A triangle shaped paper’s height is 18 cm and width is 10 cm. Find the hypotenuse of the paper.

∠A and ∠B are supplementary, and ∠A and ∠C are supplementary. Which conclusion is valid? Select one: A. ∠B and ∠C are supplement

zysi [14]

Option D is the correct answer.

Answer:

D. ∠B and ∠C are congruent.

Step-by-step explanation:

Since, ∠A and ∠B are supplementary.

Therefore,

∠A + ∠B = 180°.....(1)

Since, ∠A and ∠C are supplementary.

Therefore,

∠A + ∠C = 180°.....(2)

From equations (1) & (2)

∠A + ∠B = ∠A + ∠C

=> ∠B = ∠C

Hence, ∠B and ∠C are congruent.

4 0

3 years ago

The speed with which utility companies can resolve problems is very important. GTC, the Georgetown Telephone Company, reports it

brilliants [131]

Answer:

(a) 11.25 and 1.68

(b) 0.1651

(c) 0.3903

(d) 0.6865

Step-by-step explanation:

We are given that GTC, the Georgetown Telephone Company, reports it can resolve customer problems the same day they are reported in 75% of the cases and suppose the 15 cases reported today are representative of all complaints.

This situation can be represented through Binomial distribution as;

$P(X=r)= \binom{n}{r}p^{r}(1-p)^{n-r} ; x = 0,1,2,3,....$

where, n = number of trials (samples) taken = 15

r = number of success

p = probability of success which in our question is % of cases in

which customer problems are resolved on the same day, i.e.;75%

So, here X ~ $Binom(n=15,p=0.75)$

(a) Expected number of problems to be resolved today = E(X)

E(X) = $\mu$ = n * p = 15 * 0.75 = 11.25

Standard deviation = $\sigma$ = $\sqrt{n*p*(1-p)}$ = $\sqrt{15*0.75*(1-0.75)}$ = 1.68

(b) Probability that 10 of the problems can be resolved today = P(X = 10)

P(X = 10) = $\binom{15}{10}0.75^{10}(1-0.75)^{15-10}$

= $3003*0.75^{10} *0.25^{5}$ = 0.1651

(c) Probability that 10 or 11 of the problems can be resolved today is given by = P(X = 10) + P(X = 11)

= $\binom{15}{10}0.75^{10}(1-0.75)^{15-10}+\binom{15}{11}0.75^{11}(1-0.75)^{15-11}$

= $3003*0.75^{10} *0.25^{5} + 1365*0.75^{11} *0.25^{4}$ = 0.3903

(d) Probability that more than 10 of the problems can be resolved today is

given by = P(X > 10)

P(X > 10) = P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)

= $\binom{15}{11}0.75^{11}(1-0.75)^{15-11}+\binom{15}{12}0.75^{12}(1-0.75)^{15-12} + \binom{15}{13}0.75^{13}(1-0.75)^{15-13}+\binom{15}{14}0.75^{14}(1-0.75)^{15-14} + \binom{15}{15}0.75^{15}(1-0.75)^{15-15}$

= $1365*0.75^{11} *0.25^{4} + 455*0.75^{12} *0.25^{3}+105*0.75^{13} *0.25^{2} + 15*0.75^{14} *0.25^{1}+1*0.75^{15} *0.25^{0}$

= 0.6865

3 0

3 years ago

I run a 10km race in 56 minutes.what is my average speed in kilometres per minute

Luden [163]

Hey there! I'm happy to help!

You ran a 10 kilometer race in 56 minutes (good job!). To find how much you ran in 1 minute, you simply divide by 56.

10/56=0.178571428

Therefore, you ran around 0.18 km/minute.

Have a wonderful day! :D

7 0

3 years ago

Kinda want to cry right now.. Someone PLEASE help me.. I know there is someone out there smart enough.... 25 points and Brainlie

Svetlanka [38]

Answer:

a) Infinite solutions

b) No solutions

Step-by-step explanation:

First, know the following:

If the graphs intersect, there's only one solution.

If the graphs are parallel, there are no solutions.

If the graphs are the exact same line, there are infinite solutions.

For a):

Change the first equation into a linear one.

2x+3y=6

3y=-2x+6

$y=-\frac{2}{3} x+2$

Change the second equation into a linear one.

4x+6y=12
6y=-4x+12
$y=-\frac{2}{3} x+2$

Boom. You have two equations which are equal. As stated above, graphs on the exact same line have infinite solutions.

For b)

They are already in linear form so hurray.

$y=\frac{1}{5} x+2\\y=\frac{1}{5} x+10\\$

These lines are parallel since they have the SAME slope but a different y-intercept. As stated above, parallel lines have no solutions.

5 0

3 years ago

I might mark brainliest I just need help

Artemon [7]

Answer:

i think you have to measure your diagram with a protector to find the angles

8 0

2 years ago