All categories

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

You might be interested in

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

3 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