All categories

3 years ago

Factor completely. 25w² + 60w + 36 Express the answer in the form (aw+b)2 . Enter your answer in the box.

Mathematics

2 answers:

Ira Lisetskai [31]3 years ago

8 0

Solution:

→25 w² + 60 w + 36

You can do it by two methods, either by splitting the middle term that is term containing w

Or, directly converting it into form of perfect square.

1. Splitting the middle term

=25 w² + 30 w + 30 w + 36

= 5 w× ( 5 w + 6) +6× ( 5 w + 6)

= ( 5 w + 6)(5 w + 6)

= (5 w + 6)²

2. By perfect square method

25 w² + 60 w + 36

= (5 w)² + 2 × (5 w)× (6) + (6)²

= (5 w + 6)²→→→using the Identity, (a+b)²= a²+ b²+ 2 a b

Brut [27]3 years ago

7 0

25w²=(5w)², 36=6², 60w=2*5w*6
Recall the formula (a+b)²=a²+2ab+b²
in this case, <span>25w² + 60w + 36=(5w+6)</span>²

You might be interested in

Read 2/3 pages in an hour how long would it take to read 20 pages

alisha [4.7K]

It would take about 63 hours...

6 0

3 years ago

Let X denote the length of human pregnancies from conception to birth, where X has a normal distribution with mean of 264 days a

Kaylis [27]

Answer:

Step-by-step explanation:

Hello!

X: length of human pregnancies from conception to birth.

X~N(μ;σ²)

μ= 264 day

σ= 16 day

If the variable of interest has a normal distribution, it's the sample mean, that it is also a variable on its own, has a normal distribution with parameters:

X[bar] ~N(μ;σ²/n)

When calculating a probability of a value of "X" happening it corresponds to use the standard normal: Z= (X[bar]-μ)/σ

When calculating the probability of the sample mean taking a given value, the variance is divided by the sample size. The standard normal distribution to use is Z= (X[bar]-μ)/(σ/√n)

a. You need to calculate the probability that the sample mean will be less than 260 for a random sample of 15 women.

P(X[bar]<260)= P(Z<(260-264)/(16/√15))= P(Z<-0.97)= 0.16602

b. P(X[bar]>b)= 0.05

You need to find the value of X[bar] that has above it 5% of the distribution and 95% below.

P(X[bar]≤b)= 0.95

P(Z≤(b-μ)/(σ/√n))= 0.95

The value of Z that accumulates 0.95 of probability is Z= 1.648

Now we reverse the standardization to reach the value of pregnancy length:

1.648= (b-264)/(16/√15)

1.648*(16/√15)= b-264

b= [1.648*(16/√15)]+264

b= 270.81 days

c. Now the sample taken is of 7 women and you need to calculate the probability of the sample mean of the length of pregnancy lies between 1800 and 1900 days.

Symbolically:

P(1800≤X[bar]≤1900) = P(X[bar]≤1900) - P(X[bar]≤1800)

P(Z≤(1900-264)/(16/√7)) - P(Z≤(1800-264)/(16/√7))

P(Z≤270.53) - P(Z≤253.99)= 1 - 1 = 0

d. P(X[bar]>270)= 0.1151

P(Z>(270-264)/(16/√n))= 0.1151

P(Z≤(270-264)/(16/√n))= 1 - 0.1151

P(Z≤6/(16/√n))= 0.8849

With the information of the cumulated probability you can reach the value of Z and clear the sample size needed:

P(Z≤1.200)= 0.8849

$Z= \frac{X[bar]-Mu}{Sigma/\sqrt{n} }$

$Z*(Sigma/\sqrt{n} )= (X[bar]-Mu)$

$(Sigma/\sqrt{n} )= \frac{(X[bar]-Mu)}{Z}$

$Sigma= \frac{(X[bar]-Mu)}{Z}*\sqrt{n}$

$Sigma*(\frac{Z}{(X[bar]-Mu)})= \sqrt{n}$

$n = (Sigma*(\frac{Z}{(X[bar]-Mu)}))^2$

$n = (16*(\frac{1.2}{(270-264)}))^2$

n= 10.24 ≅ 11 pregnant women.

I hope it helps!

6 0

2 years ago

Do the three points lie on one line?

padilas [110]

Answer:

Yes. No graph paper.

Step-by-step explanation:

Let's say the 3 points are A, B, C.

If A, B and C lie on one line then $m_{AB} = m_{BC} = m_{AC}$

$m_{AB} = \frac{y_{B}-y_{A} }{x_{B}-x_{A}} = \frac{2-6}{3-5} = \frac{-4}{-2} = 2\\m_{BC} = \frac{y_{C}-y_{B} }{x_{C}-x_{B}} = \frac{8-2}{6-3} = \frac{6}{3} = 2\\\\m_{AC} = \frac{y_{C}-y_{A} }{x_{C}-x_{A}} = \frac{8-6}{6-5} = \frac{2}{1} = 2\\\\$