All categories

3 years ago

I’m need the solution ASAP

Mathematics

1 answer:

skad [1K]3 years ago

7 0

Answer:

C and D, x = 5

Step-by-step explanation:

To solve this problem, you add 9.2 to both sides then divide both sides by 5.5. The Addition Property of Equality states that if two expressions are equal to each other, and you add the same value to both sides of the equation, the equation will remain equal. This is the first step to solve the problem. The Division Property of Equality states that when we divide both sides of an equation by the same non-zero number, the two sides remain equal. That is the second step to solve the problem. To solve the problem, add 9.2 to both sides, that gets us 5.5x = 27.5. Now, we divide both sides by 5.5. That gets us x = 5.

You might be interested in

Shania is buying a car. She has narrowed her choice down to three cars: a sedan, a hatchback, and a wagon. She uses a spinner di

Bess [88]

Answer:

its D

Step-by-step explanation:

8 0

3 years ago

Read 3 more answers

PLS HELP ASAPPP I WILL GIVE BRAINLIEST

FrozenT [24]

Answer:

Yes, it is.

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

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.

(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

If y varies inversely with the square of r, what is the constant of proportionality when y - 10 and 1 = 5?

Anvisha [2.4K]

Question:

If y varies inversely with the square of r, what is the constant of proportionality when y =10 and r = 5?

Answer:

The constant of proportionality is 250

Solution:

Given that,

y varies inversely with the square of r

Which means,

$y \propto \frac{1}{r^2}\\\\y = k \times \frac{1}{r^2}$

Where, "k" is constant of proportionality

Subtitute y = 10 and r = 5

$10 = k \times \frac{1}{5^2}\\\\10 = k \times \frac{1}{25}\\\\k = 10 \times 25\\\\k = 250$

Thus the constant of proportionality is 250

6 0

3 years ago

Ms. Turner drove 825 miles in March. She drove 3 times as many miles in March as she did in January. She drove 4 times as many m

navik [9.2K]

825/3 = 275 which is how many miles she drove in January.

275*4= 1100 which is how many miles she drove in February.

Ms. Turner drove 1100 miles in February.

4 0

4 years ago