Which set of data would most likely show no correlation when graphed on a scatterplot?

LAST QUESTION! 12 Points! Will mark brainliest! Links will be ignored and deleted.

insens350 [35]

Answer:

i belive its 35/8 3.14

Step-by-step explanation:

sorry if im wrong

4 0

3 years ago

Read 2 more answers

Please help I don't understand

valina [46]

17) The line that is John is closer to certain but not exactly on certain so I'd say it's "likely"

18) Using that same example as the one above, if 0 means impossible, then 1 means certain. I'd place your label for Sara at 1.

19) Let s = sides of the square. There are 4 sides and all sides are equal so we can set up the equation 4s = 64. Divide both sides by 4. s = 16. Now add 2 to this. 16 + 2 = 18. The length of each side of the original square was 18.

20) If the scale is 1 cm : 1 meter and the height of the model is 16 cm, then the actual plane must be 16 meters.

21) 2x and 100 are supplementary angles meaning together they must add up to 180.

2x + 100 = 180
2x = 180 - 100
2x = 80
Divide both sides by 2
x = 40

22) Assuming 3x is supposed to be the right angle, right angles are 90 degrees.

3x = 90
Divide by 3 on both sides
x = 30.

6 0

3 years ago

Which of the following would be an acceptable first step in simplifying the expression

krek1111 [17]

Answer: A

Step-by-step explanation: Because it can be simplified twice.

7 0

1 year ago

A university has 750 female students and 2250 male students. Thirty per cent of female students study chemistry and 60% of male

UNO [17]

Answer:

1/7 or 14.3%

Step-by-step explanation:

30% of 750 = 225 (females who study chemistry)

60% of 2250 = 1350 (males who study chemistry)

1575 = total students studying chemistry

P (female chemistry student) = 225/1575 = 1/7 (simplified)

1/7 = 14.3%

7 0

2 years ago

Discrete R.V. Assume a student shows up for EGR 280 completely unprepared and the instructor gives a pop quiz. Assume there are

Tom [10]

Answer:

a) p=0.2

b) probability of passing is 0.01696

.

c) The expected value of correct questions is 1.2

Step-by-step explanation:

a) Since each question has 5 options, all of them equally likely, and only one correct answer, then the probability of having a correct answer is 1/5 = 0.2.

b) Let X be the number of correct answers. We will model this situation by considering X as a binomial random variable with a success probability of p=0.2 and having n=6 samples. We have the following for k=0,1,2,3,4,5,6

$P(X=k) = \binom{n}{k}p^{k}(1-p)^{n-k} = \binom{6}{k}0.2^{k}(0.8)^{6-k}$ .

Recall that $\binom{n}{k}= \frac{n!}{k!(n-k)!}$ In this case, the student passes if X is at least four correct questions, then

$P(X\geq 4) = P(X=4)+P(X=5)+P(X=6)=\binom{6}{4}0.2^{4}(0.8)^{6-4}+\binom{6}{5}0.2^{5}(0.8)^{6-5}+\binom{6}{6}0.2^{6}(0.8)^{6-6}= 0.01696$

c)The expected value of a binomial random variable with parameters n and p is $E[X] = np$ . IN our case, n=6 and p =0.2. Then the expected value of correct answers is $6\cdot 0.2 = 1.2$

5 0

3 years ago