All categories

2 years ago

Jason jogged 2/3 mile in 1/3 hour. What was his speed in miles per hour?

Mathematics

1 answer:

fenix001 [56]2 years ago

7 0

Answer:

I honestly was having the worst time trying to solve this sorry :(, for real idk if any of these are right

i got 2 answers; 1st: ≈0.5 mph

the 2nd; 2 mph

Step-by-step explanation:

Speed = $\frac{Distance}{Time}$

2/3 miles = 3520 feet

1/3 hours = 20 minutes

------Make a proportion-----

$\frac{3520ft}{20min} = \frac{1 mi=5280 ft}{1hr = 60min}$

cross multiply------

3520 * 60 = 211200

5280 * 20 = 105600

Divide $\frac{211200}{105600}= 2$

---(this is where i got confused cuz 2 doesn't make any sense)--

If you flip them, $\frac{105600}{211200}=0.5$ which makes more sense

hope i helped :(

You might be interested in

Plzzzz help me but dont ansewer if you dont know it. 100 points

guapka [62]

The first one is 2x-3y/2
The second one is 1/x+2
The third one is (2x-5y)/(2x+5y)
The fourth one is x^2/z^2

5 0

3 years ago

Read 2 more answers

3n + 9 = 27 n = ? solve for n

Debora [2.8K]

Answer:

Step-by-step explanation:

7 0

2 years ago

(-6x + 3) + (X - 7) add linear expressions

Leni [432]

-6x + 3 + x - 7...when ur adding expressions such as this, and there is a plus sign in between them, the parenthesis are not needed.

we now combine like terms...
-6x + x = -5x
3 - 7 = -4

and we are left with : -5x - 4 <==

3 0

2 years ago

Hi can somebody check if this is correct please?

olya-2409 [2.1K]

Answer:

No, it is incorrect. The line should be solid, not dotted, because it's a ≤ sign, not a < sign.

8 0

2 years ago

work for a publishing company. The company wants to send two employees to a statistics conference. To be fair, the company deci

Yuki888 [10]

Answer:

(a) S = {MR, MJ, MD, MC, RJ, RD, RC, JD, JC, DC}

(b) The probability that Roberto and John attend the conference is 0.10.

(d) The probability that John stays home is 0.60.

Step-by-step explanation:

It is provided that :

Marco (M), Roberto (R), John (J), Dominique (D) and Clarice (C) works for the company.

The company selects two employees randomly to attend a statistics conference.

(a)

There are 5 employees from which the company has to select two employees to send to the conference.

So the total number of ways to select two employees is:

${5\choose 2}=\frac{5!}{2!(5-2)!}=\frac{5\times 4\times 3!}{2\times 3!}=10$

The 10 possible samples are:

MR, MJ, MD, MC, RJ, RD, RC, JD, JC, DC

(b)

The probability of the event E is:

$P(E)=\frac{n(E)}{N}$

Here,

n (E) = favorable outcomes

N = Total number of outcomes.

The variable representing the selection of Roberto and John is, RJ.

The favorable number of outcomes to select Roberto and John is, 1.

The total number of outcomes to select 2 employees is 10.

Compute the probability that Roberto and John attend the conference as follows:

$P(RJ)=\frac{n(RJ)}{N}=\frac{1}{10}=0.10$

Thus, the probability that Roberto and John attend the conference is 0.10.

(c)

The favorable outcomes of the event where Clarice attends the conference are:

n (C) = {MC, RC, JC and DC} = 4

Compute the probability that Clarice attends the conference as follows:

$P(C)=\frac{n(C)}{N}=\frac{4}{10}=0.40$

Thus, the probability that Clarice attends the conference is 0.40.

(d)

The favorable outcomes of the event where John does not attends the conference are:

n (J') = MR, MD, MC, RD, RC, DC

Compute the probability that John stays home as follows:

$P(J')=\frac{n(J')}{N}=\frac{6}{10}=0.60$

Thus, the probability that John stays home is 0.60.

4 0

2 years ago