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3 years ago

12

Help as soon as possible please

1 answer:

marta [7]3 years ago

6 0

(3x + 5)° + (10x - 7)° = 180°

13x + (-2) = 180

13x = 180 + 2

13x = 182

x = 14

∠QRT = 3 * 14 + 5 = 42 + 5 = 47°

∠TRS = 10 * 14 - 7 = 140 - 7 = 133°

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The probability that a professor arrives on time is 0.8 and the probability that a student arrives on time is 0.6. Assuming thes

saul85 [17]

Answer:

a)0.08 , b)0.4 , C) i)0.84 , ii)0.56

Step-by-step explanation:

Given data

P(A) = professor arrives on time

P(A) = 0.8

P(B) = Student aarive on time

P(B) = 0.6

According to the question A & B are Independent

P(A∩B) = P(A) . P(B)

Therefore

${A}'$ & ${B}'$ is also independent

${A}'$ = 1-0.8 = 0.2

${B}'$ = 1-0.6 = 0.4

part a)

Probability of both student and the professor are late

P(A'∩B') = P(A') . P(B') (only for independent cases)

= 0.2 x 0.4

= 0.08

Part b)

The probability that the student is late given that the professor is on time

$P(\frac{B'}{A})$ = $\frac{P(B'\cap A)}{P(A)}$ = $\frac{0.4\times 0.8}{0.8}$ = 0.4

Part c)

Assume the events are not independent

Given Data

P $(\frac{{A}'}{{B}'})$ = 0.4

= $\frac{P({A}'\cap {B}')}{P({B}')}$ = 0.4

$P$ $({A}'\cap {B}')$ = 0.4 x P $({B}')$

= 0.4 x 0.4 = 0.16

$P({A}'\cap {B}')$ = 0.16

i)

The probability that at least one of them is on time

$P(A\cup B)$ = 1- $P({A}'\cap {B}')$

= 1 - 0.16 = 0.84

ii)The probability that they are both on time

P $(A\cap B)$ = 1 - $P({A}'\cup {B}')$ = 1 - $[P({A}')+P({B}') - P({A}'\cap {B}')]$

= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56

6 0

3 years ago

Sarah purchased a set of art supplies for $20.90 and a pair of jeans for $18.95.The sales tax rate in the state is 7%.How much c

mash [69]

Total purchase she did : $20.90+$18.95=$39.85
7% tax means that we have to 7% or (7/100) extra
7% X 39.85= 2.79
total pay: 39.85+2.79=42.64
total cash: $50
what she gets back: 50-42.64=7.36

8 0

3 years ago

En un pozo petrolìfero produce cada dia 2680 barriles de petroleo. Si cada barril contiene 159 litros de petroleo, cuantos litro

Aliun [14]

Answer:

426,120 litros se producen en un día Se producen

12,783,600 litros por mes (para un cálculo de 30 días en un mes)

Step-by-step explanation:

Aquí, debemos calcular la cantidad en litros de petróleo producido en un pozo petrolero.

Se nos dice que se producen 2.680 barriles por día y cada barril contiene 159 litros de petróleo.

La producción diaria en litros es, por lo tanto, de 159 * 2680 = 426,120 litros.

Ahora, para la producción mensual, supongamos que hay 30 días en un mes, la cantidad en litros producidos por mes sería la cantidad diaria multiplicada por la cantidad de días en un mes.

Matemáticamente, eso sería 30 * 426,120 = 12,783,600 litros

4 0

3 years ago

SOLVING EACH SYSTEM BY ELIMINATION <br><br><br> x+3y=5 x=-6y+14<br><br><br> HELP

nata0808 [166]

Answer:

$\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}$

$x=-4,\:y=3$

Step-by-step explanation:

Given the system of the equations

$x+3y=5;\:x=-6y+14$

solving by elimination method

$\begin{bmatrix}x+3y=5\\ x=-6y+14\end{bmatrix}$

$\mathrm{Arrange\:equation\:variables\:for\:elimination}$

$\begin{bmatrix}x+3y=5\\ x+6y=14\end{bmatrix}$

$x+6y=14$

$-$

$\underline{x+3y=5}$

$3y=9$

$\begin{bmatrix}x+3y=5\\ 3y=9\end{bmatrix}$

solve $3y=9$ for $y$ :

$3y=9$

$\frac{3y}{3}=\frac{9}{3}$

$y=3$

$\mathrm{For\:}x+3y=5\mathrm{\:plug\:in\:}y=3$

Solve $x+3\cdot \:3=5$ for x:

$x+3\cdot \:3=5$

$x+9=5$

$x=-4$

Therefore,

$\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}$

$x=-4,\:y=3$

6 0

2 years ago

3. Which expressions are equivalent to the given one?

Montano1993 [528]

Answer:

b, d

Step-by-step explanation:

20/-4 = -5

20 x (-1/4) = -5

4 0

1 year ago