Answer:
The answer would be A)98.
I got this by taking the 3 test scores that were provided and adding them then subtracting that by 360 (90x4 - which would give you a 90 in the end) so I had 360 - 262 which equaled 98 and that would be the score needed to have a 90 in the end.
From the information given, a right-tailed test using the z-distribution is appropriate.
At the null hypothesis, it is <u>tested if the class of statistics students is not significantly smarter than the general population</u>, that is, their mean IQ is of at most 100, hence:
At the alternative hypothesis, it is <u>tested if they are significantly smarter</u>, that is, their mean IQ is greater than 100, hence:
- We are testing if the mean is <u>greater than a value</u>, hence it is a right-tailed test.
- We have the <u>standard deviation for the population</u>, hence, the z-distribution is used.
A similar problem is given at brainly.com/question/23413408
Answer: 81%
Step-by-step explanation:
From the question, we are informed that a student received the following test scores: 71%, 89%, 72%,
84% and 83% in 5 tests and the student wants to maintain an average of 80%.
The lowest score/grade they can receive on the next test to maintain at least an 80% average first thus:
First, to make it easy we can remove the percent sign. Then we multiply 80 by 6 since we're calculating for 6 tests scores. This will be:
= 80 × 6
= 480
We then add all the 5 test scores. This will be:
= 71 + 89 + 72 + 84 + 83
= 399
We then subtract the values gotten. This will be:
= 480 - 399
= 81
This means the student must get at least 81%
Answer: he can pay for 6 people total (including him)
Step-by-step explanation:
first lets subtract the flat fee from the total amount of money he has
70 - 40 = 30
now lets divide 30 by $4.50 to see how many people he can pay for
30 / $4.50 = about 6.5
obviously you cant pay for half a person so he can invite 5 people (he also has to pay for himself)