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MrRa [10]
3 years ago
13

Y = x2 + 13 Non linear or linear

Mathematics
1 answer:
kakasveta [241]3 years ago
7 0

Answer:

linear

Step-by-step explanation:

as it has X2 +..... something in it

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Cubic yards of concrete will be required for 14 parking lot light pole standards
riadik2000 [5.3K]

Using conversion of units and volume of a cylinder, it is found that 23.17 cubic yards of concrete are needed, option D.

The volume of a cylinder of radius r and height h is given by:

V = \pi r^2h

In this problem:

  • We want the volume in cubic yards, thus, the units of radius and height have to be in yards.
  • 32 inches in diameter, thus 16 inches of radius. Since each inch has 0.0278 yards, this is a radius of 0.44 yards, thus r = 0.44.
  • Height of 8 feet, and since each feet has 0.33 yards, this is a height of 2.67 yards, thus h = 2.67
  • 14 poles, thus the volume is multiplied by 14.

Then:

V = 14\pi(0.44)^2(2.67) = 23.17

23.17 cubic yards of concrete are needed, option D.

A similar problem is given at brainly.com/question/22789697

3 0
3 years ago
I need help solving this w/ steps please
Leya [2.2K]
The answer to this question is 0.527
3 0
3 years ago
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Line p passes through A(-2, -4) and has a slope of - 1/2
Alenkasestr [34]
M = -1/2
(x₁, y₁) = (-2,-4)

Find the equation
y - y₁ = m(x - x₁)
y - (-4) = -1/2 (x - (-2))
y + 4 = -1/2 (x + 2)
2(y + 4) = -1 (x + 2)
2y + 8 = -x - 2
x + 2y = -10
8 0
3 years ago
Barak spent $20 at the arcade. this represents 10%of the money earned by working at thepizza parlor last week.
Karolina [17]
20=10%
times 10
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he earned $200 at the parlor
4 0
3 years ago
1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n
Nuetrik [128]

Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • 70% of fatalities involve an intoxicated driver, hence p = 0.7.
  • A sample of 15 fatalities is taken, hence n = 15.

The probability is:

P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)

Hence

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061

P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186

P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700

P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916

P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305

P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047

Then:

P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215

0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

A similar problem is given at brainly.com/question/24863377

5 0
3 years ago
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