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3 years ago

V=4r How would you solve for 'r'

Mathematics

2 answers:

kaheart [24]3 years ago

5 0

$r=\dfrac{V}{4}$

Fynjy0 [20]3 years ago

3 0

V = 4r

4r = V Divide both sides by 4

4r/4 = V/4

r = V/4

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Mr. Smith is paid $9. 50 per hour for fixing watches. Write an expression for the amount of money Mr. Smith earns in h hours. In

IrinaK [193]

Answer:

$9.50h, $(9.50h +2.20rh)

Step-by-step explanation:

Every hour ----- $9.50

h hours ----- h ×$9.50

Amount of money Mr Smith earns in h hours

= $9.50h

If he fixes a watch per hour, additional income= $2.20h

If he fixes r watches per hour, additional income= $2.20rh

Total income

= $(9.50h +2.20rh)

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2 years ago

Write an equation to find the number. Solve. Twenty-four less than 4 times a number is equal to 30 more than -2 times the number

Amanda [17]

4x - 24 = -2x + 30

4x + 2x = 30 + 24
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answer: the number is 9

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3 years ago

Read 2 more answers

What is the measure of XYZ?

Anuta_ua [19.1K]

<XYZ = 1/2(86 + 44)

<XYZ = 1/2 (130)

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Answer

B. 65

8 0

3 years ago

Read 2 more answers

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