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Galina-37 [17]
2 years ago
5

Catherine has $54. She plans to spend more than $20 of the money for a painting canvas. The rest will go toward paints. Each tub

e of paint costs $8.50. Assume that x represents the number of tubes of paint Catherine can buy
The inequality that represents this scenario is
1. 8.5x +20>54
2. 8.5x-20<54
3. 54-8.5x<20
4. 54- 8.5x>20
Mathematics
2 answers:
kakasveta [241]2 years ago
8 0

Answer:

4. 54- 8.5x>20

Step-by-step explanation:

Catherine only has $54, so she cannot spend more than that.

The canvas will cost at least $20, but we don't know how much exactly.

The tubes cost $8.50 each.

So, she starts with a total budget of $54, out of which she will buy paints (8.5x) and she wants to have at least $20 left for canvas.

So, we transpose those facts into the inequity:

54 - 8.5x > 20

s344n2d4d5 [400]2 years ago
8 0

Answer:

Opcion 4

54- 8.5x>20

Step-by-step explanation:

We must identify the quantities.

<u><em>Painting canvas </em></u>

more than $ 20

<u><em>Paint tubes (x): </em></u>

$ 8.50 per tube = 8.50x

<u><em>Available money: </em></u>

$ 54

Then the expenses must not exceed $ 54, and the budget for the painting canvas must be greater than $ 20

The money after buying the paint tubes is:

54-8.50x

We know that this amount must be greater than $ 20 because you need to buy the painting canvas. So  the inequality is:

54-8.50x> 20

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The accompanying data are lengths (inches) of bears. Find the percentile corresponding to 65.5 in.
Jlenok [28]

Complete question is missing, so i have attached it.

Answer:

Percentile is 74th percentile

Step-by-step explanation:

All the lengths given are;

Bear Lengths 36.5 37.5 39.5 40.5 41.5 42.5 43.0 46.0 46.5 46.5 48.5 48.5 48.5 49.5 51.5 52.5 53.0 53.0 54.5 56.8 57.5 58.5 58.5 58.5 59.0 60.5 60.5 61.0 61.0 61.5 62.0 62.5 63.5 63.5 63.5 64.0 64.0 64.5 64.5 65.5 66.5 67.0 67.5 69.0 69.5 70.5 72.0 72.5 72.5 72.5 72.5 73.0 76.0 77.5

The number of lengths (inches) of bears given are 54 in number.

We are looking for the percentile corresponding to 65.5 in.

Looking at the lengths given, since they are already arranged from smallest to highest, let's locate the position of 65.5 in.

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2 years ago
Suppose that 4% of the 2 million high school students who take the SAT each year receive special accommodations because of docum
dangina [55]

Answer:

a. 0.0122

b. 0.294

c. 0.2818

d. 30.671%

e. 2.01 hours

Step-by-step explanation:

Given

Let X represents the number of students that receive special accommodation

P(X) = 4%

P(X) = 0.04

Let S = Sample Size = 30

Let Y be a selected numbers of Sample Size

Y ≈ Bin (30,0.04)

a. The probability that 1 candidate received special accommodation

P(Y = 1) = (30,1)

= (0.04)¹ * (1 - 0.04)^(30 - 1)

= 0.04 * 0.96^29

= 0.012244068467946074580191760542164986632531806368667873050624

P(Y=1) = 0.0122 --- Approximated

b. The probability that at least 1 received a special accommodation is given by:

This means P(Y≥1)

But P(Y=0) + P(Y≥1) = 1

P(Y≥1) = 1 - P(Y=0)

Calculating P(Y=0)

P(Y=0) = (0.04)° * (1 - 0.04)^(30 - 0)

= 1 * 0.96^36

= 0.293857643230705789924602253011959679180763352848028953214976

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c.

The probability that at least 2 received a special accommodation is given by:

P (Y≥2) = 1 -P(Y=0) - P(Y=1)

= 0.294 - 0.0122

= 0.2818

d. The probability that the number among the 15 who received a special accommodation is within 2 standard deviations of the number you would expect to be accommodated?

First, we calculate the standard deviation

SD = √npq

n = 15

p = 0.04

q = 1 - 0.04 = 0.96

SD = √(15 * 0.04 * 0.96)

SD = 0.758946638440411

SD = 0.759

Mean =np = 15 * 0.04 = 0.6

The interval that is two standard deviations away from .6 is [0, 2.55] which means that we want the probability that either 0, 1 , or 2 students among the 20 students received a special accommodation.

P(Y≤2)

P(0) + P(1) + P(2)

=.

P(0) + P(1) = 0.0122 + 0.294

Calculating P(2)

P(2) = (0.04)² * (1 - 0.04)^(30 - 2(

P(2) = 0.00051

So,

P(0) +P(1) + P(2). = 0.0122 + 0.294 + 0.00051

= 0.30671

Thus it 30.671% probable that 0, 1, or 2 students received accommodation.

e.

The expected value from d) is .6

The average time is [.6(4.5) + 19.2(3)]/30 = 2.01 hours

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