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2 years ago

Integer (-1) c (+5) =

Mathematics

2 answers:

Radda [10]2 years ago

7 0

I got -13.5914091423 from a calculator I'm not sure if it's correct

LenKa [72]2 years ago

4 0

-13.5914091423 is what I got from a calculator.

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The sum of 5x2y and (2xy2 + x2y) is

allochka39001 [22]

<span>5x²y + 2xy² + x²y

Combining like terms would be
6x²y + 2xy²

The two terms are now unique and cannot be combined any further. </span><span>
</span>

6 0

2 years ago

Of all the weld failures in a certain assembly, 85% of them occur in the weld metal itself, and the remaining 15% occur in the b

Aloiza [94]

Answer:

a.) 0.1028

b.) 0.6477

c.) 0.0388

d.) 3

e.) 2.55

Step-by-step explanation:

Forming a binomial Probability distribution

n = 20

Probability of success for Weld metal failure = 85%

Probability of success for base metal failure = 15%

We use the probabilit distribution formula of combination to solve the problem.

P(x=r) = nCr * p^r * q^n-r

a.) if exactly 5 are base metal failures, then p = 15 and our solution becomes:

P(x=5) = 20C5 * 0.15^5 * 0.85^15

P(x=5) = 0.1028

b.) probability that fewer than 4 are base metal failure= P(x=0) + P(x=1) + P(x=2) + P(x=3)

P(x=0) = 20C0 * 0.15^0 * 0.85^20 = 0.0388

P(x=1) = 20C1 * 0.15¹ * 0.85^19 = 0.1368

P(x=2) = 20C2 * 0.15² * 0.85^18 = 0.2293

P(x=3) = 20C3 * 0.15³ * 0.85^17 = 0.2428

Probability that fewer than 4 are base metal failures becomes: 0.038 + 0.1368 + 0.2293 + 0.2428 = 0.6477

c.) probability that none of them are results of base metal failure = P(x=0). As earlier calculated,

P(x=0) = 0.0388

d.) mean of base metal failures = np = 20*0.15 = 3

e.) standard deviation of base metal failures = √np(1-p)

=3 * (1 - 0.15) = 3 * 0.85

= 2.55

4 0

2 years ago

A local arts council has 200 members. The council president wanted to estimate the percent of its members who have had experienc

garri49 [273]

Answer:

Yes the sample can be use to make inference

Step-by-step explanation:

The inference is possible if the conditions:

p*n > 10 and q*n > 10

where p and q are the proportion probability of success and q = 1 - p

n is sample size

Then p = 12 / 30 = 0,4 q = 1 - 0,4 q = 0,6

And p*n = 0,4 * 30 = 12 12 > 10

And q*n = 0,6 * 30 = 18 18 > 10

Therefore with that sample the conditions to approximate the binomial distribution to a Normal distribution are met

6 0

2 years ago

What is the square root of r64?

ohaa [14]

Answer is r32 because it is the square root of an exponent you must divide it.

5 0

2 years ago

Read 2 more answers

Report Error Suppose $P(x)$ is a polynomial of smallest possible degree such that: $\bullet$ $P(x)$ has rational coefficients $\

motikmotik

Answer:

We want a polynomial of smallest degree with rational coefficients with zeros in $\sqrt{7}$ , $1 - \sqrt{6}$ and -3. The last root gives us the factor (x+3). Hence, our polynomial is

$P(x) =(x+3)q(x)$

where $q$ is a polynomial with rational coefficients and roots $\sqrt{7}$ and $1 - \sqrt{6}$ . The root $\sqrt{7}$ gives us a factor $x-\sqrt{7}$ , but in order to obtain rational coefficients we must consider the factor $x^2-7$ .

An analogue idea works with $1 - \sqrt{6}$ . For convenience write $x - 1 + \sqrt{6} = ( x - 1) + \sqrt{6}$ . This gives the factor $(x-1)^2-6$ . Hence,

$P(x) = (x+3)(x^2-7)((x-1)^2-6)=x^5+x^4-18x^3-22x^2+77x+105$

Notice that $P(-1)=24$ . So, in order to satisfy the last condition we divide by 3 the whole polynomial, without altering its roots. Finally, the wanted polynomial is

$P(x) =(1/3)x^5+(1/3)x^4-6x^3-(22/3)x^2+(77/3)x+35$

Step-by-step explanation:

We must have present that any polynomial it's determined by its roots up to a constant factor. But here we have irrational ones, in order to eliminate the irrational coefficients that a factor of the type $x-\sqrt7$ will introduce in the expression, we need to multiply by its conjugate $x+\sqrt7$ . Hence, we will obtain $x^2-7$ that have rational coefficients. Finally, the last condition is given with the intention to fix the constant factor. Usually it is enough to evaluate in the point and obtain the necessary factor.

4 0

2 years ago