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2 years ago

A shirt retails for $22.40. A 10% tax is then applied to the original. What’s the final price after tax?

Mathematics

1 answer:

IrinaK [193]2 years ago

6 0

Answer:

24.64

Method:

You get 10% of the shirt cost, and it on to the 22.40

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Please ASAP help please !!! Help me anyone

worty [1.4K]

difference of squares means that both the terms are square terms. (also there must be a - symbol)

for example

y^2 - 4

square root of y^2 is y

square root of 4 is +2 as well as -2

so you would factorise it like this:

(y+2)(y-2)

1. y^4 has a square root of y^2 as y^2 × y^2 is y^4.

however, -2 doesnt have a whole number square root so it is not a difference of squares.

2. 25 has a square root of 5. m^2 has a square root of m. n^4 has a square root of n^2. so this 25m^2n^4 is a square term.

1 has a square root of +1 and -1.

therefore, this one is a difference of squares. (5mn^2 +1) (5mn^2 -1)

3. p^8 has a square root of p^4. q^4 has a square root of +q^2 and -q^2)

so it is a difference of squares. (p^4+q^2)(p^4 -q^2)

4. 16x^2 is a square term as irs square root is 4x.

however, 24 is not a square term.

therefore, it is not a difference of squares.

6 0

3 years ago

Help please! thank you so much

umka21 [38]

Answer:

4 units ...................

4 0

2 years ago

How do you convert the angle 11(pie)/12 from radians to degrees?

sergey [27]

Answer:

165°

Step-by-step explanation:

11π/12 * 180/π = 1980π/12π

Reduce & Cancel: 165°

4 0

3 years ago

Could anyone find this?

lina2011 [118]

Answer:

Step-by-step explanation:

Since the squares have areas of 32u^2. The side lengths are

$s=\sqrt{32}\\ \\ \text{By the Pythagorean Theorem}\\ \\ x^2=s^2+s^2\\ \\ x^2=32+32\\ \\ x^2=64\\ \\ x=\sqrt{64}\\ \\ x=8$

8 0

2 years ago

Consider two independent tosses of a fair coin. Let A be the event that the first toss results in heads, let B be the event that

aliina [53]

Answer with Step-by-step explanation:

We are given that two independent tosses of a fair coin.

Sample space={HH,HT,TH,TT}

We have to find that A, B and C are pairwise independent.

According to question

A={HH,HT}

B={HH,TH}

C={TT,HH}

$A\cap B=$ {HH}

$B\cap C$ ={HH}

$A\cap C$ ={HH}

P(E)= $\frac{number\;of\;favorable\;cases}{total\;number\;of\;cases}$

Using the formula

Then, we get

Total number of cases=4

Number of favorable cases to event A=2

$P(A)=\frac{2}{4}=\frac{1}{2}$

Number of favorable cases to event B=2

Number of favorable cases to event C=2

$P(B)=\frac{2}{4}=\frac{1}{2}$

$P(C)=\frac{2}{4}=\frac{1}{2}$

If the two events A and B are independent then

$P(A)\cdot P(B)=P(A\cap B)$

$P(A\cap)=\frac{1}{4}$

$P(B\cap C)=\frac{1}{4}$

$P(A\cap C)=\frac{1}{4}$

$P(A)\cdot P(B)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$

$P(B)\cdot P(C)=\frac{1}{4}$

$P(A)\cdot P(C)=\frac{1}{4}$

$P(A)\cdot P(B)=P(A\cap B)$

Therefore, A and B are independent

$P(B)\cdot P(C)=P(B\cap C)$

Therefore, B and C are independent

$P(A\cap C)=P(A)\cdot P(C)$

Therefore, A and C are independent.

Hence, A, B and C are pairwise independent.

6 0

3 years ago