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3 years ago

What is the length of leg s of the triangle below?

Mathematics

1 answer:

allochka39001 [22]3 years ago

5 0

Answer:

Step-by-step explanation:

s²+s²=(4√2)²

2s²=32

s²=16

s=4

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Algebra 1 polynomials hellllp

defon

A. 40x^2 + 240x + 200
Shape: Rectangle
Formula: A = LW
A = (10x + 10)(4x + 20)
= (10x)(4x) + (10x)(20) + (10)(4x) + (10)(20)
= 40x^2 + 200x + 40x + 200
= 40x^2 + 240x + 200
————————————————————-
2. 57,600 ft^2
Width:
4x + 20 = 160
4x = 140
x = 35
Plug-In:
10x + 10
10(35) + 10
350 + 10
360 = Length
————————————-
A = LW
A = 360 • 160
A = 57,600 ft^2

8 0

2 years ago

Is 8/6 greater than or less than 11/12

Zanzabum

$\frac{8}{6}\ \ \ \ and\ \ \ \ \frac{11}{12}\ \ \ \ \ | make\ a\ common\ denominator\ 12\\\\
\frac{16}{12}\ \ \ and\ \ \ \ \frac{11}{12}\\\\
Now\ it\ is\ visible\ that\ \frac{16}{12} \ >\ \frac{11}{12}\ so\ \frac{8}{6} \ >\ \frac{11}{12}\ 
$

8 0

4 years ago

Read 2 more answers

NEED HELP RIGHT NOW PLEASE HELP!!!!!!

Gekata [30.6K]

The rule is to subtract 6

3 0

4 years ago

Read 2 more answers

A company compiles data on a variety of issues in education. In 2004 the company reported that the national college freshman-to

nasty-shy [4]

Answer:

1) Randomization: We assume that we have a random sample of students

2) 10% condition, for this case we assume that the sample size is lower than 10% of the real population size

3) np = 500*0.66= 330 >10

n(1-p) = 500*(1-0.66) =170>10

So then we can use the normal approximation for the distribution of p, since the conditions are satisfied

The population proportion have the following distribution :

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

And we have :

$\mu_p = 0.66$

$\sigma_{p}= \sqrt{\frac{0.66(1-0.66)}{500}}= 0.0212$

Using the 68-95-99.7% rule we expect 68% of the values between 0.639 (63.9%) and 0.681 (68.1%), 95% of the values between 0.618(61.8%) and 0.702(70.2%) and 99.7% of the values between 0.596(59.6%) and 0.724(72.4%).

Step-by-step explanation:

For this case we know that we have a sample of n = 500 students and we have a percentage of expected return for their sophomore years given 66% and on fraction would be 0.66 and we are interested on the distribution for the population proportion p.

We want to know if we can apply the normal approximation, so we need to check 3 conditions:

1) Randomization: We assume that we have a random sample of students

2) 10% condition, for this case we assume that the sample size is lower than 10% of the real population size

3) np = 500*0.66= 330 >10

n(1-p) = 500*(1-0.66) =170>10

So then we can use the normal approximation for the distribution of p, since the conditions are satisfied

The population proportion have the following distribution :

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

And we have :

$\mu_p = 0.66$

$\sigma_{p}= \sqrt{\frac{0.66(1-0.66)}{500}}= 0.0212$

And we can use the empirical rule to describe the distribution of percentages.

The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".

On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:

• The probability of obtain values within one deviation from the mean is 0.68

• The probability of obtain values within two deviation's from the mean is 0.95

• The probability of obtain values within three deviation's from the mean is 0.997

8 0

3 years ago

Try this hard Math Problem if you dare!!

Dennis_Churaev [7]

Answer:

a. $(x - 3)^2 + 16$

b. $8(x -7)^2$

c. $(a^2 - 1)(7x - 6)$ or $(a+1)(a-1)(7x-6)$

d. $(x^2-4)(x^2+3)$ or $(x-2)(x+2)(x^2+3)$

e. $(a^n+b^n)(a^n-b^n)(a^{2n} +b^{2n})$

Step-by-step explanation:

$a.\ (x + 1)^2 - 8(x - 1) + 16$

Expand

$(x + 1)(x + 1) - 8(x - 1) + 16$

Open brackets

$x^2 + x + x + 1 - 8x + 8 + 16$

$x^2 + 2x + 1 - 8x + 24$

Collect Like Terms

$x^2 + 2x - 8x+ 1 + 24$

$x^2 - 6x+ 25$

Express 25 as 9 + 16

$x^2 - 6x+ 9 + 16$

Factorize:

$x^2 - 3x - 3x + 9 + 16$

$x(x -3)-3(x - 3) + 16$

$(x - 3)(x - 3) + 16$

$(x - 3)^2 + 16$

$b.\ 8(x - 3)^2 - 64(x-3) + 128$

Expand

$8(x - 3)(x - 3) - 64(x-3) + 128$

$8(x^2 - 6x+ 9) - 64(x-3) + 128$

Open Brackets

$8x^2 - 48x+ 72 - 64x+192 + 128$

Collect Like Terms

$8x^2 - 48x - 64x+192 + 128+ 72$

$8x^2 -112x+392$

Factorize

$8(x^2 -14x+49)$

Expand the expression in bracket

$8(x^2 -7x-7x+49)$

Factorize:

$8(x(x -7)-7(x-7))$

$8((x -7)(x-7))$

$8(x -7)^2$

$c.\ 7a^2x - 6a^2 - 7x + 6$

Factorize

$a^2(7x - 6) -1( 7x - 6)$

$(a^2 - 1)(7x - 6)$

The answer can be in this form of further expanded as follows:

$(a^2 - 1^2)(7x - 6)$

Apply difference of two squares

$(a+1)(a-1)(7x-6)$

$d.\ x^4 - x^2 - 12$

Express $x^4$ as $x^2$

$(x^2)^2 - x^2 - 12$

Expand

$(x^2)^2 +3x^2- 4x^2 - 12$

$x^2(x^2+3) -4(x^2+3)$

$(x^2-4)(x^2+3)$

The answer can be in this form of further expanded as follows:

$(x^2-2^2)(x^2+3)$

Apply difference of two squares

$(x-2)(x+2)(x^2+3)$

$e.\ a^{4n} -b^{4n}$

Represent as squares

$(a^{2n})^2 -(b^{2n})^2$

Apply difference of two squares

$(a^{2n} -b^{2n})(a^{2n} +b^{2n})$

Represent as squares

$((a^{n})^2 -(b^{n})^2)(a^{2n} +b^{2n})$

Apply difference of two squares

$(a^n+b^n)(a^n-b^n)(a^{2n} +b^{2n})$

6 0

3 years ago