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3 years ago

2x^3+2x^2-19x+20=0 find the roots of the polynomial equation

Mathematics

1 answer:

Alenkinab [10]3 years ago

5 0

The answer is x=-4, (3<span>±i)/2</span>

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My school's enrollment is $506$ after a $10\%$ increase. What was the school's enrollment before the increase? 25 POINTS!!!!!!!!

lana [24]

The school's enrollment before the increase IS $460.

<h3>What are percentages? </h3>

Percentage can be described as a fraction out of an amount that is usually expressed as a number out of hundred. The sign used to represent percentages is %.

<h3>What is the school enrolment before the increase? </h3>

In order to determine the required value, divide the cost of the school enrolment after the increase by 110%(100 + 10).

$506 / 1.10 = $460

To learn more about percentages, please check: brainly.com/question/25764815

8 0

2 years ago

Pls help us (I will give Brainly)

LekaFEV [45]

what do you need help with

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

An architect earn $9164 monthly determine his annual salary

Hoochie [10]

Answer:

So there is 12 months in a year and he gets paid 9164 every one of them so we just multiply

12*9164=109968

so his annual salary is 109968 dollars

Hope This Helps!!!

5 0

3 years ago

A random sample of 50 cars in the drive-thru of a popular fast food restaurant revealed an average bill of $18.21 per car. The p

mars1129 [50]

Answer:

a) $\bar{x} = 18.21$

b) 1.64

c) (16.57,19.85)

d) The sample size must be 135 or greater if the error must not exceed $1.00.

Step-by-step explanation:

We are given the following information in the question:

Sample size, n = 50

Sample mean = $18.21

population standard deviation = $5.92

a) Point estimate for the population mean cost

$\bar{x} = 18.21$

b) Margin of error =

$z_{critical}\displaystyle\frac{\sigma}{\sqrt{n}}$

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

Margin of error = $1.96\displaystyle\frac{5.92}{\sqrt{50}} = 1.64$

c) 95% Confidence interval

$\mu \pm z_{critical}\displaystyle\frac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$18.21 \pm 1.64) = (16.57,19.85)$

d) Marginal error less than $1.00

$1.96\displaystyle\frac{\sigma}{\sqrt{n}} \leq 1\\\\\sqrt{n} \geq 1.96\times 5.92\\n \geq (11.6)^2\\n \geq 134.56 \approx 135$

Thus, the sample size must be 135 or greater if the error must not exceed $1.00.

5 0

3 years ago

Jason cut three acres of grass at a farm which is 15 percent of the property how many acres are on the farm

NeX [460]

3/15=0.2
0.2=20
20 acres

5 0

3 years ago