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daser333 [38]
2 years ago
9

The price of an item has been reduced by 85% the original price was $95 what is the price of the item

Mathematics
1 answer:
Yanka [14]2 years ago
3 0

Answer:

14.25

Step-by-step explanation:

95 x 0.85 = 80.75

95-80.75 = 14.25

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Express the fraction in simplest form. 2/8
seropon [69]
It will be 1/4.
That is your answer. 
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The points A (4, 0) and B (3, 7) are two vertices of right triangle ABC . The hypotenuse of the triangle is AB point C is on the
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The point C must be in a line perpendicular tothe x-axis passing through the point B(3,7), then the oordinates of point C must be (3,0)

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3 years ago
Analyze tables use the table that shows the percents of blood types of 145 donors during a recent blood drive
Alenkinab [10]

Answer:

a. 16

b. 80

c. Blood type AB

Step-by-step explanation:

Given:

O = 45%

A = 40%

B = 11%

AB = 4%

Total number of donors = 145

a. No. of donors that had type B blood = \frac{11}{100}*145 = \frac{1,595}{100} = 15.95

Type B blood donors ≈ 16

b. No. of blood donors without blood type O = \frac{40 + 11 + 4}{100}*145 = \frac{55}{100}*145 = \frac{7,975}{100} = 79.75

Those without blood type O ≈ 80

c. Find the number of people that had type AB:

Type AB = \frac{4}{100}*145 = 5.8 ≈ 6. This is less than 10 donors.

Therefore, blood type AB had less than 10 donors.

6 0
3 years ago
There are 10 children in a class and 2 of them choose to read a books so what is the probability that a randomly selected presch
kiruha [24]

Answer: a 20% chance

Step-by-step explanation:

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We do this because if 2 chose to read and there are 10 kids.

8 0
2 years ago
Is sin theta=5/6, what are the values of cos theta and tan theta?
mina [271]

let's bear in mind that sin(θ) in this case is positive, that happens only in the I and II Quadrants, where the cosine/adjacent are positive and negative respectively.

\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{6}}\qquad \impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-5^2}=a\implies \pm\sqrt{36-25}\implies \pm \sqrt{11}=a \\\\[-0.35em] ~\dotfill

\bf cos(\theta )=\cfrac{\stackrel{adjacent}{\pm\sqrt{11}}}{\stackrel{hypotenuse}{6}} \\\\\\ tan(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{adjacent}{\pm\sqrt{11}}}\implies \stackrel{\textit{and rationalizing the denominator}~\hfill }{tan(\theta )=\pm\cfrac{5}{\sqrt{11}}\cdot \cfrac{\sqrt{11}}{\sqrt{11}}\implies tan(\theta )=\pm\cfrac{5\sqrt{11}}{11}}

5 0
3 years ago
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