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3 years ago

44 pounds is equivalent to_____^a0 kilograms.

Mathematics

1 answer:

attashe74 [19]3 years ago

3 0

Answer:

19.504

Step-by-step explanation:

divide the mass value by 2.205

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Suppose an automotive repair company wants to determine the current percentage of customers who keep up with regular vehicle mai

Nina [5.8K]

Answer: 1692

Step-by-step explanation:

Formula to find the sample size :

$n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2$

Given : Confidence level : $(1-\alpha)=0.90$

⇒ significance level = $\alpha= 0.10$

z-value for 90% confidence interval (using z-table)= $z_{\alpha/2}=1.645$

Prior estimate of the population proportion (p) of customers who keep up with regular vehicle maintenance is unknown.

Let we take p= 0.5

Margin of error : E= 2%=0.02

Now, the required sample size will be :

$n=0.5(1-0.5)(\dfrac{1.645}{0.02})^2$

Simplify , we get

$n=(0.25)(6765.0625)=1691.265625\approx1692$

Hence, the required sample size = 1692

8 0

4 years ago

Read 2 more answers

Find the slope of the line that passes through (1,6) and (4,4)

Vadim26 [7]

Answer:

y=-2/3x+20/3

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(4-6)/(4-1)

m=-2/3

y-y1=m(x-x1)

y-6=-2/3(x-1)

y=-2/3x+2/3+6

y=-2/3x+2/3+18/3

y=-2/3x+20/3

8 0

3 years ago

Prove the following DeMorgan's laws: if LaTeX: XX, LaTeX: AA and LaTeX: BB are sets and LaTeX: \{A_i: i\in I\} {Ai:i∈I} is a fam

MariettaO [177]

$X-(A\cup B)=(X-A)\cap(X-B)$

I'll assume the usual definition of set difference, $X-A=\{x\in X,x\not\in A\}$ .

Let $x\in X-(A\cup B)$ . Then $x\in X$ and $x\not\in(A\cup B)$ . If $x\not\in(A\cup B)$ , then $x\not\in A$ and $x\not\in B$ . This means $x\in X,x\not\in A$ and $x\in X,x\not\in B$ , so it follows that $x\in(X-A)\cap(X-B)$ . Hence $X-(A\cup B)\subset(X-A)\cap(X-B)$ .

Now let $x\in(X-A)\cap(X-B)$ . Then $x\in X-A$ and $x\in X-B$ . By definition of set difference, $x\in X,x\not\in A$ and $x\in X,x\not\in B$ . Since $x\not A,x\not\in B$ , we have $x\not\in(A\cup B)$ , and so $x\in X-(A\cup B)$ . Hence $(X-A)\cap(X-B)\subset X-(A\cup B)$ .

The two sets are subsets of one another, so they must be equal.

$X-\left(\bigcup\limits_{i\in I}A_i\right)=\bigcap\limits_{i\in I}(X-A_i)$

The proof of this is the same as above, you just have to indicate that membership, of lack thereof, holds for all indices $i\in I$ .

Proof of one direction for example:

Let $x\in X-\left(\bigcup\limits_{i\in I}A_i\right)$ . Then $x\in X$ and $x\not\in\bigcup\limits_{i\in I}A_i$ , which in turn means $x\not\in A_i$ for all $i\in I$ . This means $x\in X,x\not\in A_{i_1}$ , and $x\in X,x\not\in A_{i_2}$ , and so on, where $\{i_1,i_2,\ldots\}\subset I$ , for all $i\in I$ . This means $x\in X-A_{i_1}$ , and $x\in X-A_{i_2}$ , and so on, so $x\in\bigcap\limits_{i\in I}(X-A_i)$ . Hence $X-\left(\bigcup\limits_{i\in I}A_i\right)\subset\bigcap\limits_{i\in I}(X-A_i)$ .