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3 years ago

5

PLZ ANSWER QUICKLY, WILL GIVE THE BRAINLIEST AND 25 POINTS!

2 answers:

valkas [14]3 years ago

6 0

We are given that ratio of cars Sam has built to cars Tyrone has built is 3:4.

Let us assume that Sam built $3x$ cars and Tyrone built $4x$ .

Since we are given that total number of cars they built is 42. Therefore, we can set up:

$3x+4x=42\\
7x=42\\
x=6$

Therefore, Sam built $3x=3(6)=18$ cars and Tyrone built $4x=4(6)=24$ cars.

Anni [7]3 years ago

6 0

Sam has 18 cars with 6 cars per box while Tyrone has 24 cars with 6 cars per box.

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Use the similar shape concept of geometry to find the missing value (x). Round your answer to the nearest tenth if necessary.

trapecia [35]

Step-by-step explanation:

7:3

4:x

-cross multiply

7x = 12

x = 12÷7

x = 1.7

7 0

3 years ago

PLEASE HELP!!! FIND X IN THE DIAGRAM PROVIDED!! WILL GIVE BRIAIN!!

iren2701 [21]

: $\implies$ $\sf 3x⁰+45⁰+90⁰ = 180⁰$

: $\implies$ $\sf 3x⁰+135⁰ = 180⁰$

: $\implies$ $\sf 3x⁰ = 45⁰$

: $\implies$ $\sf x = \dfrac{45}{3}$

: $\implies$ $\sf x = 15⁰$

hence, last option is right

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3 years ago

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The price of an Item yesterday was $160. Today, the price rose to $248. Find the percent increase.

Tpy6a [65]

Answer:

jguifutrdduturdrfuy

Step-by-step explanation:

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2 years ago

How do you solve 6=1/12v

Veronika [31]

Answer:

72

Step-by-step explanation:

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2 years ago

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Waiting on the platform, a commuter hears an announcement that the train is running five minutes late. He assumes the arrival ti

natima [27]

Answer:

D. 91%

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Less than 15 minutes.

Event B: Less than 10 minutes.

We are given the following probability distribution:

$f(T = t) = \frac{3}{5}(\frac{5}{t})^4, t \geq 5$

Simplifying:

$f(T = t) = \frac{3*5^4}{5t^4} = \frac{375}{t^4}$

Probability of arriving in less than 15 minutes:

Integral of the distribution from 5 to 15. So

$P(A) = \int_{5}^{15} = \frac{375}{t^4}$

Integral of $\frac{1}{t^4} = t^{-4}$ is $\frac{t^{-3}}{-3} = -\frac{1}{3t^3}$

Then

$\int \frac{375}{t^4} dt = -\frac{125}{t^3}$

Applying the limits, by the Fundamental Theorem of Calculus:

At $t = 15$ , $f(15) = -\frac{125}{15^3} = -\frac{1}{27}$

At $t = 5$ , $f(5) = -\frac{125}{5^3} = -1$

Then

$P(A) = -\frac{1}{27} + 1 = -\frac{1}{27} + \frac{27}{27} = \frac{26}{27}$

Probability of arriving in less than 15 minutes and less than 10 minutes.

The intersection of these events is less than 10 minutes, so:

$P(B) = \int_{5}^{10} = \frac{375}{t^4}$

We already have the integral, so just apply the limits:

At $t = 10$ , $f(10) = -\frac{125}{10^3} = -\frac{1}{8}$

At $t = 5$ , $f(5) = -\frac{125}{5^3} = -1$

Then

$P(A \cap B) = -\frac{1}{8} + 1 = -\frac{1}{8} + \frac{8}{8} = \frac{7}{8}$

If given the train arrived in less than 15 minutes, what is the probability it arrived in less than 10 minutes?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{7}{8}}{\frac{26}{27}} = 0.9087$

Thus 90.87%, approximately 91%, and the correct answer is given by option D.

3 0

3 years ago