The rate of interest is 75 % per year
<em><u>Solution:</u></em>
Given that, Jamerra received a $3,00 car loan
she plans on paying off the loan in 2 years
<em><u>Jamerra will have paid $450 in interest</u></em>
Therefore, we get
Principal = $ 300
Number of years = 2
Simple Interest = $ 450
Rate of interest = ?
<em><u>The simple interest is given by formula:</u></em>
Where,
"p" is the principal and "n" is the number of years and "r" is the rate of Interest
<em><u>Substituting the given values we get,</u></em>
Thus rate of interest is 75 % per year
The debt-to-credit ratio of a credit account is the ratio of the balance to the credit limit:
... balance/credit limit = 245.78/3500 = 0.07022... ≈ 7.02%
The appropriate choice is ...
... A.) 7.02%
Answer:
No student liked only Nokia
Step-by-step explanation:
The information can be illustrated as shown on the Venn diagram.
The number of students who liked only Nokia is x.
The sum of all regions in the Venn diagram should be 40.
We add all to get:
Therefore no one liked only Nokia
Answer: Choice B
Two and four tenths multiplied by the difference of six and two tenths and a number is more than negative four and five tenths.
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Explanation:
2.4 = 2 + 0.4
2.4 = 2 and 4/10
2.4 = 2 and 4 tenths
2.4 = two and four tenths
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Through similar reasoning,
6.2 = six and two tenths
And also,
-4.5 = negative four and five tenths
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Notice how 6.2 - x translates into "difference of six and two tenths and a number"
We then multiply that by 2.4, aka two and four tenths.
So that's how we get the phrasing "Two and four tenths multiplied by the difference of six and two tenths and a number"
All of this is greater than -4.5 aka negative four and five tenths.
This points us to Choice B as the final answer.
Answer:
(0.4958, 0.7422)
Step-by-step explanation:
Let p be the true proportion of water specimens that contain detectable levels of lead. The point estimate for p is 
. The estimated standard deviation is given by 
. Because we have a large sample, the 90% confidence interval for p is given by 
where 
is the value that satisfies that above this and under the standard normal density there is an area of 0.05. So, the confidence interval is 
, i.e., (0.4958, 0.7422).
