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lara [203]
3 years ago
7

Charlie runs at a speed of 3 yards per second. About how many miles per hour does Charlie run

Mathematics
2 answers:
enyata [817]3 years ago
7 0
To find the answer, you first must find how many yards Charlie can run in a minute. Since there are 60 seconds in a minute, you multiply 3 times 60, which is 180. Then you must multiply this number by 60 again, because there are 60 minutes in an hour. This equals 10,800, your answer.


Charlie can run 10,800 yards in an hour.
Law Incorporation [45]3 years ago
5 0
The answer is 6.1 miles
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The rental car agency has 30 cars on the lot. 10 are in great shape, 16 are in good shape, and 4 are in poor shape. Four cars ar
Korolek [52]

Complete Question:

The rental car agency has 30 cars on the lot. 10 are in great shape, 16 are in good shape, and 4 are in poor shape. Four cars are selected at random to be inspected. Do not simplify your answers. Leave in combinatorics form. What is the probability that:

a. Every car selected is in poor shape

b. At least two cars selected are in good shape.

c. Exactly three cars selected are in great shape.

d. Two cars selected are in great shape and two are in good shape.

e. One car selected is in good shape but the other 3 selected are in poor shape.

Answer:

a

   P_A = \frac{^4 C_4}{^{30}C_{4}}

b

  P_B = \frac{[^{16} C_2 *^{14} C_2 ] +[^{16} C_3 *^{14} C_1 ] + ^{16} C_4}{^{30}C_{4}}

c

  P_C = \frac{^{10} C_3 *^{20} C_1 }{^{30}C_{4}}

d

   P_D = \frac{^{10} C_2 *^{16} C_2 }{^{30}C_{4}}

e

   P_E = \frac{^{16} C_1 *^{4} C_3 }{^{30}C_{4}}

Step-by-step explanation:

From the question we are told that

 The number of car in the parking lot is  n =  30  

  The number of cars in great shape is  k =  10

   The number of cars in good shape is r = 16

    The number of cars in poor shape is q = 4

The number of cars that were selected at random is N= 4

Considering question a

    Generally the number of way of selecting four cars that are in a poor shape from number of cars that are in poor shape is

            ^{q} C_{N}

=>        ^{4} C_{4}

Here C stands for combination.

 Generally the number of way of selecting four cars that are in a poor shape from total number of cars  in the parking lot is

          ^{n} C_{N}

=>      ^{30} C_{4}

Generally the probability that every car selected is in poor shape  is mathematically represented as

       P_A = \frac{^4 C_4}{^{30}C_{4}}

Considering question b

Generally the number of way of selecting 2 cars that are in good shape from number of cars that are in good shape is

            ^{r} C_{2}

=>        ^{16} C_{2}

Here C stands for combination.

 Generally the number of way of selecting the remaining 2 cars  from the remaining number of cars  in the parking lot is

          ^{n-r} C_{2}

=>      ^{30-16} C_{2}

=>      ^{14} C_{2}

Generally the number of way of selecting 3 cars that are in good shape from number of cars that are in good shape is

            ^{r} C_{3}

=>        ^{16} C_{3}

 Generally the number of way of selecting the remaining 1 cars  from the remaining number of cars  in the parking lot is

          ^{n-r} C_{1}

=>      ^{30-16} C_{1}

=>      ^{14} C_{1}

Generally the number of way of selecting 4 cars that are in good shape from number of cars that are in good shape is

            ^{r} C_{4}

=>        ^{16} C_{4}

Generally the probability that at least two cars selected are in good shape

       P_B = \frac{[^{16} C_2 *^{14} C_2 ] +[^{16} C_3 *^{14} C_1 ] + ^{16} C_4}{^{30}C_{4}}

Considering question c

Generally the number of way of selecting 3 cars that are in great shape from number of cars that are in great shape is      

            ^{k} C_{3}

=>        ^{10} C_{3}

 Generally the number of way of selecting the remaining 1 cars  from the remaining number of cars  in the parking lot is

          ^{n-k} C_{1}

=>      ^{30-10} C_{1}

=>      ^{20} C_{1}

Generally the probability of selecting exactly three cars selected are in great shape is

        P_C = \frac{^{10} C_3 *^{20} C_1 }{^{30}C_{4}}

Considering question d

Generally the number of way of selecting 2 cars that are in good shape from number of cars that are in good shape is

            ^{r} C_{2}

=>        ^{16} C_{2}

Generally the number of way of selecting 2 cars that are in great shape from number of cars that are in great shape is      

            ^{k} C_{2}

=>        ^{10} C_{2}

Generally the probability that two cars selected are in great shape and two are in good shape.

              P_D = \frac{^{10} C_2 *^{16} C_2 }{^{30}C_{4}}

Considering question e

Generally the number of way of selecting 1 cars that is in good shape from number of cars that are in good shape is

            ^{r} C_{1}

=>        ^{16} C_{1}

    Generally the number of way of selecting 3 cars that are in a poor shape from number of cars that are in poor shape is

            ^{q} C_{3}

=>        ^{4} C_{3}

Generally the probability that one car selected is in good shape but the other 3 selected are in poor shape is

         P_E = \frac{^{16} C_1 *^{4} C_3 }{^{30}C_{4}}

4 0
3 years ago
16
mezya [45]
The answer is the last one
8 0
2 years ago
There were 6 pizza pies at the summer barbeque at Abby's House. She shares the pizza with her 7 friends. If all 8 people get the
vodka [1.7K]

6 pies, 8 people

6 pies divided by 8 people

so 6/8

that is 0.75 or 3/4 or 75%

8 0
3 years ago
11 divided by 6 please help
Misha Larkins [42]

Answer:

1.83 (the 3 keeps going so I simplified it)

Step-by-step explanation:

11/6 = 1.8333333333 (keeps going...)

:))

7 0
3 years ago
How many integers less than 50 have exactly 2 distinct prime factors?
Vedmedyk [2.9K]

1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47

So 16 integers.

6 0
3 years ago
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