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3 years ago

13

What is the factorization of 729x15 + 1000?

2 answers:

PSYCHO15rus [73]3 years ago

6 0

Answer:

Option A.

Step-by-step explanation:

The given expression is $729x^{15} + 1000$

Now we have to factorize the given expression.

Since we know the formula of (a³ + b³) = (a + b)(a² + b² - ab)

Now we will convert the expression in this form

$a^{3}=729x^{15}=(9x^{5})^{3}$

and $b^{3}=1000=10^{3}$

Now after factorization the expression will be

$(9x^{5}+10)(81x^{10}+100-90x^{5})$

This expression matches with option A.

NeTakaya3 years ago

4 0

The answer is A. Hope that helps

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Find the equation of the line that passes through (1,4) and is parallel to 3x+y+2=0. Leave your answer in the form y=mx+c.

Mila [183]

Answer:

y = -3x + 7.

Step-by-step explanation:

First find the slope of the given line by converting to intercept form:

3x + y + 2 = 0

y = -3x - 2

So the slope is -3.

Then the line we want has a slope of -3 also (as it is parallel).

y - y1 = m(x - x1) where m is the slope and (x1, y1) is a point on the line:

m = -3, x1 = 1 and y1 = 4, so we have:

y - 4 = -3(x - 1)

y = -3x + 3 + 4

y = -3x + 7.

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2 years ago

Paul is saving for a down payment to buy a house. The account earns 13% interest compound quarterly, and he wants to have $15,00

Tatiana [17]

Answer:

The principal must be = $8991.88

Step-by-step explanation:

Formula for compound interest is:

$A = P(1 + \frac{r}{n})^{nt}$

Where A is the amount after 't' years.

P is the principal amount

n is the number of times interest is compounded each year.

r is the rate of interest.

Here, we are given that:

Amount, A = $15000

Rate of interest = 13 % compounded quarterly i.e. 4 times every year

Number of times, interest is compounded each year, n = 4

Time, t = 4 years.

To find, Principal P = ?

Putting all the given values in the formula to find P.

$15000 = P(1 + \frac{13}{400})^{4\times 4}\\\Rightarrow 15000 = P(1 + 0.0325)^{16}\\\Rightarrow 15000 = P(1.0325)^{16} \\\Rightarrow 15000 = P \times 1.66817253\\\Rightarrow P = \dfrac{15000}{1.66817253}\\\Rightarrow P \approx \$8991.88$

So, <em>the principal must be = $8991.88</em>

4 0

3 years ago

Let X be the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. The article "Methodo

Shkiper50 [21]

Answer:

a) $P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.6288$

$P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.1954=0.4335$

b) $P(4\leq X\leq 8)=0.1954+0.1563+0.1042+0.0595+0.0298=0.5452$

c) $P(X \geq 8) = 1-P(X$

d) $P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559$

Step-by-step explanation:

Let X the random variable that represent the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. We know that $X \sim Poisson(\lambda=4)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda$

$E(X)=\mu =\lambda=4$ , $Var(X)=\lambda=2$ , $Sd(X)=2$

a. Compute both P(X≤4) and P(X<4).

$P(X\leq 4)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)+P(X=4)$

Using the pmf we can find the individual probabilities like this:

$P(X=0)=\frac{e^{-4} 4^0}{0!}=e^{-4}=0.0183$

$P(X=1)=\frac{e^{-4} 4^1}{1!}=0.0733$

$P(X=2)=\frac{e^{-4} 4^2}{2!}=0.1465$

$P(X=3)=\frac{e^{-4} 4^3}{3!}=0.1954$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.9646$

$P(X< 4)=P(X\leq 3)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)$

$P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.5311=0.7692$

b. Compute P(4≤X≤ 8).

$P(4\leq X\leq 8)=P(X=4)+P(X=5)+ P(X=6)+P(X=7)+P(X=8)$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563$

$P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042$

$P(X=7)=\frac{e^{-4} 4^7}{7!}=0.0595$

$P(X=8)=\frac{e^{-4} 4^8}{8!}=0.0298$

$P(4\leq X\leq 8)=0.1954+0.1563+ 0.1042+0.0595+0.0298=0.5452$

c. Compute P(8≤ X).

$P(X \geq 8) = 1-P(X$

$P(X \geq 8) = 1-P(X$

d. What is the probability that the number of anomalies exceeds its mean value by no more than one standard deviation?

The mean is 4 and the deviation is 2, so we want this probability

$P(4\leq X \leq 6)=P(X=4)+P(X=5)+P(X=6)$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563$

$P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042$

$P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559$

4 0

3 years ago

1 is 25% of what number

lbvjy [14]

Answer:

4

25x4=100%

1x4=4

4 0

3 years ago

Read 2 more answers

James surveyed people at school and asked whether they bring their lunch to school or buy their lunch at school more often. The

Brrunno [24]

Answer:

The correct option is 4.

Step-by-step explanation:

Given information:

Bring lunch : 46 males, 254 females

Buy lunch : 176 males, 264 females

Total number of peoples is

$46+254+176+264=740$

Total number of males is

$46+176=222$

The probability of male is

$P(Male)=\frac{Males}{Total}=\frac{222}{740} =0.3$

Since probability of males is 0.3, therefore options A and C are incorrect.

Total number of persons who buys lunch is

$176+264=440$

The probability of persons who buys lunch is

$P(\text{Buys lunch})=\frac{\text{Buys lunch}}{Total}=\frac{440}{740} =\frac{22}{37}$

We need to find the probability of P(male | buys lunch).

According to the conditional probability, we get

$P(\frac{A}{B})=\frac{P(A\cap B)}{P(B)}$

P(male | buys lunch) $=\frac{P(\text{male }\cap \text{ buys lunch})}{P(\text{buys lunch})}$

P(male | buys lunch) $=\frac{\frac{176}{740}}{\frac{22}{37}}$

P(male | buys lunch) $=\frac{2}{5}=0.4$

Therefore the correct option is 4.

8 0

3 years ago

Read 2 more answers