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3 years ago

An item is regularly priced at $25. Susan bought it on sale for 55% off the regular price.

Mathematics

2 answers:

Lady_Fox [76]3 years ago

7 0

Answer:

Discount is 13.75 discounted price is 11.25

Step-by-step explanation:

musickatia [10]3 years ago

6 0

Answer: $11.25

Step-by-step explanation:

Given that item is $25 before sale, susan bought for 55% off the regular price.

55% off $25 implies subtract 55% from $25

=55/100 × 25 = $13.75

Susan therefore bought the item at $25 - $13.75 = $11.25.

I hope this helps.

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There are 500 students in a high school senior class. Of these 500 students, 300 regularly wear a necklace to

andre [41]

Answer:

1. P(N) = 3/5

2. P(R) = 2/5

3. P(N and R) = 1/4

4. P(N or R) = 3/4

Step-by-step explanation:

1. Let's review the information given to us to answer the problem correctly:

Number of students in a high school senior class = 500

Number of students that regularly wear a necklace = 300

Number of students that regularly wear a ring = 200

Number of students that wear a necklace and a ring = 125

2. What is P(N), the probability that a senior wears a necklace?

Let's recall the formula of probability:

Probability = Number of favorable outcomes/Total of possible outcomes

Substituting with the real given values:

P(N) = 300/500 = 3/5 (Dividing by 100 numerator and denominator)

P(N) = 3/5

3. What is P(R), the probability that a senior wears a ring?

Substituting with the real given values:

P(R) = 200/500 = 2/5 (Dividing by 100 numerator and denominator)

P(R) = 2/5

4. What is P(N and R), the probability that a senior wears a necklace and a ring?

Substituting with the real given values:

P(N and R) = 125/500 = 1/4 (Dividing by 125 numerator and denominator)

P(N and R) = 1/4

5. What is P(N or R), the probability that a senior wears a necklace or a ring?

P(N or R) = P(N) + P(R) - P(N and R)

Substituting with the real given values:

P(N or R) = 3/5 + 2/5 - 1/4

P(N or R) = 5/5 - 1/4

P(N or R) = 1 - 1/4

P(N or R) = 3/4

5 0

3 years ago

For the function y=3x2: (a) Find the average rate of change of y with respect to x over the interval [3,6]. (b) Find the instant

nirvana33 [79]

Answer:

The instantaneous rate of change of $y$ with respect to $x$ at the value $x = 3$ is 18.

Step-by-step explanation:

a) Geometrically speaking, the average rate of change of $y$ with respect to $x$ over the interval by definition of secant line:

$r = \frac{y(b) -y(a)}{b-a}$ (1)

Where:

$a$ , $b$ - Lower and upper bounds of the interval.

$y(a)$ , $y(b)$ - Function exaluated at lower and upper bounds of the interval.

If we know that $y = 3\cdot x^{2}$ , $a = 3$ and $b = 6$ , then the average rate of change of $y$ with respect to $x$ over the interval is:

$r = \frac{3\cdot (6)^{2}-3\cdot (3)^{2}}{6-3}$

$r = 27$

The average rate of change of $y$ with respect to $x$ over the interval $[3,6]$ is 27.

b) The instantaneous rate of change can be determined by the following definition:

$y' = \lim_{h \to 0}\frac{y(x+h)-y(x)}{h}$ (2)

Where:

$h$ - Change rate.

$y(x)$ , $y(x+h)$ - Function evaluated at $x$ and $x+h$ .

If we know that $x = 3$ and $y = 3\cdot x^{2}$ , then the instantaneous rate of change of $y$ with respect to $x$ is:

$y' = \lim_{h \to 0} \frac{3\cdot (x+h)^{2}-3\cdot x^{2}}{h}$

$y' = 3\cdot \lim_{h \to 0} \frac{(x+h)^{2}-x^{2}}{h}$

$y' = 3\cdot \lim_{h \to 0} \frac{2\cdot h\cdot x +h^{2}}{h}$

$y' = 6\cdot \lim_{h \to 0} x +3\cdot \lim_{h \to 0} h$

$y' = 6\cdot x$

$y' = 6\cdot (3)$

$y' = 18$

The instantaneous rate of change of $y$ with respect to $x$ at the value $x = 3$ is 18.

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3 years ago

Mean, median, mode or standard deviation best represents the data

pickupchik [31]

Answer:

mean

Step-by-step explanation:

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3 years ago

What is 3356983+10000004565

Leto [7]

The answer is: 1003361548

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4 years ago

10 4/5+ 8 3/4-6 1/2 simplest form

SVEN [57.7K]

The answer equals to 13 1/20 or in decimal form 13.05

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3 years ago