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Cerrena [4.2K]
2 years ago
13

Given f(x)= 17 – x^2, what is the average rate of change in f(x) over the interval [1, 5]?

Mathematics
1 answer:
Setler79 [48]2 years ago
5 0

Answer:

-6

Step-by-step explanation:

Average rate of change = \frac{f(b)-f(a)}{b-a} = \frac{(17-5^{2)}-(17-1^{2})  }{5-1} = -6

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Answer:

Hence,we need at least 136 rainfall PH values in the sample i.e

n ≥ 136

Step-by-step explanation:

We are given that:

(σ1)^2 = (σ2)^2 = Population variance = 0.25

So, E < 0.1

Confidence coefficient (c) = 0.9

n = n1 = n2

For confidence level, 1 - α = 0.9,we'll determine Z (α /2) = Z 0.05 by looking up 0.005 using the normal probability table which i have attached.

So, Z (α /2) = 1.645

The margin of error E is given as;

E = Z (α /2)√[(σ1)^2)/n1] + [(σ2)^2)/n2]

= Z (α /2)√({(σ1)^2 + (σ2)^2}/n) < 0.1

Multiply both sides by √n to get;

Z (α /2)√(σ1)^2 + (σ2)^2} < 0.1√n

Divide both sides by 0.1;

{Z (α /2)√(σ1)^2 + (σ2)^2}}/0.1 <√n

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{Z (α /2)√(σ1)^2 + (σ2)^2}}/0.1} ^2 < n

We'll now fill in the known values and solve;

n > ( 1.645 x √{(0.25 + 0.25)/0.1}^2

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7 0
3 years ago
Use repeated addition to find the solution to each multiplication problem. Change any improper fractions to mixed numbers. 4x1/3
Nata [24]

Answer:

1\frac{1}{3}

Step-by-step explanation:

Given the following question:

4\times\frac{1}{3}

To find the answer simply multiply the numerators and the denominators by each other.

4\times\frac{1}{3}
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\frac{4}{1} \times\frac{1}{3}
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=\frac{4}{3}
\frac{4}{3} =4\div3=1\frac{1}{3}
1\frac{1}{3}

Hope this helps.

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2 years ago
It is 18°C at 8 am, and 14°C by noon. The change in temperature from 8 am to noon is
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The answer to the question is B.
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2 years ago
Michael and his wife are each starting a saving plan. Michael will initially set aside $75 and then add $50.85 every week to the
lana [24]

Answer:

T = 75 + 116.9*N

Step-by-step explanation:

We have that the equation for Michael's savings is:

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T = A + B = 75 + 50.85*N + 65.95*N = 75 + 116.9*N

So the equation that relates T to N is:

T = 75 + 116.9*N

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