Answers options are 6/26 8/18 14/12 14/26 Pls help

A 95% confidence interval for a population mean is computed from a sample of size 400. Another 95% confidence interval will be c

Anna [14]

Answer:

The interval from the sample of size 400 will be approximately One -half as wide as the interval from the sample of size 100

Step-by-step explanation:

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the 95% confidence interval is dependent on the value of the margin of error at a constant sample mean or sample proportion

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

Here assume that $Z_{\frac{\alpha }{2} } \ and \ \sigma \$ is constant so

$E = \frac{k}{\sqrt{n} }$

=> $E \sqrt{n} = K$

=> $E_1 \sqrt{n}_1 = E_2 \sqrt{n}_2$

So let $n_1 = 400$ and $n_2 = 100$

=> $E_1 \sqrt{400} = E_2 \sqrt{100}$

=> $E_1 = \frac{\sqrt{100} }{\sqrt{400} } E_2$

=> $E_1 = \frac{1}{2 } E_2$

So From this we see that the confidence interval for a sample size of 400 will be half that with a sample size of 100

7 0

2 years ago

Greta wants to make 5 pounds of a nut mix using peanuts and cashews. Her budget requires the mixture to cost her $6 per pound. P

Vinil7 [7]

Answer:

3 pounds of peanuts and 2 pounds of cashews

Step-by-step explanation:

Let x be the pounds of peanuts and y be the pounds of cashews

Greta wants to make 5 pounds of a nut mix using peanuts and cashews.

So, $x+y=5$ ----- A

Cost of 1 pound of peanuts = $4

Cost of x pounds of peanuts = 4x

Cost of 1 pound of cashews = $9

Cost of y pounds of cashews = 9y

Her budget requires the mixture to cost her $6 per pound

So, cost of mixture of x pounds of peanuts and y pounds of cashews is 6(x+y)

So, $4x+9y=6(x+y)$ --- B

Plot A and B

$x+y=5$ ---Blue

$4x+9x=6(x+y)$ ---Green

Refer the attached figure

Intersection point = (x,y)=(3,2)

Hence she should use 3 pounds of peanuts and 2 pounds of cashews

6 0

3 years ago

!!!KHAN ACADEMY!!! help please

barxatty [35]

Use photomath lollllll

3 0

2 years ago

Eric's class consists of 12 males and 16 females. If 3 students are selected at random, find the probability that they

Reptile [31]

Answer:

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

Step-by-step explanation:

Let 'M' be the event of selecting males n(M) = 12

Number of ways of choosing 3 students From all males and females

$n(M) = 28C_{3} = \frac{28!}{(28-3)!3!} =\frac{28 X 27 X 26}{3 X 2 X 1 } = 3,276$

Number of ways of choosing 3 students From all males

$n(M) = 12C_{3} = \frac{12!}{(12-3)!3!} =\frac{12 X 11 X 10}{3 X 2 X 1 } =220$

The probability that all are male of choosing '3' students

$P(E) = \frac{n(M)}{n(S)} = \frac{12 C_{3} }{28 C_{3} }$

$P(E) = \frac{12 C_{3} }{28 C_{3} } = \frac{220}{3276}$

P(E) = 0.067 = 6.71%

Final answer:-

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

3 0

3 years ago

Help and please explain so I can do it on my own. There are a couple more on this sheet.

Andreas93 [3]

Answer:

(2, -10)

Step-by-step explanation:

so first i find the axis of symmetry using the equation x= -b/2a

in this problem

a=3

b= -12

c= 2

- (-12)/ 2(3) = 2

2 will be the x point for your vertex

you then plug in your x point into the original equation and solve for y

y = 3(2)^2 - 12(2) +2

y = -10

Vertex is then (2, -10)

8 0

2 years ago