All categories

3 years ago

What is 5/4= x/12 solve for proportion?

Mathematics

2 answers:

worty [1.4K]3 years ago

6 0

$\dfrac{5}{4} = \dfrac{x}{12} \\ \\4*x=5*12 \\ \\4x=60 \ \ \ /:4 \\ \\x=60:4 \\ \\x=15$

Diano4ka-milaya [45]3 years ago

5 0

$\frac{5}{4}=\frac{x}{12}\ \ \ \ \ |cross\ multiply\\\\4\cdot x=5\cdot12\\\\4x=60\ \ \ \ \ |divide\ both\ sides\ by\ 4\\\\\boxed{x=15}$

You might be interested in

What is the value of S in the equation 1/3s+14=26

Law Incorporation [45]

First, isolate 1/3s:
1/3s=12
Second, multiply both sides by 3 to get a singular value of s (just s):
s=36

4 0

3 years ago

The range of the function f(k) = k2 + 2k + 1 is {25, 64}. What is the function’s domain?

s344n2d4d5 [400]

{k^2 + 2k + 1 = 25, k^2 + 2k + 1 = 64}
{k^2 + 2k - 24 = 0, k^2 + 2k - 63 = 0}
{(k - 4)(k + 6) = 0, (k - 7)(k + 9) = 0}
{k = -6, 4, -9, 7)
Therefore, range = {-9, -6, 4, 7)

3 0

3 years ago

TO TRANSMIT INFORMATION ON THE INTERNET, LARGE FILES ARE BROKEN INTO PACKETS OF SMALLER SIZES. EACH PACKET HAS 1,500 BYTES OF IN

borishaifa [10]

Answer:

A. It would be needed 20 packets to transmit 30,000 bytes of information.

B. 45'000,000 (45 million) of bytes could be transmitted in 30,000 packets.

Step-by-step explanation:

1. Let's review the data given to us for solving the question:

Size of each packet of information = 1,500 bytes

Number of packets = P

Number of bytes of information = B

2. Let's answer the questions:

A. How many packets would be needed to transmit 30,000 bytes of information?

P = 30,000 ÷ 1,500

P = 20

It would be needed 20 packets to transmit 30,000 bytes of information

B. How much information could be transmitted in 30,000 packets?

B = 30,000 * 1,500

B = 45'000,000

45'000,000 (45 million) of bytes could be transmitted in 30,000 packets.

6 0

3 years ago

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

3 years ago

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

NeX [460]

Answer:

(a) 0.343

(b) 0.657

(d) 0.216

(e) 0.353

Step-by-step explanation:

Let P(a vehicle passing the test) = p

$p = \frac{70}{100} = 0.7$

Let P(a vehicle not passing the test) = q

q = 1 - p

q = 1 - 0.7 = 0.3

(a) P(all of the next three vehicles inspected pass) = P(ppp)

= 0.7 × 0.7 × 0.7

= 0.343

(b) P(at least one of the next three inspected fails) = P(qpp or qqp or pqp or pqq or ppq or qpq or qqq)

= (0.3 × 0.7 × 0.7) + (0.3 × 0.3 × 0.7) + (0.7 × 0.3 × 0.7) + (0.7 × 0.3 × 0.3) + (0.7 × 0.7 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.3)

= 0.147 + 0.063 + 0.147 + 0.063 + 0.147 + 0.063 + 0.027

= 0.657

= (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7)

= 0.063 + 0.063 + 0.063

= 0.189

(d) P(at most one of the next three vehicles inspected passes) = P(pqq or qpq or qqp or qqq)

= (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7) + (0.3 × 0.3 × 0.3)

= 0.063 + 0.063 + 0.063 + 0.027

= 0.216

(e) Given that at least one of the next 3 vehicles passes inspection, what is the probability that all 3 pass (a conditional probability)?

P(at least one of the next three vehicles inspected passes) = P(ppp or ppq or pqp or qpp or pqq or qpq or qqp)

= (0.7 × 0.7 × 0.7) + (0.7 × 0.7 × 0.3) + (0.7 × 0.3 × 0.7) + (0.3 × 0.7 × 0.7) + (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7)

= 0.343 + 0.147 + 0.147 + 0.147 + 0.063 + 0.063 + 0.063

= 0.973

With the condition that at least one of the next 3 vehicles passes inspection, the probability that all 3 pass is,

$= \frac{P(all\ of\ the\ next\ three\ vehicles\ inspected\ pass)}{P(at\ least\ one\ of\ the\ next\ three\ vehicles\ inspected\ passes)}$

$= \frac{0.343}{0.973}$

= 0.353

3 0

4 years ago

Read 2 more answers