Ron was vacationing in France, when his camera was stolen. As he walked into a camera store, Ron noticed that camera prices we
2 answers:
Answer:
Ron will be paying 195.65 euros for the camera.
Step-by-step explanation:
This question can be solved as a rule of three problem.
The exchange rate is $1.15 per euro, that is, 1 euro is $1.15.
Ron is willing to pay $225. How many euros is this? So:
1 euro - $1.15
x euros - $225.
Ron will be paying 195.65 euros for the camera.
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Answer:
a) growth
b) $500
c) 6% = 0.06
d)
e) $595.51
Step-by-step explanation:
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<u>Part (a)</u>
Exponential GROWTH, since interest is applied to the account each year.
<u>Part (b)</u>
Initial amount = $500
<u>Part (c)</u>
Growth factor = 6% = 0.06 (in decimal form)
<u>Part (d)</u>
General form of exponential growth function:
where:
- a = initial amount
- r = rate of growth (in decimal form)
- x = time (in years)
Given:
- a = 500
- r = 0.06
<u>Part (e)</u>
when x = 3
Answer:
We reject the null hypothesis and conclude that at least 2 proportions differ from the stated value.
Step-by-step explanation:
There are 3 types of gas listed in the question.
Thus;
n = 3
DF = n - 1
DF = 3 - 1
DF = 2
Let's state the hypotheses;
Null hypothesis; H0: P_regular = P_super unleaded = 20%; P_i leaded = 60%
Alternative hypothesis; Ha: At least 2 proportions differ from the stated value.
Observed values are;
Regular gas; O = 51
Unleaded gas; O = 261
Super Unleaded; O = 88
Total observed values = 51 + 261 + 88 = 400
We are told that super unleaded and regular were each selected 20% of the time and that unleaded gas was chosen 60% of the time.
Thus, expected values are;
Regular gas; E = 20% × 400 = 80
Unleaded gas; E = 60% × 400 = 240
Super Unleaded; E = 20% × 400 = 80
Formula for chi Square goodness of fit is;
X² = Σ[(O - E)²/E]
X² = (51 - 80)²/80) + (261 - 240)²/240) + (88 - 80)²/80)
X² = 13.15
From the chi Square distribution table attached and using; DF = 2 and X² = 13.15, we can trace the p-value to be approximately 0.001
Also from online p-value from chi Square calculator attached, we have p to be approximately 0.001 which is similar to what we got from the table.
Now, if we take the significance level to be 0.05, it means the p-value is less than it and thus we reject the null hypothesis and conclude that at least 2 proportions differ from the stated value.
