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3 years ago

If c(x)=4-2 and d(x)=x^2+5x, what is (c•d)(x)?

1 answer:

8 0

1)What is f(3) if f(x) = -5x3 + 6x2 - x - 4?
a. -74
b. -88
c. 74
d. 182
f(3) = -5(3)^3 + 6(3)^2 - 3 - 4
f(3) = -5(27) + 6(9) - 7
f(3) = -135 + 54 - 7 = -88
(b.)
2)What is f(x + 1) if f(x) = 6x3 - 3x2 + 4x - 9?
a. 6x3 + 12x2 + 4x + 2
b. 6x3 + 3x2 + 8x + 6
c. 6x3 + 21x2 + 20x + 4
d. 6x3 + 15x2 + 16x - 2
f(x + 1) = 6(x + 1)^3 - 3(x + 1)^2 + 4(x + 1) - 9
f(x + 1) = 6(x^3 + 3x^2 + 3x + 1) - 3(x^2 + 2x + 1) + 4x + 4 - 9
f(x + 1) = 6x^3 + 18x^2 + 18x + 6 - 3x^2 - 6x - 3 + 4x + 4 - 9
f(x + 1) = 6x^3 + 15x^2 + 16x - 2
(d.)
3)What is 3[f(x + 2)] if f(x) = x3 + 2x2 - 4?
a. x3 + 8x2 + 20x + 12
b. 3x3 + 12x2 + 18x + 6
c. 3x3 + 24x2 + 60x + 36
d. 3x3 + 18x2 + 24x + 60
f(x + 2) = (x + 2)^3 + 2(x + 2)^2 - 4
f(x + 2) = x^3 + 6x^2 + 12x + 8 + 2x^2 + 8x + 8 - 4
f(x + 2) = x^3 + 8x^2 + 20x + 12
3[f(x + 2)] = 3x^3 + 24x^2 + 60x + 36
(c.)
4)Use synthetic division to determine which of the following is a factor of x3 - 3x2 - 10x + 24.
a. x - 2
b. x - 3
c. x + 4

d. x + 8
2|....1....-3....-10....24
.......1.....-1.....-12....0
(x - 2) works .... (a.)
5)Use synthetic division to determine which of the following is a factor of 2x3 - 13x2 + 17x + 12.
a. x - 2
b. x - 3
c. x + 4
d. x + 6
3|....2....-13....17....12
.......2.....-7.....-4....0
(x - 3) is a factor .... (b.)
6)What is the remainder when (6x3 + 9x2 - 6x + 2) ÷ (x + 2)?
a. -4
b. 0
c. 2
d. 74
-2|....6....9....-6....2
..........6.....-3.....0....2
(c.)
7)What is the remainder when (x3 - x2 - 5x - 3) ÷ (x + 1)?
a. -8
b. 0
c. 2
d. 4
-1|....1....-1....-5....-3
.........1.....-2.....-3....0
(b.)
8)What are the factors of x3 + 2x2 - x - 2?
a. (x - 1)(x + 1)(x - 2) = (x^2 - 1)(x - 2) = x^3 - 2x^2 - x + 2
b. (x - 2)(x + 2)(x - 1)
c. (x - 2)(x + 2)(x + 1)
d. (x - 1)(x + 1)(x + 2) = (x^2 - 1)(x + 2) = x^3 + 2x^2 - x - 2
(d.)

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4 years ago

A certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder. In

lord [1]

Answer:

95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

Step-by-step explanation:

We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder.

A random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;

P.Q. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of males having blood disorder= $\frac{250}{1000}$ = 0.25

$\hat p_2$ = sample proportion of females having blood disorder = $\frac{275}{1000}$ = 0.275

$n_1$ = sample of males = 1000

$n_2$ = sample of females = 1000

$p_1$ = population proportion of males having blood disorder

$p_2$ = population proportion of females having blood disorder

Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.

So, 95% confidence interval for the difference between the population proportions, ( $p_1-p_2$ ) is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ${(\hat p_1-\hat p_2)-(p_1-p_2)}$ < $1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

P( $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ( $p_1-p_2$ ) < $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

95% confidence interval for ( $p_1-p_2$ ) =

[ $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ , $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ]

= [ $(0.25-0.275)-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ , $(0.25-0.275)+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ ]

= [-0.064 , 0.014]

Therefore, 95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

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3 years ago

How can I solve this problem?

irina [24]

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180+60=240
240/13=18.4615
X=18.4615

I hope this is correct.
(There are so many different ways to do math problems like this) I hope I could help... sorry if I didn’t.

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3 years ago

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