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snow_tiger [21]

3 years ago

13

The perimeter of a rectangle is 80 feet. Find the dimensions if the length 5 feet longer than four times the width. Then find th

e area of the rectangle.

1 answer:

nydimaria [60]3 years ago

3 0

Answer:

l=33 feet

w=7 feet

Area= 231 feet

Step-by-step explanation:

Solution in Picture is attached

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En un videojuego, Marta ha conseguido 36.450 puntos capturando 11 monedas de oro. ¿ cuántos puntos vale cada moneda de oro?

sukhopar [10]

Answer:

3313.64 puntos

Step-by-step explanation:

Podemos interpretar la pregunta anterior Matemáticamente como:

11 monedas de oro = 36.450 puntos

1 moneda de oro = x

Multiplicar cruzada

11 monedas de oro × x = 36.450 puntos × 1 monedas de oro

x = 36.450 puntos × 1 monedas de oro / 11

x = 3313.6363636 puntos

Aproximadamente

1 moneda de oro = 3313.64 puntos

3 0

3 years ago

Please help me so I can pass

Liula [17]

Answer

A

Step-by-step explanation:

i just calculated it its right

4 0

3 years ago

Which statement proves that the diagonals of square PQRS are perpendicular bisectors of each other?

MrRissso [65]

The answer is D,thank me later!

4 0

3 years ago

Read 2 more answers

Solve the system by substitution.<br> 5x+2y= 5<br> y=-2x + 3

Mamont248 [21]

5x+2(-2x+3)=5
5x-4x+6=5
x=-1

Then plug back into either to solve for y.

y=-2(-1)+3
y=2+3
y=5

Substitution means you just replace one variable with the other equation so only one variable is showing. That’s how you solve for the first variable.

3 0

3 years ago

Two events E1 and E2 are called independent if p(E1 â© E2) = p(E1)p(E2). For each of the following pairs of events, which are su

posledela

Answer: a) Independent

b) Independent

c) Dependent

Step-by-step explanation:

Since, If a coin is tossed three times,

Then, total number of outcomes, n(S) = 8

a) $E_1$ : tails comes up with the coin is tossed the first time;

$E_1$ = { TTT, THH, THT, TTH }

$E_2$ : heads comes up when the coin is tossed the second time.

$E_2$ = { THT, HHH, THH, HHT }

Thus, $n(E_1)=4$

⇒ $P(E_1)=\frac{n(E_1)}{n(S)}=\frac{4}{8}=\frac{1}{2}$

Similarly, $P(E_2)=\frac{1}{2}$

⇒ $P(E_1)\times P(E_2)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$

Since, $E_1\cap E_2$ = { THH, THT }

$n(E_1\cap E_2) = 2$

⇒ $P(E_1\cap E_2) = \frac{n(E_1\cap E_2)}{n(S)}= \frac{2}{8}=\frac{1}{4}$

Thus, $P(E_1\cap E_2)=P(E_1)\timesP(E_2)$

Therefore, $E_1$ and $E_2$ are independent events.

B) $E_1$ : the first coin comes up tails

$E_1$ = { TTT, THH, THT, TTH }

$E_2$ : two, and not three, heads come up in a row

$E_2$ = { HHT, THH }

Thus, $n(E_1)=4$

⇒ $P(E_1)=\frac{n(E_1)}{n(S)}=\frac{4}{8}=\frac{1}{2}$

Similarly, $P(E_2)=\frac{1}{4}$

⇒ $P(E_1)\times P(E_2)=\frac{1}{2}\times \frac{1}{4}=\frac{1}{8}$

Since, $E_1\cap E_2$ = { THH }

$n(E_1\cap E_2) = 1$

⇒ $P(E_1\cap E_2) = \frac{n(E_1\cap E_2)}{n(S)}= \frac{1}{8}$

Thus, $P(E_1\cap E_2)=P(E_1)\timesP(E_2)$

Therefore, $E_1$ and $E_2$ are independent events.

C) $E_1$ : the second coin comes up tails;

$E_1$ = { HTH, HTT, TTT, TTH }

$E_2$ : two, and not three, heads come up in a row

$E_2$ = { HHT, THH }

Thus, $n(E_1)=4$

⇒ $P(E_1)=\frac{n(E_1)}{n(S)}=\frac{4}{8}=\frac{1}{2}$

Similarly, $P(E_2)=\frac{1}{4}$

⇒ $P(E_1)\times P(E_2)=\frac{1}{2}\times \frac{1}{4}=\frac{1}{8}$

Since, $E_1\cap E_2$ = $\phi$

$n(E_1\cap E_2) = 0$

⇒ $P(E_1\cap E_2) = 0$

Thus, $P(E_1\cap E_2)\neq P(E_1)\timesP(E_2)$

Therefore, $E_1$ and $E_2$ are dependent events.

3 0

3 years ago