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BartSMP [9]
4 years ago
13

How do I do this qn?

Mathematics
1 answer:
denis-greek [22]4 years ago
3 0
You can take the log of the left and right hand side, and then apply the <span>logarithm rules:

log(a</span>ˣ) = x·log(a)
log(ab) = log(a) + log(b)

log(9^(x-1) * 2^(2x+2)) = log(6^(3x))
log(9^(x-1)) + log(2^(2x+2)) = 3x log(6)
(x-1) log(9) + (2x+2) log(2) - 3x log(6) = 0
x(log9 + 2log2 - 3log6) = log9 - 2log2
x = (log9 - 2log2) / (log9 + 2log2 - 3log6)

simplifying by writing log9 = 2log3 and log6 = log2+log3

x= 2(log3 - log2) / (2log3 + 2log2 - 3log2 - 3log3) = 
x= -2(log3 - log2) / (log3 + log2) = -2 log(3/2) / log(6)

So 6^x = 4/9
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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
Mr. Moses baked 11 dozen cookies for his class.
sashaice [31]

Answer:

8 cookies for each student that was present

Step-by-step explanation:

11 total dozens

10 dozens were separated throughout the present students

He has a TOTAL of 19 students but FOUR were absent

19-4=15

a dozen is 12

so 12x10=120

120 divide by 15 = 8

7 0
3 years ago
The percentage method of withholding for federal income tax​ (2003) states that a single person whose weekly​ wages, after subtr
goblinko [34]

The range that the amount withheld vary if the weekly​ wages, after subtracting withholding​ allowances, vary from $600 to $700 ​inclusive is: 76.35<x<101.35.

<h3>Range</h3>

Given

592<x1317

74.35+.25

Hence:

Lowest range

74.35+.25(600-592)

74.35+.25(8)

74.35+2

=76.35

Highest range

74.35+.25(700-592)

74.35+.25(108)

74.35+27

=101.35

Range

76.35<x<101.35

Therefore the range that the amount withheld vary if the weekly​ wages, after subtracting withholding​ allowances, vary from $600 to $700 ​inclusive is: 76.35<x<101.35.

Learn more about range here: brainly.com/question/2264373

#SPJ1

7 0
2 years ago
Question 2 : which characteristic distinguishes ieee 1394 from usb 2.0
Lorico [155]
The characteristics that distinguishes IEEE 1394 from USB 2.0 is that USB supports more devices on a single bus
3 0
3 years ago
Write an expression that shows how to multiply 7×256
rewona [7]
Hello! what I like to do is multiply by 10s and subtract the difference, so one thing you could do is 10x256-(3x256)! your final answer will be 1792, hope this helped! ;)
3 0
3 years ago
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