10) 28n4 + 16n? – 80n² = 0 Factoring quadratic

Mathematics

2 answers:

Setler79 [48]2 years ago

7 0

Answer:

2n(14n^2+3)=28n^4+16n I'm not sure if you meant to put the question mark between or not but I did it separately for -80^2=0 n would equal zero hope this helped

Step-by-step explanation:

Marysya12 [62]2 years ago

4 0

I ask a moderator to delete my answer.

I worked in LATEX and my time was over and I lost everything.

You might be interested in

Line M is represented by the following equation: x − y = 8

PIT_PIT [208]

Let's plug in (18,10) into each answer choice until we find the one that has it as a solution.

A) 2x - y = 26
2(18) - 10 = 26
36 - 10 = 26
26 = 26

Your answer is A.

I'll just do the rest to show that you the they are incorrect.

B) x + y = 18
18 + 10 = 18
28 ≠ 18

C) 2x - 2y = 36
2(18) - 2(10) = 36
36 - 20 = 36
16 ≠ 36

D) x - y = -28
18 - 10 = -28
8 ≠ -28

Your answer is A) 2x - y = 26.

6 0

3 years ago

Read 2 more answers

A chemist is using 356 milliliters of a solution of acid and water. If 14.1% of the solution is acid, how many milliliters of ac

likoan [24]

50.2 milliliters of acid

4 0

2 years ago

During the 7th examination of the Offspring cohort in the Framingham Heart Study, there were 1219 participants being treated for

AlexFokin [52]

Answer:

95% confidence interval for the proportion of the population which are on treatment is [0.3293 , 0.3607].

Step-by-step explanation:

We are given that during the 7th examination of the Offspring cohort in the Framing ham Heart Study, there were 1219 participants being treated for hypertension and 2,313 who were not on treatment.

The sample proportion is : $\hat p$ = x/n = 1219/3532 = 0.345

Firstly, the pivotal quantity for 95% confidence interval for the proportion of the population 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 = 0.345

n = sample of participants = 3532

p = population proportion

<em>Here for constructing 95% confidence interval we have used One-sample z proportion statistics.</em>

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level of

significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

<u>95% confidence interval for p</u>= [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.345-1.96 \times {\sqrt{\frac{0.345(1-0.345)}{3532} } }$ , $0.345+1.96 \times {\sqrt{\frac{0.345(1-0.345)}{3532} } }$ ]