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2 years ago

Mrs. Jackson made a line plot of the times it took students to complete a project.

Mathematics

1 answer:

Komok [63]2 years ago

8 0

This value is very different from the other values

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Kris wanted to understand whether students at her school were in favor of an extended school day. She surveyed some students and

puteri [66]

the table below:

Grade In favor Opposed Undecided Total

Grade 9 6 4 8 18

Grade 10 10 11 9 30

Grade 11 12 15 11 38

Grade 12 15 6 14 35

If the principal randomly selects a student in grade 10 from this survey,

the probability that the student is opposed to extending the school day

P ( student in grade 10 is opposed)= Number of student in opposed(grade10) / total number of students in grade 10

Number of student opposed who is in grade10 = 11

total number of students in grade 10 = 30

P( student in grade 10 is opposed)= $\frac{11}{30}$

= 0.37

the probability that the student is opposed to extending the school day= 0.37

5 0

2 years ago

andy is wrapping gifts for his family. he ties 25 inches of ribbon around each gift. which function gives the relationship betwe

Vaselesa [24]

The total length of the ribbon used is expressed as r and is a product of the number of gifts (g) and the length of ribbon tied in each gift. This relationship may be expressed by the equation,
r = (25) x (g)

8 0

2 years ago

HELP ME PLEASE find m

chubhunter [2.5K]

There is no m in the diagram showed

4 0

2 years ago

Read 2 more answers

Determine the value of y, if x is – 7.<br> y= x2 - 9

Lapatulllka [165]

Answer:

y= -23

Step-by-step explanation:

6 0

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 $6\cdot 0.2 = 1.2$

5 0

3 years ago