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user100 [1]
2 years ago
5

The price of a Year 7 mathematics textbook is $60. The store gives a discount of 20% on the textbook.

Mathematics
1 answer:
MAVERICK [17]2 years ago
7 0

Answer:

yes,you can buy the textbook

Step-by-step explanation:

determine discount

60 × 0.2(decimal form of 20%)

=12

apply discount

60-12

=48

50>48✓

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svp [43]

Answer:

Cost of per session the average rate is $45.

Step-by-step explanation:

It is given that a gym membership with two personal training session cost $125, while gym membership with five personal training sessions cost $260.

It is required to find what is the cost per session.

Step 1 of 1

It is given that a gym membership with two personal training session cost $125, while gym membership with five personal training sessions cost $260.

To find the cost of per session calculate the average rate.

Now let $f(x)$ be the cost per session use the for the average rate of change, and the input value is the number of personal traings x.

$m=\frac{f\left(x_{2}\right)-f\left(x_{1}\right)}{x_{2}-x_{1}}$$

Now substitute, $125 for $f\left(x_{2}\right), 260 for $f\left(x_{1}\right), 2$ for $x_{1}$ and 5 for $x_{2}$ then,

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Cost of per session the average rate is $45.

3 0
2 years ago
Titus is asked to prove hexagon FEDCBA is congruent to hexagon
Goshia [24]

Answer:

<u><em>(x, y) -------> ( - x , y - 10 )</em></u>

Step-by-step explanation:

Titus is not correct.

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5 0
3 years ago
A survey of athletes at a high school is conducted, and the following facts are discovered: 13% of the athletes are football pla
nekit [7.7K]

Answer:

The probability that an athlete chosen is either a football player or a basketball player is 56%.

Step-by-step explanation:

Let the athletes which are Football player be 'A'

Let the athletes which are Basket ball player be 'B'

Given:

Football players (A) = 13%

Basketball players (B) = 52%

Both football and basket ball players = 9%

We need to find probability that an athlete chosen is either a football player or a basketball player.

Solution:

The probability that athlete is a football player = P(A)= \frac{13}{100}=0.13

The probability that athlete is a basketball player = P(B)= \frac{52}{100}=0.52

The probability that athlete is both basket ball player and  football player = P(A\cap B) = \frac{9}{100}=0.09

We have to find the probability that an athlete chosen is either a football player or a basketball player P(A\cup B).

Now we know that;

P(A\cup B)= P(A) + P(B) - P(A \cap B)\\\\P(A\cup B) = 0.13+0.52-0.09=0.56\\\\P(A\cup B) = \frac{0.56}{100}=56\%

Hence The probability that an athlete chosen is either a football player or a basketball player is 56%.

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4 years ago
How dose this work out
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Can you take a better picture. That would help to answer the question.
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