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irakobra [83]
3 years ago
7

Help simplify please choose the correct answer~!!!!!

Mathematics
2 answers:
ANEK [815]3 years ago
7 0

Answer:

A

Step-by-step explanation:

Given

-4x²(5x^{4} - 3x² + x - 2)

Multiply each term in the parenthesis by - 4x²

= - 20x^{6} + 12x^{4} - 4x³ + 8x²

Bogdan [553]3 years ago
3 0

Answer:

A

Step-by-step explanation:

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In a parallel circuit, ET = 240 V, R = 330 Ω, and XL = 420 Ω. What is Z?
denis-greek [22]

Answer:

Z = 258Ω

Step-by-step explanation:

There are two methods of solving this question,

First method

Z = ( RXL)/ sqr rt ( R²+XL²)

Z= (330 Ω × 420 Ω)/ sqr rt ( 330 Ω² + 420 Ω²)

Z = 138600 Ω /534

Z = 259  Ω

THE second complete method is given in attachment below


6 0
3 years ago
The length of the 2 altitudes of a parallelogram are 4cm and 6cm. The perimeter of the parallelogram is 40cm. Find the lengths o
Afina-wow [57]
The person that answered it before is right
6 0
2 years ago
Read 2 more answers
In a certain population of the eastern thwump bird, the wingspan of the individual birds follows an approximately normal distrib
Greeley [361]

Answer:

a) P(48 < x < 58) = 0.576

b) P(X ≥ 1) = 0.9863

c) E(X) 2.88

d) P(x < 48) = 0.212

e) P(X > 2) = 0.06755

Step-by-step explanation:

The mean of the wingspan of the birds = μ = 53.0 mm

The standard deviation = σ = 6.25 mm

a) Probability of a bird having a wingspan between 48 mm and 58 mm can be found by modelling the problem as a normal distribution problem.

To solve this, we first normalize/standardize the two wingspans concerned.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ

For wingspan 48 mm

z = (48 - 53)/6.25 = - 0.80

For wingspan 58 mm

z = (58 - 53)/6.25 = 0.80

To determine the probability that the wingspan of the first bird chosen is between 48 and 58 mm long. P(48 < x < 58) = P(-0.80 < z < 0.80)

We'll use data from the normal probability table for these probabilities

P(48 < x < 58) = P(-0.80 < z < 0.80) = P(z < 0.8) - P(z < -0.8) = 0.788 - 0.212 = 0.576

b) The probability that at least one of the five birds has a wingspan between 48 and 58 mm = 1 - (Probability that none of the five birds has a wingspan between 48 and 58 mm)

P(X ≥ 1) = 1 - P(X=0)

Probability that none of the five birds have a wingspan between 48 and 58 mm is a binomial distribution problem.

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = number of birds = 5

x = Number of successes required = number of birds with wingspan between 48 mm and 58 mm = 0

p = probability of success = Probability of one bird having wingspan between 48 mm and 58 mm = 0.576

q = probability of failure = Probability of one bird not having wingspan between 48 mm and 58 mm = 1 - 0.576 = 0.424

P(X=0) = ⁵C₀ (0.576)⁰ (1 - 0.576)⁵ = (1) (1) (0.424)⁵ = 0.0137

The probability that at least one of the five birds has a wingspan between 48 and 58 mm = P(X≥1) = 1 - P(X=0) = 1 - 0.0137 = 0.9863

c) The expected number of birds in this sample whose wingspan is between 48 and 58 mm.

Expected value is a sum of each variable and its probability,

E(X) = mean = np = 5×0.576 = 2.88

d) The probability that the wingspan of a randomly chosen bird is less than 48 mm long

Using the normal distribution tables again

P(x < 48) = P(z < -0.8) = 1 - P(z ≥ -0.8) = 1 - P(z ≤ 0.8) = 1 - 0.788 = 0.212

e) The probability that more than two of the five birds have wingspans less than 48 mm long = P(X > 2) = P(X=3) + P(X=4) + P(X=5)

This is also a binomial distribution problem,

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = number of birds = 5

x = Number of successes required = number of birds with wingspan less than 48 mm = more than 2 i.e. 3,4 and 5.

p = probability of success = Probability of one bird having wingspan less than 48 mm = 0.212

q = probability of failure = Probability of one bird not having wingspan less than 48 mm = 1 - p = 0.788

P(X > 2) = P(X=3) + P(X=4) + P(X=5)

P(X > 2) = 0.05916433913 + 0.00795865476 + 0.00042823218

P(X > 2) = 0.06755122607 = 0.06755

5 0
3 years ago
What is a polynomial function in standard form with zeroes 0, 1, 4, and –1?
mr Goodwill [35]
The polynomial for this case is given by:
 (x) * (x-1) * (x + 1) * (x-4) = 0
 Rescribing we have:
 (x) * (x ^ 2-1) * (x-4) = 0
 (x) * (x ^ 3 - 4x ^ 2 - x + 4) = 0
 (x ^ 4 - 4x ^ 3 - x ^ 2 + 4x) = 0
 Therefore, the polynomial in standard form is:
 f (x) = x ^ 4 - 4x ^ 3 - x ^ 2 + 4x
 Answer:
 
a polynomial function in standard form with zeroes 0, 1, 4, and -1 is:
 
f (x) = x ^ 4 - 4x ^ 3 - x ^ 2 + 4x
3 0
3 years ago
Use the associative property to identify which expression is equal to (10)(5x)(7). 1.(7)(5x)(10) 2. 10(5x + 7) 3. (5x)(10)(7) 4.
8090 [49]
Associative property is dealing with parentahsees and orders
it is
a(bc)=(ab)c and
a+(b+c)=(a+b)+c

so we want the multiplication one

given
(10)(5x)(7)
we can arragnge these a number of ways (we should use commutative property, but whatever)

first one works, the 7 was moved to front
2nd is wrong, that is a wierd version of the distributive
3rd is right, 5x was moved to the front
4th is wack, no bueno


answer is 1st and 3rd options
6 0
3 years ago
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