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3 years ago

How do you write 11/ 4 as a percentage?

1 answer:

Ivan3 years ago

6 0

The answer is %225 First, we must convert the given fraction into a decimal number. So use a calculator or long division to convert the given fraction to a decimal number. In this case,

Now just multiply the equivalent decimal by 100 to convert it into a percent. So . Take note that we're simply moving the decimal point two spots to the right.

So the given fraction is equivalent to 275%

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Consider the graph below:

Julli [10]

It would be C) y = ( x + 1 ) ( x - 3 ) ( x - 2 ) because when you graph the points it falls to the left and rises to the right.

6 0

3 years ago

Read 2 more answers

"Suppose an object falling in the atmosphere has mass m=15kg and the drag coefficient is γ=9kg/s. Recall that the differential e

Art [367]

Answer:

a. v(t)= -6.78 $e^{-16.33t}$ + 16.33 b. 16.33 m/s

Step-by-step explanation:

The differential equation for the motion is given by mv' = mg - γv. We re-write as mv' + γv = mg ⇒ v' + γv/m = g. ⇒ v' + kv = g. where k = γ/m.Since this is a linear first order differential equation, We find the integrating factor μ(t)= $e^{\int\limits^ {}k \, dt }$ = $e^{kt}$ . We now multiply both sides of the equation by the integrating factor.

μv' + μkv = μg ⇒ $e^{kt}$ v' + k $e^{kt}$ v = g $e^{kt}$ ⇒ [v $e^{kt}$ ]' = g $e^{kt}$ . Integrating, we have

∫ [v $e^{kt}$ ]' = ∫g $e^{kt}$

v $e^{kt}$ = $\frac{g}{k}$ $e^{kt}$ + c

v(t)= $\frac{g}{k}$ + c $e^{-kt}$ .

From our initial conditions, v(0) = 9.55 m/s, t = 0 , g = 9.8 m/s², γ = 9 kg/s , m = 15 kg. k = y/m. Substituting these values, we have

9.55 = 9.8 × 15/9 + c $e^{-16.33 * 0}$ = 16.33 + c

c = 9.55 -16.33 = -6.78.

So, v(t)= 16.33 - 6.78 $e^{-16.33t}$ . m/s = - 6.78 $e^{-16.33t}$ + 16.33 m/s

b. Velocity of object at time t = 0.5

At t = 0.5, v = - 6.78 $e^{-16.33 x 0.5}$ + 16.33 m/s = 16.328 m/s ≅ 16.33 m/s

6 0

2 years ago

In a goodness-of-fit test, the null hypothesis states that the data came from a normally distributed population. The researcher

marta [7]

Answer:

Solution: C) 7.815

Hint: From Chi-square table with (k-r-1)=(6-2-1) = 3 d.f the critical value = 7.8147

8 0

3 years ago

(2p + 4) + 5 (p-1) - (p+7)

jonny [76]

Your answer will be, "6p - 8"

Thanks,

Deku ❤

6 0

3 years ago

a major metroplitan newspaper selected a simple random sample of 1,600 readers from their list of 100,000 subscribers. they aske

Arisa [49]

Answer:

The 99% confidence interval for the proportion of readers who would like more coverage of local news is (0.3685, 0.4315).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.