Please help it is due in a few mintues

In APQR, p = 690 cm, a = 290 cm and r=450 cm. Find the measure of ZQ to the

katen-ka-za [31]

9514 1404 393

Answer:

16.8°

Step-by-step explanation:

The Law of Cosines can be rearranged to give the angle measure.

angle Q = arccos((p² +r² -q²)/(2pr))

angle Q = arccos((690² +450² -290²)/(2×690×450))

angle Q = arccos(594500/621000) ≈ 16.7985°

The measure of angle Q is about 16.8°.

3 0

3 years ago

Find the area of the parallelogram that has a base of 4m and a height of 5.5m.

Sveta_85 [38]

Answer:

22m²

Step-by-step explanation:

Area of parallelogram is simply the base length x height.

In this case, base length = 4m and height = 5.5m

hence,

Area = 4 x 5.5 = 22m²

3 0

3 years ago

Read 2 more answers

PLEASE HELP TIMED TEST HELP!!!!!!!!

Lelu [443]

Answer:

Step-by-step explanation:

The equations that she wrote are

6x+8y=133

2x=y

The following statements about the system are true

1.Since the cost for each pie is $5 and $9, the first equation should be 5x+9y=133

5. The second equation is correct because there are more 6 inch pies and 2 times a number is always more.

The following statements are wrong because,

2. Because there was no relationship between the number of pipes sold and the size of each pipe.

3. x=2y shows that there are more 8 inches pipes

4. 6 and 8 are just size descriptions. It is only the cost of each size of pipe that related to the total cost

6 0

4 years ago

How many different combinations are possible if each lock contains the numbers 0 to 39, and each combination contains three dist

Georgia [21]

(e) Each license has the formABcxyz;whereC6=A; Bandx; y; zare pair-wise distinct. There are 26-2=24 possibilities forcand 10;9 and 8 possibilitiesfor each digitx; yandz;respectively, so that there are 241098 dierentlicense plates satisfying the condition of the question.3:A combination lock requires three selections of numbers, each from 1 through39:Suppose that lock is constructed in such a way that no number can be usedtwice in a row, but the same number may occur both rst and third. How manydierent combinations are possible?Solution.We can choose a combination of the formabcwherea; b; carepair-wise distinct and we get 393837 = 54834 combinations or we can choosea combination of typeabawherea6=b:There are 3938 = 1482 combinations.As two types give two disjoint sets of combinations, by addition principle, thenumber of combinations is 54834 + 1482 = 56316:4:(a) How many integers from 1 to 100;000 contain the digit 6 exactly once?(b) How many integers from 1 to 100;000 contain the digit 6 at least once?(a) How many integers from 1 to 100;000 contain two or more occurrencesof the digit 6?Solutions.(a) We identify the integers from 1 through to 100;000 by astring of length 5:(100,000 is the only string of length 6 but it does not contain6:) Also not that the rst digit could be zero but all of the digit cannot be zeroat the same time. As 6 appear exactly once, one of the following cases hold:a= 6 andb; c; d; e6= 6 and so there are 194possibilities.b= 6 anda; c; d; e6= 6;there are 194possibilities. And so on.There are 5 such possibilities and hence there are 594= 32805 such integers.(b) LetU=f1;2;;100;000g:LetAUbe the integers that DO NOTcontain 6:Every number inShas the formabcdeor 100000;where each digitcan take any value in the setf0;1;2;3;4;5;7;8;9gbut all of the digits cannot bezero since 00000 is not allowed. SojAj= 9<span>5</span>

8 0

3 years ago

A certain virus infects one in every 300 people. A test used to detect the virus in a person is positive 90% of the time when t

garik1379 [7]

Answer:

a) P[A/B] = 0,019 or P[A/B] = 1,9 %

b) P[A- /B-] = 0,9996 or P[A- /B-] = 99,96 %

Step-by-step explanation:

Bayes Theorem :

P[A/B] = P(A) * P[B/A] / P(B)

The branches of events are as follows

Condition 1 real infection 1/300 and not infection 299/300

Then

1.- 1/300 299/300

When the test is done (virus present) 0,9 (+) 0,15 (-)

2.- 299/300

When the test is done ( no virus ) 0,15 (+) 0,85 (-)

Then:

P(A) = event person infected P(B) = person test positive

a) P[A/B] = P(A) * P[B/A] / P(B)

where P(A) = 1/300 = 0,0033 P[B/A] = 0,9

Then P(A) * P[B/A] = 0,0033*0,9 = 0,00297

P(B) is ( 1/300 )*0,9 + (299/300)*0,15

P(B) = 0,0033*0,9 + 0,9966*0,15 ⇒ P(B) = 0,1524

Finally

P[A/B] = 0,00297 /0,1524

P[A/B] = 0,019 or P[A/B] = 1,9 %

b) Following sames steps:

P[A- /B-] = (299/300) * 0,85 / (299/300) * 0,85 + (1/300 * 0,1)

P[A- /B-] = 0,8471 /0,8474

P[A- /B-] = 0,9996 or P[A- /B-] = 99,96 %

6 0

3 years ago