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3 years ago

Match each quadratic equation with its roots. Tiles a2-a+11+0

1 answer:

7 0

To solve this problem you must aply the proccedure shown below:

You have the following quadratic equation: a^2-a+11=0 and to solve it you must apply the quadratic formula x=-b√(b^2-4ac)/2a

a=1
b=-1
c=11

The result is

a1=(1-i√43)/2
a2=(1+i√43)/2

The answer is:a1=(1-i√43)/2 and a2=(1+i√43)/2

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Step-by-step explanation:

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2 years ago

(1 point) A random sample of 600 movie goers in Flagstaff found 252 movie goers who had bought popcorn on their last visit. Find

garik1379 [7]

Answer-

A 95% confidence interval for the true percent of movie goers is 36.41% to 44.25%

Solution-

Given,

n = 600 (sample size)

x = 252 (number of people who bought)

Confidence interval = 95%, so z = 1.96

We know that,

$\mu = M\pm Z(SE)$

where,

M = sample mean

Z = Z statistic determined by confidence level

SE = standard error of mean

Calculating the values,

$M=\frac{x}{n} =\frac{242}{600} =0.4033$

$Z=1.96$ from the tables

$SE=\sqrt{\frac{M\times (1-M)}{n}}$

$\Rightarrow SE=\sqrt{\frac{0.4033\times (1-0.4033)}{600}}=0.02$

Putting all the values in the formula,

$\mu = M\pm Z(SE)$

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3 years ago

25. Michael is studying population changes in two

g100num [7]

Answer: Average rates of change for Bird A is 0.045 and for Bird B is 3.816.

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Average Rate of Change

Bird A:

For the interval [0,18]:

x₁ = 0 f(0) = 8.3

x₂ = 18 f(18) = 9.1

Average = $\frac{f(18) - f(0)}{x_{2} - x_{1} }$

Average = $\frac{9.1 - 8.3}{18 - 0}$

Average = 0.045

Bird B: y = 3.6(1.06)x

For the interval [0,18]:

x₁ = 0 y = 3.6(1.06)0 = 0

x₂ = 18 y = 3.6(1.06)18 = 68.7

Average = $\frac{f(18) - f(0)}{x_{2} - x_{1} }$

Average = $\frac{68.7 - 0}{18 - 0}$

Average = 3.816

The Average Rate for Change for Bird A is 0.045 and for Bird B is 3.816. This means that the population of Bird B increase in rate faster than the population of Bird A.

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2 years ago