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BartSMP [9]
3 years ago
7

Question 2 (1 point)

Mathematics
1 answer:
BaLLatris [955]3 years ago
8 0

Answer:

All options are wrong.................

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This is the question Below.
Digiron [165]

Answer:

5

Step-by-step explanation:

5

4 0
2 years ago
can someone help me solve this question
kodGreya [7K]

Answer:

D

Step-by-step explanation:

y=mx+b

m = slope

b = y intercept

m = 5

b = -8

y=5x-8

4 0
3 years ago
Read 2 more answers
Suppose that the revenue function for a certain product is given by R(x) = 17(2x + 1)−1 + 34x − 17 where x is in thousands of un
REY [17]

Answer:

Step-by-step explanation:

Given that:

R(x) = \frac{17}{2x+1} + 34x − 17

As we know that derivative of revenue function is marginal revenue function .

We will use following rules of derivative

=> dR/ dx = \frac{-17*2}{(2x+1)^{2} } + 34

=> R' (x) =   \frac{-34}{(2x+1)^{2} } + 34

=> R '(2000) =  \frac{-34}{(2*2000+1)^{2} } + 34 = 34

The revenue when 2000 units are sold is:

R(2000) = \frac{17}{2*2000+1} + 34*2000 − 17 = $69,783

3 0
3 years ago
A group of four boys (Alex, Bryan, Chris, and David) and five girls (Megan, Nancy, Olivia, Pauline, and Rebecc?
aliina [53]

Answer:

a) P = 362880 ways

b) P = 2880 ways

Step-by-step explanation:

a) We have four boys and five girls, they are going to sit together in a row of 9 theater seats, without restrictions

We have a permutation of 9 elements

P = 9!

P = 9*8*7*6*5*4*3*2*1

P = 362880 ways

b) Boys must seat together, we have two groups of people

4 boys  they can seat in 4! different ways

P₁ = 4!

P₁ = 4*3*2*1

P₁ = 24

And girls can seat in 5! dfferent ways

P₂ = 5!

P₂ = 5*4*3*2*1

P₂ = 120

To get total ways in the above mentioned condition, we have to multiply P₁*P₂

P = 24*120

P = 2880 ways

4 0
3 years ago
Rewrite each statement using the appropriate mathematical language 1. There exists a number b belonging to the set B 2. Even num
Shtirlitz [24]

Answer:

1. b ∈ B 2. ∀ a ∈ N; 2a ∈ Z 3. N ⊂ Z ⊂ Q ⊂ R 4. J ≤ J⁻¹ : J ∈ Z⁻

Step-by-step explanation:

1. Let b be the number and B be the set, so mathematically, it is written as

b ∈ B.

2. Let  a be an element of natural number N and 2a be an even number. Since 2a is in the set of integers Z, we write

∀ a ∈ N; 2a ∈ Z

3. Let N represent the set of natural numbers, Z represent the set of integers, Q represent the set of rational numbers, and R represent the set of rational numbers.

Since each set is a subset of the latter set, we write

N ⊂ Z ⊂ Q ⊂ R .

4. Let J be the negative integer which is an element if negative integers. Let the set of negative integers be represented by Z⁻. Since J is less than or equal to its inverse, we write

J ≤ J⁻¹ : J ∈ Z⁻

4 0
3 years ago
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