All categories

2 years ago

PLZ HELP I DONT KNOW ABOUT GRAPHS!!!!! (and please don’t say the wrong answer just to get points) :C

Mathematics

2 answers:

avanturin [10]2 years ago

6 0

Answer:

Point (4,100) means that the ship took 4 hours to travel 100 miles.

Step-by-step explanation:

I think it's just asking for you to name any point and explain what that points means...

--> Here's mine I guess, Point (4,100) means that the ship took 4 hours to travel 100 miles.

Alexxandr [17]2 years ago

3 0

Answer:(50,2) 50 miles can be traveled in 2 hours I think.

Sorry if im wrong :c

You might be interested in

A(n)=-14+3(n-1)(2) find A(9)

Romashka [77]

Answer: 34

Step-by-step explanation:

Substitute n with 9

-14+3(9-1)(2)

-14+3(8)(2)

= 34

7 0

1 year ago

PLEASE HELP ME FAST!!!

yawa3891 [41]

Answer:

ans -------8/5 pansnhshshsh

6 0

2 years ago

Read 2 more answers

A simple random sample of 13 four-cylinder cars is obtained, and the braking distances are measured. The mean braking distance i

Tanya [424]

G shshshusuj bshshsu. Haha yah

4 0

2 years ago

You have been assigned 40 homework assignments this marking period. You have completed 18 of them. What percent of homework assi

djyliett [7]

Answer:

45%

Step-by-step explanation:

Simply divide your partial amount (18 in this case) by your total (40), and you then multiply your answer from 18/40= .45 by 100, and you get 45. This is 45%

4 0

2 years ago

Read 2 more answers

A student is getting ready to take an important oral examination and is concerned about the possibility of having an "on" day or

Tamiku [17]

Answer:

The students should request an examination with 5 examiners.

Step-by-step explanation:

Let X denote the event that the student has an “on” day, and let Y denote the

denote the event that he passes the examination. Then,

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

The events ( $Y|X$ ) follows a Binomial distribution with probability of success 0.80 and the events ( $Y|X^{c}$ ) follows a Binomial distribution with probability of success 0.40.

It is provided that the student believes that he is twice as likely to have an off day as he is to have an on day. Then,

$P(X)=2\cdot P(X^{c})$

Then,

$P(X)+P(X^{c})=1$

⇒

$2P(X^{c})+P(X^{c})=1\\\\3P(X^{c})=1\\\\P(X^{c})=\frac{1}{3}$

Then,

$P(X)=1-P(X^{c})\\=1-\frac{1}{3}\\=\frac{2}{3}$

Compute the probability that the students passes if request an examination with 3 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{3}_{x=2}{{3\choose x}(0.80)^{x}(1-0.80)^{3-x}}]\times\frac{2}{3}+[\sum\limits^{3}_{x=2}{{3\choose x}(0.40)^{3}(1-0.40)^{3-x}}]\times\frac{1}{3}$

$=0.715$

The probability that the students passes if request an examination with 3 examiners is 0.715.

Compute the probability that the students passes if request an examination with 5 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{5}_{x=3}{{5\choose x}(0.80)^{x}(1-0.80)^{5-x}}]\times\frac{2}{3}+[\sum\limits^{5}_{x=3}{{5\choose x}(0.40)^{x}(1-0.40)^{5-x}}]\times\frac{1}{3}$

$=0.734$

The probability that the students passes if request an examination with 5 examiners is 0.734.

As the probability of passing is more in case of 5 examiners, the students should request an examination with 5 examiners.

8 0

2 years ago