Answer:
Tony make 4 half-court shots and misses 16 half-court shots.
Step-by-step explanation:
Given:
Points earned for every successful, half-court shots = 
Points deducted for missing the shots = 
or 
Total number of shots in the game =
Final points earned by Tony = 
According to the question:
Let the number of successful half-court shot be 
And the number of shots missed be 
So,
In terms of points we can re-write it as:
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
Tony make 'x' shots that is 4 and the number of shot he missed is (20-x) =(20-4)=16
Answer:
x = 4
Step-by-step explanation:
we know that 72 and (2x + 10) add up to 90
90-72 = 18
(2x + 10) = 18
2x = 8
x = 4
The ratio 4 to 7. would be 12 and 21
Answer:
7
Step-by-step explanation:
as you can see, there is a pattern in the numbers. first, you add the term with 3, then subtract the following with 2. therefore, when we get to the sixth term 9, we will subtract 2 from it, thus getting 7. then, we add 7 and 3, to get the next term, which is 10, and so on and so forth.
Answer:
So we reject the null hypothesis and accept the alternate hypothesis that rats learn slower with sound.
Step-by-step explanation:
In this data we have
Mean= u = 18
X= 38
Standard deviation = s= 6
1) We formulate the null and alternate hypothesis as
H0: u = 18 against Ha : u > 18 One tailed test .
2) The significance level alpha = ∝= 0.05 and Z alpha has a value ± 1.645 for one tailed test.
3)The test statistics used is
Z= X- u / s
z= 38-18/6= 3.333
4) The calculated value of z = 3.33 is greater than the z∝ = 1.645
5) So we reject the null hypothesis and accept the alternate hypothesis that rats learn slower with sound.
First we set the criteria for determining the true of value of the variable. That whether the rats learn in less or more than 18 trials.
Then we find the value of z for the given significance value given and the test about to be checked.
Then the test statistic is determined and calculated.
Then both value of z and z alpha re compared. If the test statistics falls in the rejection region reject the null hypothesis and conclude alternate hypothesis is true.
The figure shows that the calulated z value lies outside the given z values
