Ask question

Ask question

All categories

3 years ago

13

fie abc un triunghi cu masura unghiului a de 90 de grade ab este 6 cm bc=10 si ca=8 calculati;sin b; tg b; sin c; tg c; cos b; c

tg b; cos c; ctg c

1 answer:

lions [1.4K]3 years ago

6 0

SinB = cosC = AC / BC = 8/10 = 4/5;
tgB = ctg C = AC/AB = 8/6 = 4/3, because we use T. Pitagora for calculating AB;
sinC = cosB = AB/ BC = 6/10 = 3/5;
tgC = ctgB = AB/ AC = 6/8 = 3/4.

You might be interested in

The increasing annual cost (including tuition, room, board, books, and fees) to attend college has been widely discussed (Time).

NeX [460]

Answer:

(a) PRIVATE COLLEGES

Sample mean is $42.5 thousand

Sample standard deviation is $6.65 thousand

PUBLIC COLLEGES

Sample mean is $22.3 thousand

Sample standard deviation is $4.34 thousand

(b) Point estimate is $20.2 thousand. The mean annual cost to attend private colleges ($42.5 thousand) is more than the mean annual cost to attend public colleges ($22.3 thousand)

(c) 95% confidence interval of the difference between the mean annual cost of attending private and public colleges is $19.2 thousand to $21.2 thousand

Step-by-step explanation:

(a) PRIVATE COLLEGES

Sample mean = Total cost ÷ number of colleges = (51.8+42.2+45+34.3+44+29.6+46.8+36.8+51.5+43) ÷ 10 = 425 ÷ 10 = $42.5 thousand

Sample standard deviation = sqrt[summation (cost - sample mean)^2 ÷ number of colleges] = sqrt([(51.8-42.5)^2 + (42.2-42.5)^2 + (45-42.5)^2 + (34.3-42.5)^2 + (44-42.5)^2 + (29.6-42.5)^2 + (36.8-42.5)^2 + (51.5-42.5)^2 + (43-42.5)^2] ÷ 10) = sqrt (44.24) = $6.65 thousand

PUBLIC COLLEGES

Sample mean = (20.3+22+28.2+15.6+24.1+28.5+22.8+25.8+18.5+25.6+14.4+21.8) ÷ 12 = 267.6 ÷ 12 = $22.3 thousand

Sample standard deviation = sqrt([(20.3-22.3)^2 + (22-22.3)^2 + (28.2-22.3)^2 + (15.6-22.3)^2 + (24.1-22.3)^2 + (28.5-22.3)^2 + (22.8-22.3)^2 + (25.8-22.3)^2 + (18.5-22.3)^2 + (25.6-22.3)^2 + (14.4-22.3)^2 + (21.8-22.3)^2] ÷ 12) = sqrt (18.83) = $4.34 thousand

(b) Point estimate = mean annual cost of attending private colleges - mean annual cost of attending public colleges = $42.5 thousand - $22.3 thousand = $20.2 thousand.

This implies the the mean annual cost of attending private colleges is greater than the mean annual cost of attending public colleges

(c) Confidence Interval = Mean + or - Margin of error (E)

E = t×sd/√n

Mean = $42.5 - $22.3 = $20.2 thousand

sd = $6.65 - $4.34 = $2.31 thousand

n = 10+12 = 22

degree of freedom = 22-2 = 20

t-value corresponding to 20 degrees of freedom and 95% confidence level is 2.086

E = 2.086×$2.31/√22 = $1.0 thousand

Lower bound = Mean - E = $20.2 thousand - $1.0 thousand = $19.2 thousand

Upper bound = Mean + E = $20.2 thousand + $1.0 thousand = $21.2 thousand

95% confidence interval is $19.2 thousand to $21.2 thousand

6 0

3 years ago

I just want to know if I’m right

Svetradugi [14.3K]

Answer:

correct

Step-by-step explanation:

supplementary is when 2 angles add up to 180

8 0

3 years ago

Read 2 more answers

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

4 0

2 years ago

Please help and explain this

IRINA_888 [86]

Multiply each price by the discount into a decimal (move the decimal place over 2 on the percents) then subtract that from the regular price and that's sale price

3 0

3 years ago

Read 2 more answers

Bryan’s monthly electric bill is determined by adding a flat administration fee to the product of the number of kilowatt hours o

Sophie [7]

We will form the equations for this problem:
(1) 1100*y + z = 113
(2) 1500*y + z = 153
z = ? Monthly administration fee is notated with z, and that is the this problem's question.
Number of kilowatt hours of electricity used are numbers 1100 and 1500 respectively.
Cost per kilowatt hour is notated with y, but its value is not asked in this math problem, but we can calculate it anyway.
The problem becomes two equations with two unknowns, it is a system, and can be solved with method of replacement:
(1) 1100*y + z = 113
(2) 1500*y + z = 153
----------------------------
(1) z = 113 - 1100*y [insert value of z (right side) into (2) equation instead of z]:
(2) 1500*y + (113 - 1100*y) = 153
-------------------------------------------------
(1) z = 113 - 1100*y
(2) 1500*y + 113 - 1100*y = 153
------------------------------------------------
(1) z = 113 - 1100*y
(2) 400*y + 113 = 153
------------------------------------------------
(1) z = 113 - 1100*y
(2) 400*y = 153 - 113
------------------------------------------------
(1) z = 113 - 1100*y
(2) 400*y = 40
------------------------------------------------
(1) z = 113 - 1100*y
(2) y = 40/400
------------------------------------------------
(1) z = 113 - 1100*y
(2) y = 1/10
------------------------------------------------
if we insert the obtained value of y into (1) equation, we get the value of z:
(1) z = 113 - 1100*(1/10)
(1) z = 113 - 110
(1) z = 3 dollars is the monthly fee.

6 0

3 years ago

Read 2 more answers