H(h+4) monomial times polynomial

Mathematics

1 answer:

Fofino [41]2 years ago

4 0

Distribute the h:

h(h+4)
h*h + 4*h
h² + 4h

Hope this helps!

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Consider a population list x with μx=10 and SDx = 1. A second population list, y, with μy=10 and SDy=2, is added to the first li

Basile [38]

Answer:

The combined standard deviation is 1.58114.

Step-by-step explanation:

The formula to compute the combined standard deviations of two different data sets is:

$SD_{c} =\sqrt{\frac{n_{X}S^{2}_{X}+n_{2}S^{2}_{Y}+n_{X}(\mu_{X}-\mu_{c})^{2}+n_{Y}(\mu_{Y}-\mu_{c})^{2}}{n_{X}+n_{Y}}$

Here $\mu_{c}$ is the combined mean given by:

$\mu_{c}=\frac{n_{X}\mu_{X}+n_{Y}\mu_{Y}}{n_{X}+n_{Y}}$

It is provided that the sample size is same for both the data sets, i.e. $n_{X} = n_{Y}=n$

Compute the combined mean as follows:

$\mu_{c}=\frac{n_{X}\mu_{X}+n_{Y}\mu_{Y}}{n_{X}+n_{Y}}\\=\frac{(n\times10)+(n\times10)}{n+n}}\\=\frac{20n}{2n}\\ =10$

Compute the combined standard deviation as follows:

$SD_{c} =\sqrt{\frac{n_{X}S^{2}_{X}+n_{2}S^{2}_{Y}+n_{X}(\mu_{X}-\mu_{c})^{2}+n_{Y}(\mu_{Y}-\mu_{c})^{2}}{n_{X}+n_{Y}}}\\=\sqrt{\frac{(n\times1^{2})+(n\times2^{2})+(n(10-10))+(n(10-10))}{n+n}}\\=\sqrt{\frac{n+4n}{2n} } \\=\sqrt{\frac{5n}{2n} } \\=\sqrt{\frac{5}{2}} \\=1.58114$

Thus, the combined standard deviation is 1.58114.

3 0

2 years ago

Can someone please help me i will give brainliest

Ronch [10]

Answer:

50 out of 100 or 50%

Step-by-step explanation:

the data represents between quartile 3 and quartile 1 (known as the Inter-Quartile Range, IQR) is, theoretically, supposed to represent 50% of all data items

4 0

2 years ago

Can someone pls help me with number 28.

alexandr402 [8]

Answer:

$f(x) = \frac{3}{x} \\\\g(x) = x - 5$