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2 years ago

Write a polynomial function of least degree with integral coefficients that has the given zeros of 5 and -2i

Mathematics

1 answer:

vova2212 [387]2 years ago

5 0

Hello :
f(x) = (x-5)(x+2i)
= x² +2ix -5x-10i
f(x) = x² +(2i-5)x -10i
<span>coefficients : a =1 b = 2i-5 c =-10i</span>

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HELP ME PLSSS Which of the following number sentences is an example of the associative property of multiplication?

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Answer:

The first or third one

Step-by-step explanation:

Associative property is where you can switch the place of the digits and it still have the same answer in the end.

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3 years ago

Grocery store sella kiwis for $4.25 a pound and Asian pears for $2.25 a pound. Write an equation in standard form using integers

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3 years ago

Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1

alexandr1967 [171]

Answer:

$a) \simeq 0.3012$ $b) \simeq 0.0494$ c) $\simeq 0.2438$

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, $\frac{1.2}{4}$

= 0.3 collisions per month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

P(X =x) = $\frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}
$

for x ∈ N ∪ {0}

= 0 otherwise --------------------------------------(1)

here, $\lambda = 0.3$ collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

P(X = 0) = $\frac{e^{-0.3}\times {\0.3}^{0}}{0!}
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$\simeq 0.7408182207$ ------------------------------------(2)

Now, since we are calculating this for 4 months,

so, P(No collision in 4 month period)

= $0.7408182207^{4}$

$\simeq 0.3012$ -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

P(X =1) = $\frac{e^{-0.3}\times {\0.3}^{1}}{1!}
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$\simeq 0.2222454662$ ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

= $0.2222454662^{2}$

$\simeq 0.0494$ -----------------------------------------(5)

1 collision in 6 months period means

$\frac{1}{6}$ collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6] (which is to be estimated)

= $\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}$

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

$\simeq 0.6543894283$ -------------------------------------------(6)

So,

P(1 collision in 6 month period)

= $0.6543894283^{6}$

$\simeq 0.0785267444$ ------------------------------------------------(7)

So,

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= $(P(X =0)^{6}$

$\simeq 0.1652988882$ ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

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7 0

2 years ago

Couldn’t figure this out help please

WARRIOR [948]

(B)

Step-by-step explanation:

Rewrite the equations into their standard forms. The first one can be rewritten as

$10x - 12y = -5$

and the 2nd can be rewritten as

$3x + 5y = -1$

Solving this system either by substitution or elimination, we get

$x = -\dfrac{37}{86}\:\:\text{and}\:\:y= \dfrac{25}{86}$

If you add x + y, you'll get a negative number.

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2 years ago

Read 2 more answers

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vovikov84 [41]

Answer:

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using the natural log (e), we were able to give the power of 5 to 2x and then take it out from the parantheses

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2 years ago