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2 years ago

PLEASE ANSWER MY RECENT! ITS FOR 47 POINTS AND SUPER EASY! I JUST NEED ADVICE AND NO ONE HAS ANSWERD YET!

Mathematics

2 answers:

Anettt [7]2 years ago

5 0

How do i look at your last question

Ann [662]2 years ago

5 0

Answer:

Step-by-step explanation:

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Question 1:

Morgarella [4.7K]

Answer:

Step-by-step explanation:

KE = ½mv²

KE₁₀ = ½(1100)10² = 55000 = 55 kJ

KE₃₀ = ½(1100)30² = 495000 = 495 kJ

9 times greater

3 0

2 years ago

If f(x) = 2(34) + 1, what is the value of f(2)? <br>

larisa86 [58]

I believe that f(2) is 34.5

f(2) = 2(34) + 1
f(2) = 68 + 1
f(2) = 69
f = 69/2
f = 34.5

I AM NOT SURE

3 0

2 years ago

2/3*x+3/4=8 what is x

matrenka [14]

Answer: x=10 7/8

Step-by-step explanation:

Subtract 3/4 from both sides. Multiply both sides by 3. Divide both sides by 2 (basically just isolate x on one side of the equal sign).

7 0

3 years ago

A tree casts a shadow of 3 meters, while a meter stick casts a shadow of 0.75 meters. How tall is the tree?

Lunna [17]

3.75 meters I would presume.

Explanation: since a single meter casts a shadow of 0.75 meters that means for each meter, 0.25 meters are missing. So I added 0.75 to three, because there was 3 meters and 3 time 0.25 was 0.75 .

6 0

3 years ago

A manufacturing company has two retail outlets. It is known that 30% of all potential customers buy

aleksandrvk [35]

Answer:

(a) A y P(A) = 0.4 (b) $\bar{B}$ y $P(\bar{B})$ =0.5 (c) $\bar{A}$ ∪ $\bar{B}$ y P( $\bar{A}$ ∪ $\bar{B}$ ) = 0.9 (d) $\bar{A}$ ∩ $\bar{B}$ y P( $\bar{A}$ ∩ $\bar{B}$ )=0.2

Step-by-step explanation:

A was defined as the event that a potential customer, randomly chosen, buys from outlet 1 in the original problem statement. We know that B denotes the event that a randomly chosen customer buys from outlet 2. So

P( $A\cap \bar{B}$ ) = 0.3, P( $B\cap \bar{A}$ ) = 0.4 and P( $A\cap B$ ) = 0.1

(a) P(A) = P( $A\cap (B\cup\bar{B})$ ) = P( $A\cap B$ ) + P( $A\cap \bar{B}$ ) = 0.1 + 0.3 = 0.4

(b) P(B) = P( $B\cap (A\cup\bar{A})$ ) = P( $B\cap A$ ) + P( $B\cap \bar{A}$ ) = 0.1 + 0.4 = 0.5

P( $\bar{B}$ ) = 1-P(B) = 1-0.5 = 0.5

(c) The customer does not buy from outlet 1 is the complement of A, i.e., $\bar{A}$ , and the customer does not buy from outlet 2 is the complement of B, i.e., $\bar{B}$ , so, the customer does not buy from outlet 1 or does not buy from outlet 2 is $\bar{A}$ ∪ $\bar{B}$ and P( $\bar{A}$ ∪ $\bar{B}$ ) = P( $(A\cap B)^{c}$ ) by De Morgan's laws

P( $(A\cap B)^{c}$ ) = 1-P(A∩B)=1-0.1=0.9

(d) The customer does not buy from outlet 1 is the complement of A, and the customer does not buy from outlet 2 is the complement of B, so we have that the statement in (d) is equivalent to $\bar{A}$ ∩ $\bar{B}$ and P( $\bar{A}$ ∩ $\bar{B}$ ) = P( $(AUB)^{c}$ ) by De Morgan's laws, and

P( $(AUB)^{c}$ ) = 1-P(A∪B)=1-[P(A)+P(B)-P(A∩B)]=1-[0.4+0.5-0.1]=1-0.8=0.2

8 0

2 years ago