Ask question

Ask question

All categories

2 years ago

15

A(t)=(t−k)(t−3)(t−6)(t+3)a, left parenthesis, t, right parenthesis, equals, left parenthesis, t, minus, k, right parenthesis, le

ft parenthesis, t, minus, 3, right parenthesis, left parenthesis, t, minus, 6, right parenthesis, left parenthesis, t, plus, 3, right parenthesis is a polynomial function of ttt, where kkk is a constant. given that a(2)=0a(2)=0a, left parenthesis, 2, right parenthesis, equals, 0, what is the absolute value of the product of the zeros of aaa?

2 answers:

Vladimir [108]2 years ago

5 0

Answer:

The absolute value of the product of the zeros of A(t) is 108

Step-by-step explanation:

We are given

$A(t)=(t-k)(t-3)(t-6)(t+3)$

we have

A(2)=0

we can use it and find k

we can plug t=2 , A(t)=0

$0=(2-k)(2-3)(2-6)(2+3)$

so, we get

$0=(2-k)$

$k=2$

now, we can plug it back

$A(t)=(t-2)(t-3)(t-6)(t+3)$

now, we can find zeros

$A(t)=(t-2)(t-3)(t-6)(t+3)=0$

$(t-2)=0$

$t=2$

$(t-3)=0$

$t=3$

$(t-6)=0$

$t=6$

$(t+3)=0$

$t=-3$

so, zeros are

$t=-3,t=2,t=3,t=6$

now, we can find it's product

$=-3\times 2\times 3\times 6$

$=-108$

now, we can find it's absolute value

$=|-108|$

$=108$

Pie2 years ago

5 0

Answer:

its 11

Step-by-step explanation:

You might be interested in

You have a function that takes in an X value and produces a Y value. The Y value equals 24 times the X value, plus 4?More. What'

guapka [62]

Answer:

100

Step-by-step explanation:

The function takes in an X value and produces a Y value.

The Y value equals 24 times the X value plus 4 more.

This means that:

Y = 24X + 4

When the X value equals 4, the Y value will be:

Y = 24(4) + 4

Y = 96 + 4

Y = 100

When the X value is 4, the Y value is 100.

7 1

2 years ago

Read 2 more answers

hree TAs are grading a final exam. There are a total of 60 exams to grade. (a) How many ways are there to distribute the exams a

nalin [4]

Answer:

a. 205320

b. 34220

c. 60! / (35)! (25)! + 60!/ (40)!(20)! + 60!/ (45)! (15)!

Step-by-step explanation:

a) The number of ways to dustribute exams among the TA's is:

n / (n - r)!

n= number of things to choose from

r= Choosing r number

60P3= 60! / (60 - 3)!

(60)(59)(58)(57)! / (57)!

=205320

B) The number of ways to dustribute the exams among the TA's is:

n! /(n - r)! r!

60C3= 60! /(60 - 3)! 3!

= 60!/ 57! 3!

= 60 × 59 × 58 / 3 × 2 × 1

= 34220

C) The required number of ways is:

60C25 + 60C20 + 60C15

= 60! / (35)! (25)! + 60!/ (40)!(20)! + 60!/ (45)! (15)!

6 0

2 years ago

What is the word form of 34/100

valkas [14]

34/100
= 0.34

3 tenths, 4 hundreths

3 0

2 years ago

Read 2 more answers

Two samples each of size 20 are taken from independent populations assumed to be normally distributed with equal variances. The

Harlamova29_29 [7]

Answer:

$t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923$

The degrees of freedom are

$df=20+20-2=38$

And the p value is given by:

$p_v =2*P(t_{38}>2.923) =0.0058$

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

Step-by-step explanation:

When we have two independent samples from two normal distributions with equal variances we are assuming that

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

And the statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:

$S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

We have the following data given:

$n_1 =20$ represent the sample size for group 1

$n_2 =20$ represent the sample size for group 2

$\bar X_1 =43.5$ represent the sample mean for the group 1

$\bar X_2 =40.1$ represent the sample mean for the group 2

$s_1=4.1$ represent the sample standard deviation for group 1

$s_2=3.2$ represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

$\S^2_p =\frac{(20-1)(4.1)^2 +(20 -1)(3.2)^2}{20 +20 -2}=13.525$

And the deviation would be just the square root of the variance:

$S_p=3.678$

The statistic is givne by:

$t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923$

The degrees of freedom are

$df=20+20-2=38$

And the p value is given by:

$p_v =2*P(t_{38}>2.923) =0.0058$

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

6 0

2 years ago

5x680 in the Distributive property

AysviL [449]

It is (5x600)+(5x80)=3400

5 0

2 years ago