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galina1969 [7]
3 years ago
13

(Picture) MULTIPLYING MONOMIALS AND BINOMIALS

Mathematics
1 answer:
Sauron [17]3 years ago
3 0

Answer:

Option C is correct, i.e. 2x² +7x -4.

Step-by-step explanation:

Given is (2x-1)(x+4)

Applying FOIL method:-

F- multiply First two, O- multiply Outer two, I- multiply Inner two, L- multiply Last two.

So, (2x-1)(x+4) = 2x*x +2x*4 -1*x -1*4

(2x-1)(x+4) = 2x² +8x -1x -4

(2x-1)(x+4) = 2x² +7x -4

Hence, option C is correct, i.e. (2x-1)(x+4) = 2x² +7x -4.

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There are 34 dogwood trees and 39 maple trees currently in the park. Park workers will plant more dogwood trees today. When the
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33

Step-by-step explanation:

take the total and then subtract by 34 to get 33

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This is the last one I just really need help I am slow
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Drag each label to the correct location on the image.
Molodets [167]

Answer:

1 x (4 x 2) = 8

Step-by-step explanation:

1 x (4 x 2) = 8

7 0
2 years ago
Find the zeroes of the polynomial function f(x)=x^4-5x^3+11x^2-25x+30.
katen-ka-za [31]

Step-by-step explanation:

x⁴ − 5x³ + 11x² − 25x + 30

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7 0
3 years ago
In an article regarding interracial dating and marriage recently appeared in a newspaper. Of 1719 randomly selected adults, 311
Bingel [31]

Answer:

Step-by-step explanation:

Hello!

The parameter of interest in this exercise is the population proportion of Asians that would welcome a person of other races in their family. Using the race of the welcomed one as categorizer we can define 3 variables:

X₁: Number of Asians that would welcome a white person into their families.

X₂: Number of Asians that would welcome a Latino person into their families.

X₃: Number of Asians that would welcome a black person into their families.

Now since we are working with the population that identifies as "Asians" the sample size will be: n= 251

Since the sample size is large enough (n≥30) you can apply the Central Limit Theorem and approximate the variable distribution to normal.

Z_{1-\alpha /2}= Z_{0.975}= 1.965

1. 95% CI for Asians that would welcome a white person.

If 79% would welcome a white person, then the expected value is:

E(X)= n*p= 251*0.79= 198.29

And the Standard deviation is:

V(X)= n*p*(1-p)= 251*0.79*0.21=41.6409

√V(X)= 6.45

You can construct the interval as:

E(X)±Z₁₋α/₂*√V(X)

198.29±1.965*6.45

[185.62;210.96]

With a 95% confidence level, you'd expect that the interval [185.62; 210.96] contains the number of Asian people that would welcome a White person in their family.

2. 95% CI for Asians that would welcome a Latino person.

If 71% would welcome a Latino person, then the expected value is:

E(X)= n*p= 251*0.71= 178.21

And the Standard deviation is:

V(X)= n*p*(1-p)= 251*0.71*0.29= 51.6809

√V(X)= 7.19

You can construct the interval as:

E(X)±Z₁₋α/₂*√V(X)

178.21±1.965*7.19

[164.08; 192.34]

With a 95% confidence level, you'd expect that the interval [164.08; 192.34] contains the number of Asian people that would welcome a Latino person in their family.

3. 95% CI for Asians that would welcome a Black person.

If 66% would welcome a Black person, then the expected value is:

E(X)= n*p= 251*0.66= 165.66

And the Standard deviation is:

V(X)= n*p*(1-p)= 251*0.66*0.34= 56.3244

√V(X)= 7.50

You can construct the interval as:

E(X)±Z₁₋α/₂*√V(X)

165.66±1.965*7.50

[150.92; 180.40]

With a 95% confidence level, you'd expect that the interval [150.92; 180.40] contains the number of Asian people that would welcome a Black person in their family.

I hope it helps!

5 0
3 years ago
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