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2 years ago

An adidas hat that has a regular price of $15.50 and is discounted 25%

Mathematics

2 answers:

netineya [11]2 years ago

5 0

(A) The price of the hat after the discount is $11.63
(B) The tax is $1.16
(C) The total cost of the hat is $16.66

lutik1710 [3]2 years ago

3 0

A) $11.63
B) 7.5% tax on something that is $11.63 would be $.87225
C) total cost with sale and tax would be $12.50

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Combine like terms to create an equivalent expression 3.4 - 2.8d + 2.8d - 1.3

zalisa [80]

Answer:

3.4 - 2.8d + 2.8d - 1.3 = 2.1

Step-by-step explanation:

The given expression is 3.4 -2.8d + 2.8d -1.3

Let's see the definition of like terms.

Like terms are the terms having the same variable and the same exponents.

Examples: -3xy, 2xy and 4y, 5y and -3, 2.

Now let's identify the like terms from the given expression.

3.4 -2.8d + 2.8d -1.3

Here the like terms are -2.8d, +2.8d and 3.4, -1.3

3.4 -2.8d + 2.8d -1.3

= -2.8d + 2.8d + 3.4 - 1.3 [-2.8d + 2.8d = 0] and 3.4 -1.3 = 2.1

= 0 + 2.1

=2.1

The answer is 2.1

6 0

2 years ago

Read 2 more answers

3∙(a+x), if a=8; x=−10

Oksi-84 [34.3K]

Your answer should be -6

6 0

3 years ago

Lizzie rolls two dice. What is the probability that the sum of the dice is:

zhenek [66]

Answer:

$A.\ \dfrac{1}{3}\\B.\ \dfrac{5}{12}\\C.\ \dfrac{7}{36}\\$

Step-by-step explanation:

Total outcomes possible: 36

A. Divisible by 3

Possible options are:

3, 6, 9 and 12.

Possible outcomes for 3 are: {(1,2), (2,1)} Count 2

Possible outcomes for 6 are: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

Possible outcomes for 9 are: {(3,6), (4,5), (5,4),(6,3)} Count 4

Possible outcomes for 12 are: {(6,6)} Count 1

Total count = 2 + 5 + 4 + 1 = 12

Probability of an event E can be formulated as:

$P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}$

$P(A) = \dfrac{12}{36} = \dfrac{1}{3}$

B. Less than 7:

Possible sum can be 2, 3, 4, 5, 6

Possible cases for sum 2: {(1,1)} Count 1

Possible cases for sum 3: {(1,2), (2,1)} Count 2

Possible cases for sum 4: {(1,3), (3,1), (2,2)} Count 3

Possible cases for sum 5: {(1,4), (2,3), (3,2),(4,1)} Count 4

Possible cases for sum 6: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

Total count = 1 + 2 + 3 + 4 + 5 = 15

$P(B) = \dfrac{15}{36} = \dfrac{5}{12}$

C. Divisible by 3 and less than 7:

$P(A \cap B) = \dfrac{n(A\cap B)}{\text{Total Possible outcomes}}$

Here, common cases are:

Possible outcomes for 3 are: {(1,2), (2,1)} Count 2

Possible outcomes for 6 are: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

$P(A \cap B) = \dfrac{7}{\text{36}}$

5 0

2 years ago

Prove that the roots of x2+(1-k)x+k-3=0 are real for all real values of k

masha68 [24]

Answer:

Roots are not real

Step-by-step explanation:

To prove : The roots of x^2 +(1-k)x+k-3=0x

+(1−k)x+k−3=0 are real for all real values of k ?

Solution :

The roots are real when discriminant is greater than equal to zero.

i.e. b^2-4ac\geq 0b

−4ac≥0

The quadratic equation x^2 +(1-k)x+k-3=0x

+(1−k)x+k−3=0

Here, a=1, b=1-k and c=k-3

Substitute the values,

We find the discriminant,

D=(1-k)^2-4(1)(k-3)D=(1−k)

−4(1)(k−3)

D=1+k^2-2k-4k+12D=1+k

−2k−4k+12

D=k^2-6k+13D=k

−6k+13

D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))

For roots to be real, D ≥ 0

But the roots are imaginary therefore the roots of the given equation are not real for any value of k.

6 0

2 years ago

at davidson's bike rentals it cost $42 to rent a bike for 9 hours. how many hours of bike use does a customer get per dollar?

Alona [7]

Answer:

3/14 of an hour

Step-by-step explanation:

9 hours/42 dollars

5 0

2 years ago