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3 years ago

A 48$ shirts cost $22 more than one half the cost of a pair of pants. How much does the pair of pants cost?

Mathematics

1 answer:

Natali5045456 [20]3 years ago

6 0

Answer:

Step-by-step explanation:

48-22=26

26•2=52

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Which multiplication problem is represented by the model?

maksim [4K]

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3 years ago

How do I solve this .... question in the image. <br><br>thank you

dezoksy [38]

Answer:

Below

Step-by-step explanation:

● x-20 = y+20 (1)

● 2(y-22) = x+22 (2)

This is a system of simulataneous equations

Let's simplify the expressions first

● x -20 = y + 20 (1)

Add 20 to both sides

● x -20 + 20 = y+20 +20

● x = y + 40 (1)

● 2(y-22) = x+22 (2)

● 2y - 44 = x +22

Substrat 22 from both sides

● 2y-44-22 = x+22-22

● 2y -66 = x (2)

This is the new system:

● x = y+40 (1)

● x = 2y-66 (2)

Substract (2) from (1)

● x-x = y+40-(2y-66)

● y+40-2y+66 = 0

● -y +106 = 0

● y = 106

Replace y with 106 in (1)

● x = y +40

● x = 106+40

● x = 146

So the solutions are (146,106)

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3 years ago

6. What is the product of 3/5 x 2 1/3 written in simplest form?

Aleksandr [31]

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3 years ago

Read 2 more answers

at a particular high school, 42% of the students participate in sports and 25% of the students participate in drama. if 53% of t

baherus [9]

The probability that a student participates in both sports and drama is $0.14$ .

<h3>What is the formula for P(AUB), where A and B are any two events?</h3>

If $A$ and $B$ are any two events, then the probability of the joint event $(A\cup B)$ is given by the following formula: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Given that 42% of the students participate in sports and 25% of the students participate in drama and 53% of the students participate in either sports or drama.

Suppose $S$ denotes that "a student participates in sports" and $D$ denotes that "a student participates in drama".
So, we have $P(S)=\frac{42}{100}=0.42$ , $P(D)=\frac{25}{100}=0.25$ , $P(S\cup D)=\frac{53}{100}=0.53$ .

We want to find the probability that a student participates in both sports and drama i.e., we want to find $P(S\cap D)$ .

By the above formula, we obtain:

$P(S\cup D)=P(S)+P(D)-P(S\cap D)\\\Longrightarrow P(S\cap D)=P(S)+P(D)-P(S\cup D)\\\Longrightarrow P(S\cap D)=0.42+0.25-0.53\\\therefore P(S\cap D)=0.14=\frac{14}{100}$