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2 years ago

The power 9^2 is equivalent to 81. What is the value of 9^-2?

2 answers:

8 0

Since it’s a negative exponent you have to use the reciprocal, so this would be 1/9^2. remember you have to apply the exponent to both the 1 and 9. since 1^2 is equal to one still the answer would just be 1/81, or 0.012345679

lidiya [134]2 years ago

5 0

Answer:

0.01234567901

Step-by-step explanation:

pls make me brainliest

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What’s the answer to 4x+3=15

Anastasy [175]

Answer: 4.500

Step-by-step explanation:

4x - 18 = 2 • (2x - 9)

2 = 0

2x-9 = 0

2x = 9

x = 9/2 = 4.500

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3 years ago

What is the absolute value of Point C labelled on the number line? A number line with point A at coordinate -2.75, point B at co

lina2011 [118]

Answer:

2.25

Step-by-step explanation:

The value of point C is already positive, so nothing changes.

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2 years ago

9 more than 2 multiplied by f

Andrej [43]

(2+9)*f
= 11*f

The final answer is 11f~

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3 years ago

You decide to go to dinner with three friends. Before the check comes, the group decides to split the check evenly. The total ch

Vika [28.1K]

Answer:

$\$38.67$

Step-by-step explanation:

Let

x------> The money each friend must pay, including the tip

we know that

$100\%+15\%=115 \%=115/100=1.15$

$x=\$134.50*(1.15)/4$

$x=\$38.67$

4 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%.

5 0

3 years ago