im ading and subtracting rational expression.my question is 5 is the numerator 2x -1 is the denominator + 6x +7 and 1 -4x squared

Answers

Answer 1

Answer: It's my method of solving this type of account, i get the denominators and multiply them by the fractions, then you cut the denominators, any questions just ask, i don't think that it was clarifying:

(2x-1) . (1-4x^2) . (5/2x-1) + (2x-1) . (1-4x^2) . (6x/1-4x^2) =
-> (1-4x^2) . 5 + (2x-1) . 6x =
-> 5 - 20x^2 + 12x^2 - 6x =
-> -8x^2 - 6x + 5 = 0

Delta = (-6)^2 - 4.(-8).5
Delta = 36 + 160 = 196

x' = - (-6) + sqroot(196) / 2.(-8)
x' = 6 + 16 / -16
x' = 22/-16 (simplifying by 2) = -11/8

x'' = 6 - 16 / -16 = -10/-16 (simplifying by 2)
x'' = 5/8

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A boat leaves the harbor entrance and travels 28 miles in the direction N 43° E. The captain then turns the boat 90° and travels another 15 miles in the direction S 47° E. At that time, how far is the boat from the harbor entrance, and what is the bearing of the boat from the harbor entrance?

A company that manufactures toothpaste is studying five different package designs. Assuming that one design is just as likely to be selected by a consumer as any other design, what selection probability would you assign to each of the package designs? We would assign a probability of to the design 1 outcome, to design 2, to design 3, to design 4, and to design 5. In an actual experiment, 100 consumers were asked to pick the design they preferred. The following data were obtained. Design Number of Times Preferred 1 10 2 5 3 30 4 40 5 15 Do the data confirm the belief that one design is just as likely to be selected as another? Explain. Yes, the sum of the assigned probabilities is 1. No, a probability of about 0.20 would be assigned using the relative frequency method if selection is equally likely. Yes, the average of the assigned probabilities is 0.20. No, a probability of about 0.50 would be assigned using the relative frequency method if selection is equally likely.

Answers

Answer:

Correct option: "No, a probability of about 0.20 would be assigned using the relative frequency method if selection is equally likely."

Step-by-step explanation:

The assumption made is that all the 5 different packages are equally likely, i.e. the probability of selecting a package is $(1)/(5)=0.20$ .

The probability distribution is shown below.

According to the probability distribution:

The probability of a person preferring design 1 is,

$P(X=1)=0.10$

The probability of a person preferring design 2 is,

$P(X=2)=0.05$

The probability of a person preferring design 3 is,

$P(X=3)=0.30$

The probability of a person preferring design 4 is,

$P(X=4)=0.40$

The probability of a person preferring design 1 is,

$P(X=5)=0.15$

So it can be seen that the probability of preferring any of the 5 designs are not same.

Thus, the designs are not equally likely.

The correct option is "No, a probability of about 0.20 would be assigned using the relative frequency method if selection is equally likely."

The selection Probability determined using the relative frequency method do not match the assigned probabilities, suggesting that the data do not confirm the belief that one design is as likely to be selected as another.

The given data can be used to calculate the relative frequencies of each package design selected by the consumers.

To determine the selection probabilities using the relative frequency method, divide the number of times a design was preferred by the total number of consumers.

For example, for design 1, the selection probability would be 10/100 = 0.1.

Similarly, for design 2, the selection probability would be 5/100 = 0.05.

The selection probabilities for designs 3, 4, and 5 would be 0.3, 0.4, and 0.15 respectively.

Comparing these probabilities to the assigned probabilities, it can be observed that the assigned probabilities do not match the observed relative frequencies, indicating that the data do not confirm the belief that one design is just as likely to be selected as another.

Learn more about Probability here:

brainly.com/question/22962752

#SPJ3

Please help as soon as you can :(Alicia rolls two fair number cubes numbered from 1 to 6. She first defines the sample space, as shown below:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5,) (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5,) (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5,) (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5,) (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5,) (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5,) (6, 6)
Based on the sample spaces, what is the probability of getting a total of 7?

A.) 4/36
B.) 5/36
C.) 6/36
D.) 8/36

Hope you can help!

Answers

The answer is C
hope this helped...

Answer:

6/36

(C.)

I hope this helps you if it hasn't already! Bye.

Who can possibly beat me?

A. Kakarot
B. Gohan
C. Broly
D. Frieza

Answers

Answer:

Step-by-step explanation:

Answer c

Explanation

A high school auditorium seats 110 people. The school play has 106 people in attendance leaving 4 seats empty.because the order in which
There are
the seats are chosen
ways that 4 seats can be left empty in the auditorium. This is a
important.

Answers

Answer:

5773185

Step-by-step explanation:

There are 110 seats

110 ways to choose the first empty seat

Now there are 109 seats

109 ways to choose the next empty seat

Now there are 108 seats

108 ways to choose the next empty seat

Now there are 107 seats

110*109*108*107=138556440

Now the order of the empty seats doesn't matter so we need to divide by 4!

138556440/ 4!

138556440/ 24

5773185

Final answer:

In this mathematics problem, we are asked to determine the number of ways that 4 seats can be left empty in a high school auditorium that seats 110 people. We can use the concept of combinations to solve this.

Explanation:

In this problem, we are asked to determine the number of ways that 4 seats can be left empty in a high school auditorium that seats 110 people. To solve this, we can use the concept of combinations. The total number of ways to choose 4 seats out of 110 is represented by the combination formula: C(110, 4). To calculate this, we can use the formula: C(n, r) = n! / (r!(n - r)!), where n is the total number of seats and r is the number of seats left empty. Plugging in the values, we have C(110, 4) = 110! / (4!(110 - 4)!).

Using a calculator, we can simplify this expression and calculate the answer.