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3 years ago

Explain how to model represents 5 divided by 1/3 = 15

Mathematics

1 answer:

nexus9112 [7]3 years ago

6 0

Answer:

Step-by-step explanation:

5 ÷ 1/3

= 5 × 3/1

= 15

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Fin the sum -11 + (-6) =

lesantik [10]

It is - 16 because adding two negatives give you a negative

7 0

3 years ago

The estimated daily living costs for an executive traveling to various major cities follow. The estimates include a single room

Alexandra [31]

Answer:

$\bar x = 260.1615$

$\sigma = 70.69$

The confidence interval of standard deviation is: $53.76$ to $103.25$

Step-by-step explanation:

Given

$n =20$

See attachment for the formatted data

Solving (a): The mean

This is calculated as:

$\bar x = \frac{\sum x}{n}$

So, we have:

$\bar x = \frac{242.87 +212.00 +260.93 +284.08 +194.19 +139.16 +260.76 +436.72 +355.36 +.....+250.61}{20}$

$\bar x = \frac{5203.23}{20}$

$\bar x = 260.1615$

$\bar x = 260.16$

Solving (b): The standard deviation

This is calculated as:

$\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}$

$\sigma = \sqrt{\frac{(242.87 - 260.1615)^2 +(212.00- 260.1615)^2+(260.93- 260.1615)^2+(284.08- 260.1615)^2+.....+(250.61- 260.1615)^2}{20 - 1}}$ $\sigma = \sqrt{\frac{94938.80}{19}}$

$\sigma = \sqrt{4996.78}$

$\sigma = 70.69$ --- approximated

Solving (c): 95% confidence interval of standard deviation

We have:

$c =0.95$

So:

$\alpha = 1 -c$

$\alpha = 1 -0.95$

$\alpha = 0.05$

Calculate the degree of freedom (df)

$df = n -1$

$df = 20 -1$

$df = 19$

Determine the critical value at row $df = 19$ and columns $\frac{\alpha}{2}$ and $1 -\frac{\alpha}{2}$

So, we have:

$X^2_{0.025} = 32.852$ ---- at $\frac{\alpha}{2}$

$X^2_{0.975} = 8.907$ --- at $1 -\frac{\alpha}{2}$

So, the confidence interval of the standard deviation is:

$\sigma * \sqrt{\frac{n - 1}{X^2_{\alpha/2} }$ to $\sigma * \sqrt{\frac{n - 1}{X^2_{1 -\alpha/2} }$

$70.69 * \sqrt{\frac{20 - 1}{32.852}$ to $70.69 * \sqrt{\frac{20 - 1}{8.907}$

$70.69 * \sqrt{\frac{19}{32.852}$ to $70.69 * \sqrt{\frac{19}{8.907}$

$53.76$ to $103.25$

8 0

3 years ago

A particle moves according to a law of motion s = f(t), t ? 0, where t is measured in seconds and s in feet.

Usimov [2.4K]

Answer:

a) $\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

b) $v(t=3) = 3(3)^2 -18(3) +15=-12$

c) $t =1s, t=5s$

d) $[0,1) \cup (5,\infty)$

e) $D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

f) $a(t) = \frac{dv}{dt}= 6t -18$

g) The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

Step-by-step explanation:

For this case we have the following function given:

$f(t) = s = t^3 -9t^2 +15 t$

Part a: Find the velocity at time t.

For this case we just need to take the derivate of the position function respect to t like this:

$\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

Part b: What is the velocity after 3 s?

For this case we just need to replace t=3 s into the velocity equation and we got:

$v(t=3) = 3(3)^2 -18(3) +15=-12$

Part c: When is the particle at rest?

The particle would be at rest when the velocity would be 0 so we need to solve the following equation:

$3t^2 -18 t +15 =0$

We can divide both sides of the equation by 3 and we got:

$t^2 -6t +5=0$

And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

$(t-5)*(t-1) =0$

And for this case we got $t =1s, t=5s$

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

$t^2 -6t +5 >0$

$(t-5) *(t-1)>0$

So then the intervals positive are $[0,1) \cup (5,\infty)$

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

$D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6$

And if we replace we got:

$D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

Part f: Find the acceleration at time t.

For this case we ust need to take the derivate of the velocity respect to the time like this:

$a(t) = \frac{dv}{dt}= 6t -18$

Part g and h

The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

5 0

3 years ago

Convert 0.025 days to seconds

EleoNora [17]

Answer:

2,160 seconds

Step-by-step explanation:

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3 years ago

What is 48,100,000,000,000 in scientific notation?

Over [174]

It's D. scientific notation is in the form of n x 10^m, with n any number between 1 to 10.

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3 years ago