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Andru [333]
3 years ago
6

Suppose r(x) and t(x) are two functions with the same domain, and let h(x)=r(x)+t(x).

Mathematics
1 answer:
Sphinxa [80]3 years ago
4 0

Answer:

If the maximum of function r(x) and t(x) occur at the same point c in domain P = max(r(x)+t(x)) = M+N

In general P  ≤  M+N

Step-by-step explanation:

If the maximum of function r(x) and t(x) occur at the same point c in domain then M=r(c) and N=t(c) So in this case P = max(r(x)+t(x)) = M+N

In general P  ≤  M+N

by definition of maximum  

r(x)≤M,t(x)≤N   for all x in domain

=>  r(x)+t(x)≤M+N   for all x in domain

=> max(r(x)+t(x)) <= M+N

=> P  ≤  M+N

Thus we get in general the relationship is P  ≤  M+N

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82%

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To save money, a local charity organization wants to target its mailing requests for donations to individuals who are most suppo
mars1129 [50]

Answer:

The 80% confidence interval for difference between two means is (0.85, 1.55).

Step-by-step explanation:

The (1 - <em>α</em>) % confidence interval for difference between two means is:

CI=(\bar x_{1}-\bar x_{2})\pm t_{\alpha/2,(n_{1}+n_{2}-2)}\times SE_{\bar x_{1}-\bar x_{2}}

Given:

\bar x_{1}=M_{1}=6.1\\\bar x_{2}=M_{2}=4.9\\SE_{\bar x_{1}-\bar x_{2}}=0.25

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*Use a <em>t</em>-table for the critical value.

Compute the 80% confidence interval for difference between two means as follows:

CI=(6.1-4.9)\pm 1.397\times 0.25\\=1.2\pm 0.34925\\=(0.85075, 1.54925)\\\approx(0.85, 1.55)

Thus, the 80% confidence interval for difference between two means is (0.85, 1.55).

3 0
3 years ago
Evaluate m9+4 when m=3.
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Answer:


Step-by-step explanation:


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Draw the image of the following segment after a dilation centered at the origin with a scale factor of 2/3
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Answer:

Step-by-step explanation:

Let the ends of the given segment are A and B.

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Coordinates of B → (12, 12)

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(x, y) → (kx, ky)

If k = \frac{2}{3}

(x, y) → (\frac{2}{3}x, \frac{2}{3}y)

By this rule coordinates of the image points of A and B will be,

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