1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
GrogVix [38]
2 years ago
13

a student gets 11 out of 16 problems correct on a homework assignment. what percent of the assignment did the student get corret

?. thanks if someone answers this​
Mathematics
1 answer:
Greeley [361]2 years ago
7 0

Answer:

68.75 Percent

Step-by-step explanation:

11/16 =0.6875

0.6875 into percent is 68.75

You might be interested in
P(X< ) 1-P(X> ) A softball pitcher has a 0.626 probability of throwing a strike for each curve ball pitch. If the softball
Mashutka [201]

Answer:

19.49% probability that no more than 16 of them are strikes

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 30, p = 0.626

So

\mu = E(X) = np = 30*0.626 = 18.78

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{30*0.626*0.374} = 2.65

What is the probability that no more than 16 of them are strikes?

Using continuity correction, this is P(X \leq 16 + 0.5) = P(X \leq 16.5), which is the pvalue of Z when X = 16.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{16.5 - 18.78}{2.65}

Z = -0.86

Z = -0.86 has a pvalue of 0.1949

19.49% probability that no more than 16 of them are strikes

7 0
3 years ago
Question 5
krek1111 [17]

Answer: 3/4 or 75%

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Help me solve this question!
MakcuM [25]

Answer:

1st quartile is 136

The second quartile is also the median 142

The third quartile is 162

The interquartile range is the difference between the 3rd and first quartile or 26

Step-by-step explanation:

5 0
2 years ago
What is the length of the hypotenuse of a right triangle whose legs have lengths of 5 and 12?
ruslelena [56]
It would be 13... if I did my math right.
6 0
3 years ago
Read 2 more answers
In a certain community, 30% of the families own a dog, and 20% of the families that own a dog also own a cat. It is also known t
Komok [63]

Answer:

What is the probability that a randomly selected family owns a cat? 34%

What is the conditional probability that a randomly selected family doesn't own a dog given that it owns a cat? 82.4%

Step-by-step explanation: We can use a Venn (attached) diagram to describe this situation:

Imagine a community of 100 families (we can assum a number, because in the end, it does not matter)

So, 30% of the families own a dog = .30*100 = 30

20% of the families that own a dog also own a cat = 0.2*30 = 6

34% of all the families own a cat = 0.34*100 = 34

Dogs and cats: 6

Only dogs: 30 - 6 = 24

Only cats: 34 - 6 = 28

Not cat and dogs: 24+6+28 = 58; 100 - 58 = 42

What is the probability that a randomly selected family owns a cat?

34/100 = 34%

What is the conditional probability that a randomly selected family doesn't own a dog given that it owns a cat?

A = doesn't own a dog

B = owns a cat

P(A|B) = P(A∩B)/P(B) = 28/34 = 82.4%

5 0
3 years ago
Read 2 more answers
Other questions:
  • Which statement below shows the best use of friendly numbers to estimate the value of 27 x 4
    6·2 answers
  • What is this SSS,AAS,ASA,SAS
    9·1 answer
  • This is urgent can someone please help.......
    5·2 answers
  • The cost, c(x), for a taxi ride
    11·1 answer
  • Please help quickly :D
    6·1 answer
  • Bess and Carolina and their 14 classmates write their names on cards. The teacher, Miss Scott, picks two cards without replacing
    8·1 answer
  • A student said that the relation{(-2,3),(4,4),(4,5),(5, -8)} is a function. Do you agree? Why or why not?
    14·1 answer
  • In a population of 128 people, 45 have curly hair, 66 have straight hair, and 17 have no hair. What is the probability of random
    11·1 answer
  • You have a job earning $12.50 per hour, and you receive a raise so that you earn $13.25 per hour. What is the percent change in
    6·1 answer
  • Which equation could represent a linear combination of the system?
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!