Given:
A student says that the graph of the equation 
is the same as the graph of 
, only translated upwards by 8 units.
To find:
Whether the student is correct or not.
Solution:
Initial equation is
Equation of after transformation is
Now,
...(i)
The translation is defined as
...(ii)
Where, a is horizontal shift and b is vertical shift.
If a>0, then the graph shifts a units left and if a<0, then the graph shifts a units right.
If b>0, then the graph shifts b units up and if b<0, then the graph shifts b units down.
From (i) and (ii), we get
Therefore, the graph of translated left by 8 units. Hence, the student is wrong.
Answer:
-$13.5
Step-by-step explanation:
Let x be a random variable of a count of player gain.
- We are told that if the die shows 3, the player wins $45.
- there is a charge of $9 to play the game
If he wins, he gains; 45 - 9 = $36
If he looses, he has a net gain which is a loss = -$9
Thus, the x-values are; (36, -9)
Probability of getting a 3 which is a win is P(X) = 1/6 since there are 6 numbers on the dice and probability of getting any other number is P(X) = 5/6
Thus;
E(X) = Σ(x•P(X)) = (1/6)(36) + (5/6)(-9)
E(X) = (1/6)(36 - (5 × 9))
E(X) = (1/6)(36 - 45)
E(X) = -9/6 = -3/2
E(X) = -3/2
This represents -3/2 of $9 = -(3/2) × 9 = - 27/2 = -$13.5
Answer:
7
Step-by-step explanation:
Answer:
Claim : men weigh of wild jackalopes is 69.9
The null hypothesis : H0 : μ = 69.9
Alternative hypothesis : H1 : μ ≠ 69.9
Test statistic = −2.447085
P value = 0.0174
Conclusion :
Fail to Reject the Null hypothesis
Step-by-step explanation:
From the question given :
The claim is that : mean weight of wild jackalopes is still the same as 10 years with a mean weight of 69.9 lbs.
The null hypothesis : H0 : μ = 69.9
Alternative hypothesis : H1 : μ ≠ 69.9
Using calculator :
Sample mean (x) = 66
Sample standard deviation (s) = 12.345
The test statistic t :
(x - μ) / (s/√n)
n = sample size = 60
(66 - 69.9) / (12.345 / √60)
t = −2.447085
P value at α 0.01, df = 59 is 0.0174
Since the p value is > 0.01, the result is not significant at 0.01. Therefore, we fail to reject the Null
