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3 years ago

Explain why EACH sign doesn't belong.

Mathematics

2 answers:

Ainat [17]3 years ago

4 0

45 doesn’t belong because it’s the only one that says speed limit.
40 doesn’t belong because it’s the only one that says km/h.
70 doesn’t belong because it’s the middle number of of the four including in it.
100 doesn’t belong because it’s the only one that has a 3 digit.

ANTONII [103]3 years ago

3 0

Answer:

because none of them say MPH

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3 years ago

Suppose that \nabla f(x,y,z) = 2xyze^{x^2}\mathbf{i} + ze^{x^2}\mathbf{j} + ye^{x^2}\mathbf{k}. if f(0,0,0) = 2, find f(1,1,1).

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The simplest path from (0, 0, 0) to (1, 1, 1) is a straight line, denoted $C$ , which we can parameterize by the vector-valued function,

$\mathbf r(t)=(1-t)(\mathbf i+\mathbf j+\mathbf k)$

for $0\le t\le1$ , which has differential

$\mathrm d\mathbf r=-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

Then with $x(t)=y(t)=z(t)=1-t$ , we have

$\displaystyle\int_{\mathcal C}\nabla f(x,y,z)\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}\nabla f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r$

$=\displaystyle\int_{t=0}^{t=1}\left(2(1-t)^3e^{(1-t)^2}\,\mathbf i+(1-t)e^{(1-t)^2}\,\mathbf j+(1-t)e^{(1-t)^2}\,\mathbf k\right)\cdot-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

$\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)(t^2-2t+2)\,\mathrm dt$

Complete the square in the quadratic term of the integrand: $t^2-2t+2=(t-1)^2+1=(1-t)^2+1$ , then in the integral we substitute $u=1-t$ :

$\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)((1-t)^2+1)\,\mathrm dt$

$\displaystyle=-2\int_{u=0}^{u=1}e^{u^2}u(u^2+1)\,\mathrm du$

Make another substitution of $v=u^2$ :

$\displaystyle=-\int_{v=0}^{v=1}e^v(v+1)\,\mathrm dv$

Integrate by parts, taking

$r=v+1\implies\mathrm dr=\mathrm dv$

$\mathrm ds=e^v\,\mathrm dv\implies s=e^v$

$\displaystyle=-e^v(v+1)\bigg|_{v=0}^{v=1}+\int_{v=0}^{v=1}e^v\,\mathrm dv$

$\displaystyle=-(2e-1)+(e-1)=-e$

So, we have by the fundamental theorem of calculus that

$\displaystyle\int_C\nabla f(x,y,z)\cdot\mathrm d\mathbf r=f(1,1,1)-f(0,0,0)$

$\implies-e=f(1,1,1)-2$

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3 years ago

The average score of all pro golfers for a particular course has a mean of 70 and a standard deviation of 3.0 (Think of these as

bija089 [108]

Answer:

0.477 is the probability that the average score of the 36 golfers was between 70 and 71.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 70

Standard Deviation, σ = 3

Sample size, n = 36

Let the average score of all pro golfers follow a normal distribution.

Formula:

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

P(score of the 36 golfers was between 70 and 71)

$\text{Standard error of sampling} = \displaystyle\frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{36}} = \frac{1}{2}$

$P(70 \leq x \leq 71) = P(\displaystyle\frac{70 - 70}{\frac{3}{\sqrt{36}}} \leq z \leq \displaystyle\frac{71-70}{\frac{3}{\sqrt{36}}}) = P(0 \leq z \leq 2)\\\\= P(z \leq 2) - P(z \leq 0)\\= 0.977 - 0.500 = 0.477= 47.7\%$

$P(70 \leq x \leq 71) = 47.7\%$

0.477 is the probability that the average score of the 36 golfers was between 70 and 71.

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