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Tju [1.3M]
2 years ago
3

The price of a stock had a change of -3 dollars explain how you would use a number line to find the absolute value of -3.

Mathematics
2 answers:
sukhopar [10]2 years ago
5 0

Plot –3 on a number line. Find the distance the point is from 0. –3 is 3 units from zero. The absolute value is the distance from 0. The absolute value of –3 is 3.

stira [4]2 years ago
4 0

Starting at 0, count the number of whole numbers between 0 and -3. The amount counted is the abolsute value of -3. I've included an image showing ths process. From the number line shown, its clear the abolsute value is 3.

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A study of immunizations among school‑age children in California found that some areas had rates of unvaccinated school‑age chil
Rom4ik [11]

Answer:

Probability that none of the 20 children in such a classroom would be unvaccinated is 0.055.

Step-by-step explanation:

We are given that a classroom of 20 children in one such area where 13.5% of children are unvaccinated.

If there are no siblings in the classroom, we are willing to consider the vaccination status of the 2020 unrelated children to be independent.

The above situation can be represented through binomial distribution;

P(X=r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r} ; x = 0,1,2,3,......

where, n = number of trials (samples) taken = 20 children

            r = number of success = none of the 20 children

            p = probability of success which in our case is probability that

                   children are unvaccinated, i.e; p = 13.5%

<u><em>Let X = Number of children that are unvaccinated</em></u>

So, X ~ Binom(n = 20, p = 0.135)

Now, Probability that none of the 20 children in such a classroom would be unvaccinated is given by = P(X = 0)

           P(X = 0)  =  \binom{20}{0} \times 0.135^{0} \times (1-0.135)^{20-0}

                          =  1\times 1 \times 0.865^{20}

                          =  0.055

<em>Hence, the probability that none of the 20 children in such a classroom would be unvaccinated is 0.055.</em>

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2 years ago
A sled is being held at rest on a slope that makes an angle theta with the horizontal. After the sled is released, it slides a d
Alenkasestr [34]

Answer:

μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

d₂ = (d₁*Sin θ) / μ

Step-by-step explanation:

a) We apply The work-energy theorem

W = ΔE

W = - Ff*d

Ff = μ*N = μ*m*g

<em>Distance 1:</em>

- Ff*d₁ = Ef - Ei

⇒  - (μ*m*g*Cos θ)*d₁ = (Kf+Uf) - (Ki+Ui) = (Kf+0) - (0+Ui) = Kf - Ui

Kf = 0.5*m*vf² = 0.5*m*v²

Ui = m*g*h = m*g*d₁*Sin θ

then

- (μ*m*g*Cos θ)*d₁ = 0.5*m*v² - m*g*d₁*Sin θ  

⇒   - μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ   <em>(I)</em>

 

<em>Distance 2:</em>

<em />

- Ff*d₂ = Ef - Ei

⇒  - (μ*m*g)*d₂ = (0+0) - (Ki+0) = - Ki

Ki = 0.5*m*vi² = 0.5*m*v²

then

- (μ*m*g)*d₂ = - 0.5*m*v²

⇒   μ*g*d₂ = 0.5*v²     <em>(II)</em>

<em />

<em>If we apply (I) + (II)</em>

- μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ

μ*g*d₂ = 0.5*v²

 ⇒ μ*g (d₂ - Cos θ*d₁) = v² - g*d₁*Sin θ   <em>  (III)</em>

Applying the equation (for the distance 1) we get v:

vf² = vi² + 2*a*d = 0² + 2*(g*Sin θ)*d₁   ⇒   vf² = 2*g*Sin θ*d₁ = v²

then (from the equation <em>III</em>) we get

μ*g (d₂ - Cos θ*d₁) = 2*g*Sin θ*d₁ - g*d₁*Sin θ

⇒  μ (d₂ - Cos θ*d₁) = Sin θ * d₁

⇒   μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

b)

If μ is a known value

d₂ = ?

We apply The work-energy theorem again

W = ΔK   ⇒   - Ff*d₂ = Kf - Ki

Ff = μ*m*g

Kf = 0

Ki = 0.5*m*v² = 0.5*m*2*g*Sin θ*d₁ = m*g*Sin θ*d₁

Finally

- μ*m*g*d₂ = 0 - m*g*Sin θ*d₁   ⇒   d₂ = Sin θ*d₁ / μ

3 0
2 years ago
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