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Liono4ka [1.6K]
3 years ago
9

Solve the equation or formula for the indicated variable. S = 5 r 2 t, for t

Mathematics
1 answer:
Masja [62]3 years ago
5 0
<u>Answer</u>
t = (sr⁻₂)/5   or

t = (1/5)sr⁻²  or 

t = s/(5r²)



<u>Explanation</u>
S=5r²t

Dividing both sides by 5r²;
t = s/(5r²)
 = (1/5)sr⁻²
 = (sr⁻₂)/5 
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<h2>Answer:</h2>

This is a 2-step equation:

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The coordinates of point D are (7, 4) and the coordinates of point E are (1, -3).
mylen [45]

Answer:

  -6/7

Step-by-step explanation:

The slope of DE is ...

 (change in y)/(change in x) = (y2 -y1)/(x2 -x1) = (-3-4)/(1-7) = -7/-6 = 7/6

The slope of the line perpendicular to DE is the negative reciprocal of this:

  perpendicular slope = -1/slope = -1/(7/6) = -6/7

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8 0
3 years ago
Production of a gas flow meter takes place in two distinct operations. Measurements (n = 15) of the time required for the first
Karolina [17]

Answer:

a) P(a > 80) = 0.323

b) The 95% confidence interval = (73.40, 80.60)

c) The 95% confidence interval expresses that the mean of the distribution can always be found in the given range, with a 95% confidence level.

Step-by-step explanation:

X ~ (45, 4)

Y ~ (32, 2.5)

(X+Y) ~ (77, 6.5)

Let a = (X+Y)

a) Probability that the time required to complete both of those steps will exceed 80 min = P(a > 80)

This is a normal distribution problem

We then standardize 80 min time

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (a - μ)/σ = (80 - 77)/6.5 = 0.46

To determine the probability the time required to complete both of those steps will exceed 80 min = P(a > 80) = P(z > 0.46)

We'll use data from the normal probability table for these probabilities

P(a > 80) = P(z > 0.46) = 1 - P(z ≤ 0.46) = 1 - 0.677 = 0.323

b) 95% confidence interval for the expected total time required to produce one flow meter.

We need to obtain the margin of error

Margin of error = (critical value) × (standard error of the sample)

Critical value for a 95% confidence interval = critical value for a significance level of 5% = t(15-1, 0.05/2) = 2.145 (using the t-score since information on the population mean and standard deviation isn't known)

Standard error for the sample of sum of times = (standard deviation of the sum of times)/√n = (6.5/√15) = 1.678

Margin of error = 2.145 × 1.678 = 3.60

Limits of the confidence interval = (Sample mean ± margin of error)

Lower limit of the confidence interval = (Sample mean - margin of error) = 77 - 3.60 = 73.40

Upper limit of the confidence interval = 77 + 3.60 = 80.60

The confidence interval = (73.40, 80.60)

4 0
3 years ago
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