HELP PLEASEEE ASAAAAPPPPPPPPPPPP I WILL GIVE BRAINLY TO THE FIRST ONE!!!!!!!!

Heyyyy help please please please

Viefleur [7K]

Answer:

% Decrease = 19.2 ; 19.2%

Step-by-step explanation:

% Decrease = (Original Price - New price)/ Original price × 100

% Decrease = (78 - 63)/78 × 100

% Decrease = 15/78 × 100

% Decrease = 19.23

8 0

2 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

Are these calculated correctly?

blsea [12.9K]

14. Was the perimeter calculated correctly?

Length = 4 yards

Breadth = 1 *2/5 yards = 7/5 yards

We know that,

Perimeter of rectangle = 2 ( l + b )

= 2 ( 4 + 7 / 5 )

= 2 ( 20 + 7 / 5 )

= 2 × 27/5

= 54 / 5

= 1 * 4/5

No ...

8 0

3 years ago

Noah rides his bike 175 miles a week. If he rides his bike the same number of miles for 15 weeks, how many miles will noah ride?

qwelly [4]

Multiply 175 and 15 to get 2625 miles.

3 0

4 years ago

I need help solving this equation: Solve x^2-4+1=0

Mamont248 [21]

Answer:

X=3/2 or 1.732

Step-by-step explanation:

X^2-4+1=0

Combine like terms

-4 + 1 = -3

X^2-3=0

add 3 to both sides

X^2= 3

X= 3/2

4 0

3 years ago