All categories

3 years ago

PLEASE HELPPP!!!!

Mathematics

1 answer:

Salsk061 [2.6K]3 years ago

8 0

40% of $3,200 is $ 1,280

william payed patio town $1,280 for the materials

You might be interested in

Please help me with this

timama [110]

Answer:

choice 1

Step-by-step explanation:

angle 1 = angle 7 (exterior alternate angle)

angle 3=angle 5 (interior alternate angle)

3 0

1 year ago

Question 2 In the fall, a community service club sold 10 shirts and 20 hats for a total of $490. In the spring, the club sold 20

Stells [14]

To answer the question above, we let s be the cost of a shirt and h be the cost of each hat such that the equations that would best represent the given are:
10s + 20h = 490
20s + 30h = 860
The system of equations is the answer to the first question and the values of s and h are $25 and $12, respectively.

8 0

3 years ago

If J is the centroid of cde, de=52, fc=15, he=14, find each missing measure

const2013 [10]

Answer:

$DG = 26$

$GE = 26$

$DF = 15$

$CH = 14$

$CE = 28$

Step-by-step explanation:

The figure has been attached, to complement the question.

$DE = 52$

$FC = 15$

$HE = 14$

Given that J is the centroid, it means that J divides sides CD, DE and CE into two equal parts respectively and as such the following relationship exist:

$DF = FC$

$CH = HE$

$DG = GE$

Solving (a): DG

If $DG = GE$ , then

$DE = DG + GE$

$DE = DG + DG$

$DE = 2DG$

Make DG the subject

$DG = \frac{1}{2}DE$

Substitute 52 for DE

$DG = \frac{1}{2} * 52$

$DG = 26$

Solving (b): GE

If $DG = GE$ , then

$GE = DG$

$GE = 26$

Solving (c): DF

$DF = FC$

So:

$DF = 15$

Solving (d): CH

$CH = HE$

$CH = 14$

Solving (e): CE

If $CH = HE$ , then

$CE = CH + HE$

$CE = 14 + 14$

$CE = 28$

7 0

3 years ago

Fill in the missing work and justification for step 5 when solving 4(x + 1) - 8.

belka [17]

Answer:

multiplication property of equality

Step-by-step explanation:

8 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