There were 462 people sitting in the audience. Round the number of people to the nearest hundred.

Julia wrote a linear equation in standard form as –6x – 3y = 10. Arthur noticed that the equation was not written in standard fo

Scrat [10]

When an equation is in standard form, x isn't negative so she should divide everything by a negative 1 to change it into a positive equation.

8 0

3 years ago

Read 2 more answers

Question

hichkok12 [17]

Answer:

Area: 201.06

Circumference:50.26

Step-by-step explanation:

Area of a circle: pi*r^2

8^2=8*8=64

64*pi=201.06193 Rounded to the neares hundredth=201.06

_________________________________________________

Circumference of a circle: 2pi * r

2*8=16

16*pi=50.2654825 Rounded to the nearest hundredth = 50.26

4 0

2 years ago

n the Question ax + bx, a is a decimal and bis a fraction. How do you decide whether to write a as a fraction or b as a decimal?

ASHA 777 [7]

It’s a number hjcjckchcjghh Igh do

3 0

3 years ago

fiona is heating water for a science experiment.the temperature of the water increases 4/5 of a degree every 2/5 of a minute.how

Scilla [17]

Answer:

4/5:2/5 = ratio of degrees to time

2/5*2.5 = 1 min

4/5*2.5 = 10/5 = 2

2 degrees every min

Step-by-step explanation:

3 0

3 years ago

Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive

LenaWriter [7]

Answer:

(a) The probability that all the next three vehicles inspected pass the inspection is 0.343.

(b) The probability that at least 1 of the next three vehicles inspected fail is 0.657.

(c) The probability that exactly 1 of the next three vehicles passes is 0.189.

(d) The probability that at most 1 of the next three vehicles passes is 0.216.

(e) The probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

Step-by-step explanation:

Let X = number of vehicles that pass the inspection.

The probability of the random variable X is P (X) = 0.70.

(a)

Compute the probability that all the next three vehicles inspected pass the inspection as follows:

P (All 3 vehicles pass) = [P (X)]³

$=(0.70)^{3}\\=0.343$

Thus, the probability that all the next three vehicles inspected pass the inspection is 0.343.

(b)

Compute the probability that at least 1 of the next three vehicles inspected fail as follows:

P (At least 1 of 3 fails) = 1 - P (All 3 vehicles pass)

$=1-0.343\\=0.657$

Thus, the probability that at least 1 of the next three vehicles inspected fail is 0.657.

(c)

Compute the probability that exactly 1 of the next three vehicles passes as follows:

P (Exactly one) = P (1st vehicle or 2nd vehicle or 3 vehicle)

= P (Only 1st vehicle passes) + P (Only 2nd vehicle passes)

+ P (Only 3rd vehicle passes)

$=(0.70\times0.30\times0.30) + (0.30\times0.70\times0.30)+(0.30\times0.30\times0.70)\\=0.189$

Thus, the probability that exactly 1 of the next three vehicles passes is 0.189.

(d)

Compute the probability that at most 1 of the next three vehicles passes as follows:

P (At most 1 vehicle passes) = P (Exactly 1 vehicles passes)

+ P (0 vehicles passes)

$=0.189+(0.30\times0.30\times0.30)\\=0.216$

Thus, the probability that at most 1 of the next three vehicles passes is 0.216.

(e)

Let X = all 3 vehicle passes and Y = at least 1 vehicle passes.

Compute the conditional probability that all 3 vehicle passes given that at least 1 vehicle passes as follows:

$P(X|Y)=\frac{P(X\cap Y)}{P(Y)} =\frac{P(X)}{P(Y)} =\frac{(0.70)^{3}}{[1-(0.30)^{3}]} =0.3525$

Thus, the probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

7 0

3 years ago