty percent of a radioactive substance decays in ten years by what percent does the substance decay each year
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Answer:

2%

Step-by-step explanation:

20% takes 10 years to decay

To find per year decay, we divide the percentage by the number of years, so we get unit rate of years {per year]. Thus we have:

20%/10 = 2% per year

It's that simple.

Hence, the percentage by which the substance decays each year is 2%

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3 years ago
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A random sample of 100 people from City A has an average IQ of 120 with a SD of 18. Independently of this, a random sample of 15
Sloan [31]

Answer:

z=\frac{(120-116)-0}{\sqrt{\frac{18^2}{100}+\frac{15^2}{150}}}}=1.837

p_v =P(z>1.837)=1-P(Z  

Comparing the p value with a significance level for example \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the average IQ on city A is signficantly higher than city B at 5% of singificance.  

Step-by-step explanation:

\bar X_{A}=120 represent the mean for sample 1

\bar X_{B}=116 represent the mean for sample 2

s_{A}=18 represent the sample standard deviation for 1  

s_{B}=15 represent the sample standard deviation for 2  

n_{A}=100 sample size for the group 2  

n_{B}=150 sample size for the group 2  

\alpha Significance level provided

z would represent the statistic (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if residents of City A smarter on average, the system of hypothesis would be:  

Null hypothesis:\mu_{A}-\mu_{B}\leq 0  

Alternative hypothesis:\mu_{A} - \mu_{B}> 0  

We don't have the population standard deviation's, but the sample sizes are large enough we can apply a z test to compare means, and the statistic is given by:  

z=\frac{(\bar X_{A}-\bar X_{B})-\Delta}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}} (1)

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.

With the info given we can replace in formula (1) like this:  

z=\frac{(120-116)-0}{\sqrt{\frac{18^2}{100}+\frac{15^2}{150}}}}=1.837

P value

Since is a one right tailed test the p value would be:  

p_v =P(z>1.837)=1-P(Z  

Comparing the p value with a significance level for example \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the average IQ on city A is signficantly higher than city B at 5% of singificance.  

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