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dlinn [17]
3 years ago
15

Analyze the following table of values and select the true statement.

Mathematics
1 answer:
Yuri [45]3 years ago
7 0
The relationship is linear becuase it all equals each other
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Easy:

((18-x) -5) + ((18+x)➗2)
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One tank is filling at a rate of 5/8 gallon per 7/10 hour. A second tank is filling at a rate of 5/9 gallon per 2/3 hour. Which
anzhelika [568]
First recognize that the unit rate we're finding is gallons per hour, or \frac{g}{h}.

rate for Tank 1 = x gallons per hour = <em>a</em> gallons / <em>b</em> hours

T_{1}=\frac{\frac{5}{8}}{\frac{7}{10}}\\\\T_{1}=\frac{5}{8}*\frac{10}{7}\\\\T_{1}=\frac{50}{56}\\\\\\Fill\ rate\ of\ Tank\ 1=\frac{25}{28}


rate for Tank 2 = y gallons per hour = <em>c</em> gallons / <em>d</em> hours

T_{2}=\frac{\frac{5}{9}}{\frac{2}{3}}\\\\T_{2}=\frac{5}{9}*\frac{3}{2}\\\\T_{2}=\frac{15}{18}\\\\T_{2}=\frac{15}{18}\\\\\\Fill\ rate\ of\ Tank\ 2=\frac{5}{6}
3 0
3 years ago
What is the written form of 30.55
kirza4 [7]
Thirty and fifty-five hundredth.
4 0
3 years ago
A company makes rubber rafts. 12% of them develop cracks within the first month of operation. 27 new rafts are randomly sampled
rewona [7]

The probability that the number of tested rafts that develop cracks is no more than 3 is <u>.00006</u>.

The true proportion, p for the population is given to 0.12.

Thus, the mean, μ, for the sample = np = 27*0.12 = 3.24.

The sample size, n, given to us is 27.

Thus, the standard deviation, s, for the sample can be calculated using the formula, s = √{p(1 - p)}/n.

s = √{0.12(1 - 0.12)}/27 = √0.003911 = 0.0625389.

We are asked to calculate the probability that the number of tested rafts that develop cracks is no more than 3, that is, we need to calculate P(X ≤3).

P(X ≤ 3)

= P(Z ≤ {(3 - 3.24)/0.0625389) {Using the formula z = (x - μ)/s}

= P(Z ≤ -3.8376114706)

= .00006 {From table}.

Thus, the probability that the number of tested rafts that develop cracks is no more than 3 is <u>.00006</u>.

Learn more about sampling distributions at

brainly.com/question/15507495

#SPJ4

8 0
1 year ago
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