Write the following algebraically, using x as your

Jermiah works at an appliance store. He reacantly sold 12 clothes dryers and 36 other appliances. Of the next 100 appliances he

Viefleur [7K]

12/36 = x/100
Then cross multiply
1200 = 36x
Divide by 36
x = 33.33

Can’t sell 0.33 of an appliance, so your answer is 33.

4 0

2 years ago

200 meters

Katyanochek1 [597]

Answer:

520 m

Step-by-step explanation:

Perimeter is the distance around a shape. To find the perimeter, you add up all the sides.

200 + 60 + 60 + 200 = 520 m

7 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

2 years ago

Plz help me NOW I will mark you brainliest

puteri [66]

Let x be the percentage of p.
Therefore,
$m\% \: \: of \: \: n = x\% \: \: of \: \: p \\ = > \frac{mn}{100} = \frac{px}{100} \\ = > mn = \frac{px}{100} \times 100 \\ = > mn = px \\ = > \frac{mn}{p} = x$

Answer:

$\frac{mn}{p}$

Hope you could get an idea from here.

Doubt clarification - use comment section.

8 0

2 years ago

Yu narukami fight me rigth now⇆⇆π⇒ωω↑ω⊇→→∧∉⇔⇒ππ∠⇄㏑∡⇄⇆→←βαβ∀∀⊅⊄ФФω≅

Andru [333]

Nani!? Omae wa mou shindeiru

3 0

2 years ago

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