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Alekssandra [29.7K]
4 years ago
15

Irving cannot remember the correct order of the five digits in his ID number. He does remember that the ID number contains the d

igits 1, 4, 3, 7, 6. What is the probability that the first three digits of Irving's ID number will all be odd numbers?
Mathematics
2 answers:
MakcuM [25]4 years ago
6 0

Answer: The probability is 0.10

Step-by-step explanation:

The ID number has five digits, and the digits can be 1, 4, 3, 7 and 6, and I will assume that each digit appears only once.

Then if we want to calculate the probability that the first 3 digits will be odd is:

Suppose that we have 5 slots, we want that in the first two slots to have odd numbers.

In our set, we have only 3 odd numbers {1, 3 and 7}

Then if we want an odd number in the first digit, we have 3 options

If we want an odd number in the second digit, we have two options (because we already selected one in the first selection)

If we want an odd number in the first digit, we have only one option.

For the fourth digit we have one of the two remaining even options, so we have 2 options.

For the fifth digit, we have only one digit.

The number of combinations is equal to the product of the number of options in each selection:

c = 3*2*1*2*1 = 12

Now, the total number of combinations is:

For the first digit we have 5 options

for the second digit we have 4 options.

for the third digit we have 3 options.

for the fourth digit we have 2 options.

for the fifth digit we have 1 options.

The number of combinations is:

C = 5*4*3*2*1 = 120.

Then the probability that the first 3 digits are odd numbers, is equal to the quotient between number of combination that start with 3 odd digits and the total number of combinations:

P = c/C = 12/120 = 0.10

atroni [7]4 years ago
4 0

Answer: 1/10

Step-by-step explanation:

given:

numbers contained in the i.d

1,4,3,7,6

1. permutations of 5 possible outcome

T = 5

= 5 * 4 * 3 * 2 *1

= 120 times.

2. permutations  of  3 odd numbers

( 1,3 and 7 )

T = 3

= 3 * 2 * 1 * 2 * 1

= 12

probability of of first three digits being odd numbers

P = 12 / 120

= 1 / 10

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Of 580580 samples of seafood purchased from various kinds of food stores in different regions of a country and genetically compa
Lubov Fominskaja [6]

Answer:

a) The 99% confidence interval would be given (0.204;0.296).

b) We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).  

c) No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

Step-by-step explanation:

Part a

Data given and notation  

n=580 represent the random sample taken    

X represent the seafood sold in the country that is mislabeled or misidentified by the people

\hat p=0.25 estimated proportion of seafood sold in the country that is mislabeled or misidentified by the people

\alpha=0.01 represent the significance level (no given, but is assumed)    

p= population proportion of seafood sold in the country that is mislabeled or misidentified by the people

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.58

And replacing into the confidence interval formula we got:

0.25 - 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.204

0.25 + 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.296

And the 99% confidence interval would be given (0.204;0.296).

Part b

We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).  

Part c

A government spokesperson claimed that the sample size was too​ small, relative to the billions of pieces of seafood sold each​ year, to generalize. Is this criticism​ valid?

No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

4 0
4 years ago
Find the percent of each number
iragen [17]
1321 x .03 = 36.93
631 x .45 = 283.95
100 x .01 = 10
350 x .25 = 62.5
7 0
4 years ago
Carolina is mowing lawns for a summer job. For every mowing job, she charges an initial fee plus $6 for each hour of work. Her t
MrRa [10]
Well first the initial fee would be the x, which is what we have to solve for. Each hour of work is $6, and she does four hours of work. This is equal to $32. The equation would look as follows...

x + 6(4) = 32

Once you have the equation start solving. Use the Pemdas method. First you have to multiply, so now the equation looks as follows...

x + 24 = 32

Now you have to get the x by itself. You do that by subtracting 24 from both sides, because what you do to one side, you do to the other. That will get you to the answer of...

x = 8

In conclusion, Carolina has an initial fee of $8.

Hope this helped.
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4 years ago
Avenue A is perpendicular to North Street. What is the relationship between Avenue A and South Street?
MAVERICK [17]
I need more information
6 0
3 years ago
(a) What is the actual cost incurred in producing the 1051st and the 2191st disc? (Round your answers to the nearest cent.)
Evgen [1.6K]

This question is incomplete, the complete question is;

The total weekly cost (in dollars) incurred by Lincoln Records in pressing x compact discs is given by the following function;

C(x) = 2000 + 2x -0.0001x²  (0 ≤ x ≤ 6000) .

(a) What is the actual cost incurred in producing the 1051st and the 2191st disc? (Round your answers to the nearest cent.)

(b) What is the marginal cost when x = 1050 and 2190? (Round your answers to the nearest cent.)

Answer:

a)

Actual cost for 1051st disc is $1.79

Actual cost for 2191st disc is $1.56

b)

the marginal cost are;

C'(1050) = $1.79

C'(2190) =  $1.56

Step-by-step explanation:

Given the data in the question;

C(x) = 2000 + 2x -0.0001x²

a)

What is the actual cost incurred in producing the 1051st and the 2191st disc? (Round your answers to the nearest cent.)

Actual cost for 1051st disc will be;

⇒ C(1051) - C(1050)  

= [ 2000 + 2(1051) - 0.0001×(1051)²] - [ 2000 + 2(1050) - 0.0001×(1050)² ]

= [ 2000 + 2102 - 110.4601 ] - [ 2000 + 2100 - 110.25 =

= 3991.5399 - 3989.75

= $ 1.7899 ≈ $1.79

Actual cost for 2191st disc will be;

⇒ C(2191) - C(2190)  

= [ 2000 + 2(2191) - 0.0001×(2191)²] - [ 2000 + 2(2190) - 0.0001×(2190)² ]

= [ 2000 + 4382 - 480.0481 ] - [ 2000 + 4380 - 479.61 ]

= 5901.9519 - 5900.39

= $ 1.5619 ≈ $1.56

Therefore,

Actual cost for 1051st disc is $1.79

Actual cost for 2191st disc is $1.56

b)

What is the marginal cost when x = 1050 and 2190? (Round your answers to the nearest cent.)

Marginal cost is; C'(x) = 2 - 0.0002x

so;

C'(1050) = 2 - 0.0002(1050) = 2 - 0.21 = $1.79

C'(2190) = 2 - 0.0002(2190) = 2 - 0.438 = $1.562 ≈ $1.56

Therefore; the marginal cost is;

C'(1050) = $1.79

C'(2190) =  $1.56

3 0
3 years ago
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