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Sveta_85 [38]
3 years ago
13

The selling price of a shirt before sales tax 10$.the finial price of a shirt including sales tax is 12.5 .what is the rate of t

he sales tax applied
Mathematics
1 answer:
maw [93]3 years ago
7 0
$11.25 because 10 times 1.125 is equal to 11.25
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Please help me i am desperate please (1/3)
Sergeu [11.5K]

I think we're not supposed to do more than 3 problems in one answer; certainly not five pages. Does the 1/3 mean there are 10 more to come?

I'll do the first page.

6)

Right triangle, opposite side of 10, adjacent side of 21,

tan x = opp/adj = 10/21

x = arctan 10/21 = 25.46°

Answer: 25

7)

cos x =   adj / hyp = 10/14

x = arccos 10/14 = 44.42°

Answer: 44

8)

tan x = opp/adj = 12/24 = 1/2

x = arctan 1/2 = 26.57°

Answer: 27

9)

tan x = 31/32

x = arctan 31/32 = 44.09°

Answer: 44

10)

x = arctan 10/27 = 20.32°

Answer: 20

8 0
3 years ago
Simplify <br> (7y+4)(8y+2)
lesantik [10]

Answer:

56y^2+46y+8

Explanation:

Hope this helps you.

6 0
1 year ago
9x^3+6x^2-3x Please!!! HELP!!! need it done asap
Igoryamba

Step-by-step explanation:

9x^3+6x^2-3x

Cannot be simplified because there are no like terms but can be factored as

=3x(3x−1)(x+1)

8 0
3 years ago
A professor pays 25 cents for each blackboard error made in lecture to the student who pointsout the error. In a career ofnyears
marta [7]

Answer:

(a) The probability that <em>Y</em>₂₀ exceeds 1000  is 3.91 × 10⁻⁶.

(b) <em>n</em> = 28.09

Step-by-step explanation:

The random variable <em>Y</em>ₙ is defined as the total numbers of dollars paid in <em>n</em> years.

It is provided that <em>Y</em>ₙ can be approximated by a Gaussian distribution, also known as Normal distribution.

The mean and standard deviation of <em>Y</em>ₙ are:

\mu_{Y_{n}}=40n\\\sigma_{Y_{n}}=\sqrt{100n}

(a)

For <em>n</em> = 20 the mean and standard deviation of <em>Y</em>₂₀ are:

\mu_{Y_{n}}=40n=40\times20=800\\\sigma_{Y_{n}}=\sqrt{100n}=\sqrt{100\times20}=44.72\\

Compute the probability that <em>Y</em>₂₀ exceeds 1000 as follows:

P(Y_{n}>1000)=P(\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}>\frac{1000-800}{44.72})\\=P(Z>  4.47)\\=1-P(Z

**Use a <em>z </em>table for probability.

Thus, the probability that <em>Y</em>₂₀ exceeds 1000  is 3.91 × 10⁻⁶.

(b)

It is provided that P (<em>Y</em>ₙ > 1000) > 0.99.

P(Y_{n}>1000)=0.99\\1-P(Y_{n}

The value of <em>z</em> for which P (Z < z) = 0.01 is 2.33.

Compute the value of <em>n</em> as follows:

z=\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}\\2.33=\frac{1000-40n}{\sqrt{100n}}\\2.33=\frac{100}{\sqrt{n}}-4\sqrt{n}  \\2.33=\frac{100-4n}{\sqrt{n}} \\5.4289=\frac{(100-4n)^{2}}{n}\\5.4289=\frac{10000+16n^{2}-800n}{n}\\5.4289n=10000+16n^{2}-800n\\16n^{2}-805.4289n+10000=0

The last equation is a quadratic equation.

The roots of a quadratic equation are:

n=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}

a = 16

b = -805.4289

c = 10000

On solving the last equation the value of <em>n</em> = 28.09.

8 0
3 years ago
The starting and stopping point for Amir’s run this morning are 0.5 inch apart on the map. If each inch on the map represents 3
Aleonysh [2.5K]
Amir ran 1.5 miles in the morning.
7 0
3 years ago
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