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3 years ago

PLEASE HELP WITH THIS ONE QUESTION

Mathematics

1 answer:

oksian1 [2.3K]3 years ago

7 0

Answer:

x = 0 , -2 , 1 +i√3 ; 1 - i√3

Step-by-step explanation:

2x⁴ + 16x = 0

2x(x³ + 8) = 0

2x(x³ + 2³) = 0

2x (x + 2)(x² -2x + 2²) = 0 {a³ + b³ = (a + b)*(a² -ab +b²}

2x = 0

x = 0

x + 2 = 0

x = -2

x² - 2x + 4 = 0

a = 1 ; b = -2 and c = 4

D = b² - 4ac = (-2)² - 4*1*4= 4 - 16 = -12

$\sqrt{D} = \sqrt{-12}=\sqrt{2*2*3i^{2}}=2i\sqrt{3}$

root = (-b ± √D)/2a

= (-(-2) ± 2i√3)/2*1

= (2 ± 2i√2)/2

$= \dfrac{2(1 + i\sqrt{3})}{2} \ ; \dfrac{2(1+i\sqrt{3})}{2}\\\\\\= 1 + i\sqrt{3} \ ; \ 1 - i\sqrt{3}$

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The amount of money spent on textbooks per year for students is approximately normal.

Contact [7]

Answer:

(A) A 95% confidence for the population mean is [$332.16, $447.84] .

(B) If the confidence level in part (a) changed from 95% to 99%, then the margin of error for the confidence interval would increase.

(C) If the sample size in part (a) changed from 19 to 22, then the margin of error for the confidence interval would decrease.

(D) A 99% confidence interval for the proportion of students who purchase used textbooks is [0.363, 0.477] .

Step-by-step explanation:

We are given that 19 students are randomly selected the sample mean was $390 and the standard deviation was $120.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = $390

s = sample standard deviation = $120

n = sample of students = 19

$\mu$ = population mean

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.101 < $t_1_8$ < 2.101) = 0.95 {As the critical value of t at 18 degrees of

freedom are -2.101 & 2.101 with P = 2.5%}

P(-2.101 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.101) = 0.95

P( $-2.101 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.101 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.101 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.101 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.101 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.101 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $\$390-2.101 \times {\frac{\$120}{\sqrt{19} } }$ , $\$390+2.101 \times {\frac{\$120}{\sqrt{19} } }$ ]

= [$332.16, $447.84]

(A) Therefore, a 95% confidence for the population mean is [$332.16, $447.84] .

(B) If the confidence level in part (a) changed from 95% to 99%, then the margin of error for the confidence interval which is $Z_(_\frac{\alpha}{2}_) \times \frac{s}{\sqrt{n} }$ would increase because of an increase in the z value.

(C) If the sample size in part (a) changed from 19 to 22, then the margin of error for the confidence interval which is $Z_(_\frac{\alpha}{2}_) \times \frac{s}{\sqrt{n} }$ would decrease because as denominator increases; the whole fraction decreases.

(D) We are given that to estimate the proportion of students who purchase their textbooks used, 500 students were sampled. 210 of these students purchased used textbooks.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion students who purchase their used textbooks = $\frac{210}{500}$ = 0.42

n = sample of students = 500

p = population proportion

Here for constructing a 99% confidence interval we have used a One-sample z-test statistics for proportions

So, 99% confidence interval for the population proportion, p is ;

P(-2.58 < N(0,1) < 2.58) = 0.99 {As the critical value of z at 0.5%

level of significance are -2.58 & 2.58}

P(-2.58 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.58) = 0.99

P( $-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

99% confidence interval for p = [ $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.42 -2.58 \times {\sqrt{\frac{0.42(1-0.42)}{500} } }$ , $0.42 +2.58 \times {\sqrt{\frac{0.42(1-0.42)}{500} } }$ ]

= [0.363, 0.477]

Therefore, a 99% confidence interval for the proportion of students who purchase used textbooks is [0.363, 0.477] .

8 0

3 years ago

The temperature at 6pm was 0 degrees Fahrenheit at 10 pm the temperature was -11.2 degrees Fahrenheit write an expression that y

lana66690 [7]

The answer is the temperature goes down 2.8° every hour and the expression is x=-11.2/4 because 4 is how many hours between 6pm and 10pm and its -11.2 because that was how much the temperature dropped from 6pm to 10pm

8 0

3 years ago

A food truck sells tacos, burritos, and drinks.

zubka84 [21]

Answer: Choice B) The probability of buying a taco OR a drink is 45%

The uppercase P means "probability". Using parenthesis with the P tells us "probability of whatever event happens"

P(A) means Probability of event A happening

P(B) means probability of event B happening

P(A or B) means Probability of event A or event B (or both) happening

You then replace A with "customer buys taco" and B with "customer buys drink" to get the answer mentioned above.

6 0

3 years ago