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3 years ago

An object is moving at a speed of 230 feet per week. Express this speed in inches per

Mathematics

2 answers:

olga2289 [7]3 years ago

7 0

I agree :) with the first person comment

Dafna11 [192]3 years ago

5 0

You Americans are killing me with your inches per hamburgers and eagles per ar-15

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My Notes The displacement from equilibrium of an oscillating weight suspended by a spring is given by y(t) = 1 4 cos(5t) where y

ikadub [295]

Answer:

y(0) = 0.25 feet

$y(\frac{1}{4}) = 0.0789 \text{ feet}$

$y(\frac{1}{2}) =-0.2003 \text{ feet}$

Step-by-step explanation:

We are given the following information in the question:

The displacement from equilibrium of an oscillating weight suspended by a spring =

$y(t) = \displaystyle\frac{1}{4} \cos(5t)$

where y is the displacement in feet and t is the time in seconds.

Here, cos is in radians.

1) t = 0

$y(0) = \displaystyle\frac{1}{4} \cos(5(0)) = \frac{1}{4} \cos(0) = \frac{1}{4}(1) = \frac{1}{4}$

y(0) = 0.25 feet

2) t = $\frac{1}{4}$

$y(\displaystyle\frac{1}{4}) = \displaystyle\frac{1}{4} \cos(5(\frac{1}{4})) = \frac{1}{4} \cos(1.25) = \frac{1}{4}(0.31532236) =0.07883059$

$y(\frac{1}{4}) = 0.0789 \text{ feet}$

3) t = $\frac{1}{2}$

$y(\displaystyle\frac{1}{2}) = \displaystyle\frac{1}{4} \cos(5(\frac{1}{2})) = \frac{1}{4} \cos(2.5) = \frac{1}{4}(-0.80114362) = -0.200285905$

$y(\frac{1}{2}) =-0.2003 \text{ feet}$

The negative sign indicates the opposite direction of displacement.

3 0

3 years ago

HELP PLEASE DUE TODAY

N76 [4]

1 A. 5

1 B. 12.9

2. For this one, u can write any number less than 4.7, which could be 3, 2, -100, 1.21, etc.

3. For this one, u can write any number greater than (positive!) 2.6, such as 10, 30, 3.1415, etc.

hope this helps!

6 0

2 years ago

Simplify: (3) + (-4)

nikitadnepr [17]

Answer:

it should -1 because you have to evaluate the number

3 0

2 years ago

What is the range of the function f(x) = 2x +4 when the domain is {-2, 0, 3}?

Vladimir79 [104]

Look at the attached picture..

Hope it will be helpful to you...

7 0

2 years ago

Read 2 more answers

A professor finds that the average SAT score among all students attending his college is 1150 ± 150 (μ ± σ). He polls his class

solniwko [45]

Answer:

Step-by-step explanation

Hello!

Be X: SAT scores of students attending college.

The population mean is μ= 1150 and the standard deviation σ= 150

The teacher takes a sample of 25 students of his class, the resulting sample mean is 1200.

If the professor wants to test if the average SAT score is, as reported, 1150, the statistic hypotheses are:

H₀: μ = 1150

H₁: μ ≠ 1150

α: 0.05

$Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } } ~~N(0;1)$

$Z_{H_0}= \frac{1200-1150}{\frac{150}{\sqrt{25} } } = 1.67$

The p-value for this test is 0.0949

Since the p-value is greater than the level of significance, the decision is to reject the null hypothesis. Then using a significance level of 5%, there is enough evidence to reject the null hypothesis, then the average SAT score of the college students is not 1150.

I hope it helps!

7 0

3 years ago