I need help please and thank you

Given the following polynomial find the maximum possible number of turning points. f(x)=x^11+x^10+x^12+x^9

Turning points are inflection points

think
1st degree (linear) has no turning/inflecion points
2nd degree (quadratic, parabola) has 1 turning/inflection point
so
nth degree has n-1 turning/inflection point

this is 11th degree since highest power is 12
12-1=11
11 turning oints

5 0

3 years ago

Can yoy please help solve yhis problem.2/5b+1=11

shusha [124]

Answer:

(2/5)*b+1 = -11 // + 11

(2/5)*b+1+11 = 0

2/5*b+12 = 0 // - 12

2/5*b = -12 // : 2/5

b = -12/2/5

b = -30

Step-by-step explanation:

hope this helped :)

5 0

2 years ago

Read 2 more answers

YOUR TURN

Ivahew [28]

Answer:

$Width: \frac{3in}{8ft} = \frac{5in}{x} (cross multiply)$

= $40$ ÷3 ≈ 13.3ft

$Width: \frac {3in}{8ft} = \frac{6.5in}{x}$

= 17.3

$A=e$ × $w$

$= 13.3$ × $17.3$ ≈ $230.09$ ≈ $230.1$

8 0

2 years ago

www.g A bag contains 3 white counters, 10 black counters, and 4 green counters. What is the probability of drawing (a) a white c

erastovalidia [21]

Answer:

41.18% probability of drawing a white counter or a green counter

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

In this question:

There are 3+10+4 = 17 counters.

Of those, 3+4 = 7 are white or green

7/17 = 0.4118

41.18% probability of drawing a white counter or a green counter

4 0

2 years ago

To test the belief that sons are taller than their fathers, a student randomly selects 13 fathers who have adult male children.

KATRIN_1 [288]

Answer:

1) B. The differences are normally distributed or the sample size is large

C. The sample size mus be large

E. The sampling method results in an independent sample

2) The null hypothesis H₀: $\bar x_1$ = $\bar x_2$

The alternative hypothesis Hₐ: $\bar x_1$ < $\bar x_2$

Test statistic, t = -0.00693

p- value = 0.498

Do not reject Upper H₀ because, the P-value is greater than the level of significance. There is sufficient evidence to conclude that sons are the same height as their fathers at 0.10 level of significance

Step-by-step explanation:

1) B. The differences are normally distributed or the sample size is large

C. The sample size mus be large

E. The sampling method results in an independent sample

2) The null hypothesis H₀: $\bar x_1$ = $\bar x_2$

The alternative hypothesis Hₐ: $\bar x_1$ < $\bar x_2$

The test statistic for t test is;

$t=\dfrac{(\bar{x}_1-\bar{x}_2)}{\sqrt{\dfrac{s_{1}^{2} }{n_{1}}-\dfrac{s _{2}^{2}}{n_{2}}}}$

The mean

Height of Father, h₁, Height of Son h₂

72.4, 77.5

70.6, 74.1

73.1, 75.6

69.9, 71.7

69.4, 70.5

69.4, 69.9

68.1, 68.2

68.9, 68.2

70.5, 69.3

69.4, 67.7

69.5, 67

67.2, 63.7

70.4, 65.5

$\bar x_1$ = 69.6

s₁ = 1.58

$\bar x_2$ = 69.9

s₂ = 3.97

n₁ = 13

n₂ = 13

$t=\dfrac{(69.908-69.915)}{\sqrt{\dfrac{3.97^{2}}{13}-\dfrac{1.58^{2} }{13}}}$

(We reversed the values in the square root of the denominator therefore, the sign reversal)

t = -0.00693

p- value = 0.498 by graphing calculator function

P-value > α Therefore, we do not reject the null hypothesis

Do not reject Upper H₀ because, the P-value is greater than the level of significance. There is sufficient evidence to conclude that sons are the same height as their fathers at 0.10 lvel of significance

8 0

3 years ago