An airplane travels 2248 miles in 4 hours.

Chloe’s 5 ft 4 inches tall. There are 2.54 centimeters in 1 inch . What is Chloe’s height in centimeters ?

Lerok [7]

162.56 :) :P :) :P.......

6 0

3 years ago

Suppose there are 4 defective batteries in a drawer with 10 batteries in it. A sample of 3 is taken at random without replacemen

SSSSS [86.1K]

Answer:

a.) 0.5

b.) 0.66

c.) 0.83

Step-by-step explanation:

As given,

Total Number of Batteries in the drawer = 10

Total Number of defective Batteries in the drawer = 4

⇒Total Number of non - defective Batteries in the drawer = 10 - 4 = 6

Now,

As, a sample of 3 is taken at random without replacement.

a.)

Getting exactly one defective battery means -

1 - from defective battery

2 - from non-defective battery

So,

Getting exactly 1 defective battery = ⁴C₁ × ⁶C₂ = $\frac{4!}{1! (4 - 1 )!}$ × $\frac{6!}{2! (6 - 2 )!}$

= $\frac{4!}{(3)!}$ × $\frac{6!}{2! (4)!}$

= $\frac{4.3!}{(3)!}$ × $\frac{6.5.4!}{2! (4)!}$

= $4$ × $\frac{6.5}{2.1! }$

= $4$ × $15$ = 60

Total Number of possibility = ¹⁰C₃ = $\frac{10!}{3! (10-3)!}$

= $\frac{10!}{3! (7)!}$

= $\frac{10.9.8.7!}{3! (7)!}$

= $\frac{10.9.8}{3.2.1!}$

= 120

So, probability = $\frac{60}{120} = \frac{1}{2} = 0.5$

b.)

at most one defective battery :

⇒either the defective battery is 1 or 0

If the defective battery is 1 , then 2 non defective

Possibility = ⁴C₁ × ⁶C₂ = 60

If the defective battery is 0 , then 3 non defective

Possibility = ⁴C₀ × ⁶C₃

= $\frac{4!}{0! (4 - 0)!}$ × $\frac{6!}{3! (6 - 3)!}$

= $\frac{4!}{(4)!}$ × $\frac{6!}{3! (3)!}$

= 1 × $\frac{6.5.4.3!}{3.2.1! (3)!}$

= 1× $\frac{6.5.4}{3.2.1! }$

= 1 × 20 = 20

getting at most 1 defective battery = 60 + 20 = 80

Probability = $\frac{80}{120} = \frac{8}{12} = 0.66$

c.)

at least one defective battery :

⇒either the defective battery is 1 or 2 or 3

If the defective battery is 1 , then 2 non defective

Possibility = ⁴C₁ × ⁶C₂ = 60

If the defective battery is 2 , then 1 non defective

Possibility = ⁴C₂ × ⁶C₁

= $\frac{4!}{2! (4 - 2)!}$ × $\frac{6!}{1! (6 - 1)!}$

= $\frac{4!}{2! (2)!}$ × $\frac{6!}{1! (5)!}$

= $\frac{4.3.2!}{2! (2)!}$ × $\frac{6.5!}{1! (5)!}$

= $\frac{4.3}{2.1!}$ × $\frac{6}{1}$

= 6 × 6 = 36

If the defective battery is 3 , then 0 non defective

Possibility = ⁴C₃ × ⁶C₀

= $\frac{4!}{3! (4 - 3)!}$ × $\frac{6!}{0! (6 - 0)!}$

= $\frac{4!}{3! (1)!}$ × $\frac{6!}{(6)!}$

= $\frac{4.3!}{3!}$ × 1

= 4×1 = 4

getting at most 1 defective battery = 60 + 36 + 4 = 100

Probability = $\frac{100}{120} = \frac{10}{12} = 0.83$

3 0

2 years ago

What is the quadratic function that is created with roots at 2 and 4 and a vertex at (3, 1)?

Arada [10]

Hello,

y=k*(x-2)(x-4)
and is passing throught (3,1)
==>1=k*(3-2)(3-4)==>k=-1

y=-(x-2)(x-4) is an answer

4 0

3 years ago

PLEASE help me and PLEASE show the work PLEASE.

Dafna1 [17]

20(4×-2)

________

(4 - (-2))²

simplified;

20(-8)

_____

(6)²

simplified;

-160

______

36

simplified;

-4 16/36

simplified;

-4 4/9

3 0

2 years ago

Country Financial, a financial services company, uses surveys of adults age and older to determine if personal financial fitness

Yakvenalex [24]

Complete Question

Country financials, a financial services company, uses surveys of adults age 18 and older to determine if personal financial fitness is changing over time. In February 2012, a sample of 1000 adults showed 410 indicating that their financial security was more that fair. In Feb 2010, a sample of 900 adults showed 315 indicating that their financial security was more than fair.

a

State the hypotheses that can be used to test for a significant difference between the population proportions for the two years.

b

What is the sample proportion indicating that their financial security was more that fair in 2012?In 2010?

c

Conduct the hypothesis test and compute the p-value.At a .05 level of significance what is your conclusion?

Answer:

a

The null hypothesis is $H_o : p_1 = p_2$

The alternative hypothesis is $H_a : p_1 \ne p_2$

b

in 2012 $\r p_1 =0.41$

in 2010 $\r p_2 =0.35$

c

The p-value is $p-value = 0.0072$

The conclusion is

There is sufficient evidence to conclude that the proportion of those indicating that financial security is more fair in Feb 2010 is different from the proportion of those indicating that financial security is more fair in Feb 2012.

Step-by-step explanation:

From the question we are told that

The sample size in 2012 is $n_1 = 1000$

The number that indicated that their finance was more than fail is k = 410

The sample size in 2010 is $n_2 = 900$

The number that indicated that their finance was more than fail is u = 315

The level of significance is $\alpha = 0.05[/ex]The null hypothesis is [tex]H_o : p_1 = p_2$

The alternative hypothesis is $H_a : p_1 \ne p_2$

Generally the sample proportion for 2012 is mathematically represented as

$\r p_1 = \frac{k}{n_1}$

=> $\r p_1 = \frac{410}{1000}$

=> $\r p_1 =0.41$

Generally the sample proportion for 2010 is mathematically represented as

$\r p_2 = \frac{u}{n_2}$

=> $\r p_2 = \frac{315}{900}$

=> $\r p_2 =0.35$

Generally the pooled sample proportion is mathematically represented as

$\r p = \frac{k + u }{n_1 + n_2}$

=> $\r p = \frac{410 + 315 }{1000+ 900}$

=> $\r p = 0.3816$

Generally the test statistics is mathematically represented as

$z = \frac{( \r p_1 - \r p_2 ) - 0}{\sqrt{\r p (1 - \r p ) ( \frac{1}{n_1} +\frac{1}{n_2} )} }$

$z = \frac{( 0.41 -0.35 ) - 0}{\sqrt{0.3816 (1 - 0.3816 ) ( \frac{1}{1000} +\frac{1}{900} )} }$

$z = 2.688$

Generally the p- value is mathematically represented as

$p-value = 2 P(Z >2.688 )$

From the z -table

$P(Z > 2.688) = 0.0036$

So

$p-value = 2 *0.0036$

$p-value = 0.0072$

So from the p-value obtained we see that p-value < $\alpha$ so we reject the null hypothesis

Thus there is sufficient evidence to conclude that the proportion of financial security is more fair in Feb 2010 is different from the proportion of financial security is more fair in Feb 2012.

5 0

3 years ago