Maya's test score average decreased by 18 points this semester.

Mathematics

1 answer:

ololo11 [35]2 years ago

8 0

Answer: -18

Step-by-step explanation:

It is decreasing so the now will become negative since it is same as losing some points.

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A random sampling of a company's monthly operating expenses for n=36 months produced a sample mean of $5474 and a stan-dard devi

lilavasa [31]

Answer with explanation:

Mean of the sample(m) = $ 5474

Standard deviation of the sample (S)=764

Number of observation(n)=36

$Z_{90 \text{Percent}}=Z_{0.09}=0.5359$

$z_{score}=\frac{\Bar x-\mu}{\frac{S}{\sqrt{n}}}\\\\0.5359=\frac{5474- \mu}{\frac{764}{\sqrt{36}}}\\\\0.5359=\frac{5474- \mu}{\frac{764}{6}}\\\\764 \times 0.5359=6 \times (5474- \mu)\\\\409.4276=32844-6 \mu\\\\6 \mu=32844 -409.4276\\\\ 6 \mu=32434.5724\\\\ \mu=\frac{32434.5724}{6}\\\\ \mu=5405.76$

So, Mean Monthly Expenses of Population =$ 5405.76, which is 90% upper confidence bound for the company's mean monthly expenses.

4 0

3 years ago

(1+x^2)dy/dx+2xy-4x^2=0 bu using Bernoulli's Equation

erastovalidia [21]

Not of Bernoulli type, but still linear.

$(1+x^2)\dfrac{\mathrm dy}{\mathrm dx}+2xy=4x^2$

There's no need to find an integrating factor, since the left hand side already represents a derivative:

$\dfrac{\mathrm d}{\mathrm dx}[(1+x^2)y]=(1+x^2)\dfrac{\mathrm dy}{\mathrm dx}+2xy$

So, you have

$\dfrac{\mathrm d}{\mathrm dx}[(1+x^2)y]=4x^2$

and integrating both sides with respect to $x$ yields

$(1+x^2)y=\displaystyle\int4x^2\,\mathrm dx$
$(1+x^2)y=\dfrac43x^3+C$
$y=\dfrac{4x^3}{3(1+x^2)}+\dfrac C{1+x^2}$