All categories

2 years ago

What is 1.4+3.75+6 please awnser

Mathematics

2 answers:

patriot [66]2 years ago

7 0

11.15 is the answer.

fiasKO [112]2 years ago

5 0

Thi is the answer i got 11.15

You might be interested in

ANSWERS IN MY QUESTION :)

shutvik [7]

The answers are - 3 and 5.

8 0

3 years ago

Read 2 more answers

In the exercise, X is a binomial variable with n = 5 and p = 0.2. Compute the given probability. Check your answer using technol

Ronch [10]

Answer:

0.0512

Step-by-step explanation:

We are told in the question that X isa Binomial variable. Hence we solve this question, using the formula for Binomial probability.

The formula for Binomial Probability = P(x) = (n!/(n - x) x!) × p^x × q ^n - x

In the above question,

x = 3

n = 5

p = probability for success = 0.2

q = probability for failures = 1 - 0.2 = 0.8

P(x) = (n!/(n - x) x!) × p^x × q ^n - x

P(3) = (5!/(5 - 3) 3!) × 0.2^3 × 0.8^5-3

P(3) =( 5!/2! × 3!) × 0.2³ × 0.8²

P(3) = 10 × 0.008 × 0.64

P(3) = 0.0512

6 0

3 years ago

Measure these angles. a) b)

svetoff [14.1K]

Are the angles john cena? cuz i cant see anything

3 0

2 years ago

You are looking at your power bill for the month, you pay 12 cents per kilowatt-hour. Your power bill came out to $74.80 how man

vesna_86 [32]

Answer:

623.33 KWH

Divide 74.80 by .12

5 0

2 years ago

(Y+3)(y^2-3y+9) A. Y^3+27 B. Y^3-27 C. Y^3-6y^2+27 D. Y^3+6y^2+27

KonstantinChe [14]

Answer:

A) y^3+27

Step-by-step explanation:

There are two ways of solving this problem:

1. Recognizing this as the factored form of the sum of perfect cubes

2. Distribute and add the like terms.

1. In order to distribute we must multiply y by y^2-3y+9, and then 3 by y^2-3y+9:

$(y+3)(y^2-3y+9)=y(y^2-3y+9)+3(y^2-3y+9)$

$y(y^2-3y+9)+3(y^2-3y+9)=y^3-3y^2+9y+3y^2-9y+27$

After we add the positive and negative 3y^2 and 9y, they will cancel out and be gone entirely:

$y^3-3y^2+9y+3y^2-9y+27=y^3+27$

2. You know how you can factor the difference of perfect squares?

As an example:

$a^2-b^2=(a+b)(a-b)$

Well, not many people know this but you can actually factor both the sum and difference of perfect cubes:

$a^3+b^3=(a+b)(a^2-ab+b^2)$

$a^3-b^3=(a-b)(a^2+ab+b^2)$

Because we have these identities, we can easily establish here that we have the sum of perfect cubes, and that (y+3)(y^2-3y+9)= y^3+3^3 = y^3+27

6 0

2 years ago