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Naddik [55]
3 years ago
10

$29.99 markdown 33 and one third

Mathematics
1 answer:
Sholpan [36]3 years ago
5 0
Original Price: $29.99
Mark down: 33 1/3%
Mark down Price: $20.00

29.99 * 33 1/3 = 29.99 * 0.3333 = 9.99 value of mark down.
29.99 - 9.99 = 20.00 new price.

The problem is not really clear. We can assume that $29.99 is the mark down price. We need to look for the original price.

Mark down price: $29.99
Mark down rate : 33 1/3%
Original Price    : $44.98

29.99 / ((100% - 33 1/3%) / 100%) = 29.99 / ( 66 2/3% / 100%)
29.99 / 0.6667 = 44.98
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Giving brainliest, please answer as soon as possible.
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Answer:

Step-by-step explanation:

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Each going from side midpoint to opposite vertex.

here is a drawing of all of them filled in:

4 0
3 years ago
I need help for this, thanks
gulaghasi [49]

(a) Yes all six trig functions exist for this point in quadrant III. The only time you'll run into problems is when either x = 0 or y = 0, due to division by zero errors. For instance, if x = 0, then tan(t) = sin(t)/cos(t) will have cos(t) = 0, as x = cos(t). you cannot have zero in the denominator. Since neither coordinate is zero, we don't have such problems.

---------------------------------------------------------------------------------------

(b) The following functions are positive in quadrant III:

tangent, cotangent

The following functions are negative in quadrant III

cosine, sine, secant, cosecant

A short explanation is that x = cos(t) and y = sin(t). The x and y coordinates are negative in quadrant III, so both sine and cosine are negative. Their reciprocal functions secant and cosecant are negative here as well. Combining sine and cosine to get tan = sin/cos, we see that the negatives cancel which is why tangent is positive here. Cotangent is also positive for similar reasons.

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3 years ago
For the trinomial x2 + 10x + 21, we want to find two binomial factors
ratelena [41]
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7 0
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The average score of all pro golfers for a particular course has a mean of 70 and a standard deviation of 3.0 (Think of these as
bija089 [108]

Answer:

0.477 is  the probability that the average score of the 36 golfers was between 70 and 71.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 70

Standard Deviation, σ = 3

Sample size, n = 36

Let the average score of all pro golfers follow a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

P(score of the 36 golfers was between 70 and 71)

\text{Standard error of sampling} = \displaystyle\frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{36}} = \frac{1}{2}

P(70 \leq x \leq 71) = P(\displaystyle\frac{70 - 70}{\frac{3}{\sqrt{36}}} \leq z \leq \displaystyle\frac{71-70}{\frac{3}{\sqrt{36}}}) = P(0 \leq z \leq 2)\\\\= P(z \leq 2) - P(z \leq 0)\\= 0.977 - 0.500 = 0.477= 47.7\%

P(70 \leq x \leq 71) = 47.7\%

0.477 is  the probability that the average score of the 36 golfers was between 70 and 71.

8 0
3 years ago
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Answer:

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