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2 years ago

Help please I need help with 15,16,17,18

Mathematics

1 answer:

N76 [4]2 years ago

4 0

For 15 its just the shape, if you draw a shape based on the dots, compare them. Like left one is a bit taller and wider.

16: for left one median is 8 i think. Right one is 9.5 i think
17: for left one, 13 is an outlier, for right one, 11.5 is outliner. You just write the beginning and ends including and not including those outliers.
18: people in group A were kids and people in group B were adults because their shoe sizes are much higher.

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396.8

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3 years ago

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If we let p and t be the masses of the paper and textbook, respectively, the equations that would best represent the given in this item are:
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3 years ago

Which of the lines is perpendicular to the line shown in the graph

Nitella [24]

Answer:

The line passing through (-8, 10) and (-1, 4).

Step-by-step explanation:

Two lines are perpendicular if the product of their slopes is -1. The slope of the line in the picture is $7\over 6$ , so we should find a line with slope of $-6\over 7$ .

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3 years ago

Any easier way to solve this than plugging each x value in?

Evgesh-ka [11]

Solving The Problem From Eaxh Side Is Easier

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3 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 $6\cdot 0.2 = 1.2$

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4 years ago