Which value is equivalent to 93.982 ОА 911 o B 924 O c 8111 c

A bank sampled its customers to determine the proportion of customers who use their debit card at least once each month. A sampl

Zinaida [17]

Answer:

(0.084,0.396)

Step-by-step explanation:

The 99% confidence interval for the proportion of customers who use debit card monthly can be constructed as

$p-z_{\frac{\alpha }{2}} \sqrt{\frac{pq}{n} }$

$p=\frac{x}{n}$

$p=\frac{12}{50}$

$p=0.24$

$q=1-p=1-0.24=0.76$

$\frac{\alpha }{2} =\frac{\0.01 }{2}=0.005$

$p-z_{\frac{\alpha }{2}} \sqrt{\frac{pq}{n} }$

$0.24-z_{0.005} \sqrt{\frac{0.24*0.76}{50} }$

$0.24-2.58(0.0604)$

$0.24-0.155832$

By rounding to three decimal places we get,

$0.084$

The 99% confidence interval for the proportion of customers who use debit card monthly is (0.084,0.396).

3 0

2 years ago

Suppose we describe the weather as either hot (Upper H) or cloudy (Upper C). Answer parts (a) and (b) below.

Ket [755]

Answer:

{HH,CC,HC,CH}

Step-by-step explanation:

We are given that

H denotes hot and cloudy denotes C.

We have to find the total possible outcomes for the weather on two consecutive days.

The possible cases in two consecutive days

Both days are hot=HH

Both days are cloud=CC

First day is hot other day cloudy=HC

First day is cloudy other day is hot=CH

Total possible cases=HH,CC,HC,CH

Therefor, the total outcomes for the weather on two consecutive days={HH,CC,HC,CH}

4 0

3 years ago

All vectors are in Rn. Check the true statements below:

Oduvanchick [21]

Answer:

A), B) and D) are true

Step-by-step explanation:

A) We can prove it as follows:

$Proy_{cv}y=\frac{(y\cdot cv)}{||cv||^2}cv=\frac{c(y\cdot v)}{c^2||v||^2}cv=\frac{(y\cdot v)}{||v||^2}v=Proy_{v}y$

B) When you compute the product Ax, the i-th component is the matrix of the i-th column of A with x, denote this by Ai x. Then, we have that $||Ax||=\sqrt{(A_1 x)^2+\cdots (A_n x)^2}$ . Now, the colums of A are orthonormal so we have that (Ai x)^2=x_i^2. Then $||Ax||=\sqrt{(x_1)^2+\cdots (x_n)^2}=||x||$ .

C) Consider $S=\{(0,2),(2,0)\}\subseteq \mathbb{R}^2$ . This set is orthogonal because $(0,2)\cdot(2,0)=0(2)+2(0)=0$ , but S is not orthonormal because the norm of (0,2) is 2≠1.

D) Let A be an orthogonal matrix in $\mathbb{R}^n$ . Then the columns of A form an orthonormal set. We have that $A^{-1}=A^t$ . To see this, note than the component $b_{ij}$ of the product $A^t A$ is the dot product of the i-th row of $A^t$ and the jth row of $A$ . But the i-th row of $A^t$ is equal to the i-th column of $A$ . If i≠j, this product is equal to 0 (orthogonality) and if i=j this product is equal to 1 (the columns are unit vectors), then $A^t A=I$

E) Consider S={e_1,0}. S is orthogonal but is not linearly independent, because 0∈S.

In fact, every orthogonal set in R^n without zero vectors is linearly independent. Take a orthogonal set $\{u_1,u_2\cdots u_p\}$ and suppose that there are coefficients a_i such that $a_1u_1+a_2u_2\cdots a_nu_n=0$ . For any i, take the dot product with u_i in both sides of the equation. All product are zero except u_i·u_i=||u_i||. Then $a_i||u_i||=0$ then $a_i=0$ .