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PtichkaEL [24]
3 years ago
10

During a two-hour period, the temperature in a city dropped from a high of 51°F to a low of −11°F. What was the range of the tem

peratures during this period?
Mathematics
1 answer:
Likurg_2 [28]3 years ago
6 0

you have to take 51 plus 11 and add it up. then your answer should be 62 degrees different.

You might be interested in
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
Elizabeth quiere organizar la fiesta de su cumpleaños y pretende invitar a 20 de sus amigos. Para los preparativos de la fiesta
alexandr1967 [171]

no entiendo que estas diciendo

3 0
3 years ago
Which side lengths form a right triangle?
Eva8 [605]

Answer:

abswer is b no

Step-by-step explanation:

hope it help

3 0
3 years ago
HELP ASAP! GIVING BRAINLIEST!
bearhunter [10]

Answer: Approximately 19.18 Cubic Centimeters

Step-by-step explanation:

V=πr^2h/3

=π·2^2·4.58/3

≈19.18466

Possibly answer B?

5 0
4 years ago
Find the value of x
iVinArrow [24]

Answer:

x = 121

Step-by-step explanation:

Let the measure of the altitude of the given triangle be h units.

By geometric mean theorem:

h =  \sqrt{23 \times x}  \\  \\ h =  \sqrt{23x}  \\  \\ now \: by \: pythagors \: theorem \\  \\  {(x + 11)}^{2}  =  {h}^{2}  +  {x}^{2}  \\  \\   \cancel{{x}^{2}}  + 22x + 121 =  {( \sqrt{23x} )}^{2}  +  \cancel{{x}^{2}} \\  \\ 22x + 121 = 23x \\  \\ 22x - 23x =  - 121 \\  \\  - x =  - 121 \\  \\ x = 121

4 0
3 years ago
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