5. Lauren has $50 to spend at the Strawberry Festival. She spends $ 10 on admission and $12 on strawberry shortcake. She wants t
1 answer:
For this case we have that Lauren's initial amount of money was $ 50, if he spends $10 and $12 on admission and strawberry pie respectively we have:
$50 - ($10 + $12) = $50- $22 = $28
That is, she has $ 28 left
n: Let the variable that represents the amount of games Lauren can play.
Then, it must be fulfilled that:
Thus, Lauren can play 37 times the game (at most)
ANswer:
Thus, Lauren can play 37 times the game (at most)
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There are two ways to work this out: normal variables or using "imaginary" numbers.
Normal variables:
![(7+2i)(3-i)\\(7*3)+[7*(-i)]+(3*2i)+[2i*(-i)]\\21-7i+6i-2i^{2}\\\\21-i-2i^{2}](https://tex.z-dn.net/?f=%20%287%2B2i%29%283-i%29%5C%5C%287%2A3%29%2B%5B7%2A%28-i%29%5D%2B%283%2A2i%29%2B%5B2i%2A%28-i%29%5D%5C%5C21-7i%2B6i-2i%5E%7B2%7D%5C%5C%5C%5C21-i-2i%5E%7B2%7D)
Imaginary numbers:
Using the result from earlier:
Now since
, then the expression becomes:
Normal variables:
Imaginary numbers:
Using the result from earlier:
Now since
Answer:
t = 2.2450
d. 0.264
Step-by-step explanation:
The null hypothesis is:
Alternative hypothesis;
The pooled variance t-Test would have been determined if the population variance are the same.
The t-test statistics can be computed as:
t = 2.2450
Degree of freedom
df = (8-1)+(8-1)
df = 7 + 7
df = 14
At df = 14 and ∝ = 0.05;
Decision Rule: To reject the null hypothesis if the t-test is greater than the critical value.
Conclusion: We reject and there is sufficient evidence to conclude that the test scores for contact address s less than Noncontact athletes.
To calculate r²
The percentage of the variance is;