Help please !!!!!!!!!!!!

Mathematics

1 answer:

e-lub [12.9K]3 years ago

7 0

Is this from k12? i am a k12 student, and the answer should be 2.2 units but I could be wrong.

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Louisa ran at an average speed of five miles per hour along an entire circular park path. Calvin ran along the same path in the

docker41 [41]

Answer:

15 miles

Step-by-step explanation:

Let $x$ be the miles in the circular park path, $t_{L}$ the time Louisa takes to finish and $t_{C}$ the time Calvin takes to finish both in hours.

Then $x$ , the longitude is equal to the velocity times the time used to finish. So

$x=5t_{L}$

$x=6t_{C}$

And the difference between Louisa's time and Calvin' time is 30 minutes, half an hour. So:

$t_{C}=t_{L}-0.5$

Three equations, three unknowns, the system can be solved.

Equalizing the equation with x :

$5t_{L}=6t_{C}$

In this last equation replace $t_{C}$ with the other equation and solve:

$5t_{L}=6(t_{L}-0.5)\\ 5t_{L}=6t_{L}-3\\ 3=6t_{L}-5t_{L}\\ 3=t_{L}\\ t_{L}=3$

With Louisa's time find x:

$x=5t_{L}\\ x=5(3)\\ x=15$

7 0

2 years ago

I need the answer for this problem please

Savatey [412]

Answer:

Step-by-step explanation:

Find all probabilities:

A. False

$Pr(\text{red shirt}|\text{large shirt})=\dfrac{\text{number red large shirts}}{\text{number large shirts}}=\dfrac{42}{77}=\dfrac{6}{11}\\ \\Pr(\text{large shirt})=\dfrac{\text{number large shirts}}{\text{number shirts}}=\dfrac{77}{165}=\dfrac{7}{15}$

B. True

$Pr(\text{blue shirt}|\text{large shirt})=\dfrac{\text{number blue large shirts}}{\text{number large shirts}}=\dfrac{35}{77}=\dfrac{5}{11}\\ \\Pr(\text{blue shirt})=\dfrac{\text{number blue shirts}}{\text{number shirts}}=\dfrac{75}{165}=\dfrac{5}{11}$

C. False

$Pr(\text{shirt is medium and blue})=\dfrac{\text{number medium and blue shirts}}{\text{number shirts}}=\dfrac{48}{165}=\dfrac{16}{55}\\ \\Pr(\text{medium shirt})=\dfrac{\text{number medium shirts}}{\text{number shirts}}=\dfrac{88}{165}=\dfrac{8}{15}$

D. False

$Pr(\text{large shirt}|\text{red shirt})=\dfrac{\text{number red large shirts}}{\text{number red shirts}}=\dfrac{42}{90}=\dfrac{7}{15}\\ \\Pr(\text{red shirt})=\dfrac{\text{number red shirts}}{\text{number shirts}}=\dfrac{90}{165}=\dfrac{6}{11}$