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3 years ago

The water level of a lake rose by 1 2/3 in. during a 3 1/3 week-long wet spell. Simplify the complex

Mathematics

1 answer:

MissTica3 years ago

8 0

Given that,

The water level of a lake rose by 1 2/3 in. during a 3 1/3 week-long wet spell. We need to find the average rate at which the water level changed every week. It is a problem based on the concept of ratio. So,

$R=\dfrac{1\dfrac{2}{3}}{3\dfrac{1}{3}}\\\\=\dfrac{\dfrac{5}{3}}{\dfrac{10}{3}}\\\\=\dfrac{5}{3}\times \dfrac{3}{10}\\\\=\dfrac{1}{2}$

So, the average rate at which the water level changed every week is 1/2 inches/week.

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Step-by-step explanation:

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What is the midpoint of a line segment with endpoints (-2, 6) and (4, 3) in the

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(1, 9/2) or (1, 4.5)

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3 years ago

Consider the equation y=-2x+5. Describe everything you know about this graph compared to the parent linear function y=x.

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2 years ago

Engineers want to design seats in commercial aircraft so that they are wide enough to fit 9090% of all males. (Accommodating 1

Molodets [167]

Answer:

Hip breadths less than or equal to 16.1 in. includes 90% of the males.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 14.5

Standard Deviation, σ = 1.2

We are given that the distribution of hip breadths is a bell shaped distribution that is a normal distribution.

Formula:

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

We have to find the value of x such that the probability is 0.10.

P(X > x)

$P( X > x) = P( z > \displaystyle\frac{x - 14.5}{1.2})=0.10$

$= 1 -P( z \leq \displaystyle\frac{x - 14.5}{1.2})=0.10$

$=P( z \leq \displaystyle\frac{x - 14.5}{1.2})=0.90$

Calculation the value from standard normal z table, we have,

$P(z < 1.282) = 0.90$

$\displaystyle\frac{x - 14.5}{1.2} = 1.282\\x = 16.0384 \approx 16.1$

Hence, hip breadth of 16.1 in. separates the smallest 90% from the largest 10%.

That is hip breaths greater than 16.1 in. lies in the larger 10%.

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3 years ago