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hichkok12 [17]
2 years ago
8

Suppose a 5-digit number is formed using the digits from 1 to 9 (without replacement). What is the probability that it will be a

n even number? ​
Mathematics
1 answer:
TiliK225 [7]2 years ago
8 0

By finding the probability that the unit digit is even, we will see that the probability that the 5-digit number will be even is P = 4/9.

So we have a 5-digit number, let's say that it is:

abcde

Now, we have 5 "slots" where we can put numbers from 1 to 9 (without replacement).

We want to find the probability that the number will be even, so we only care for the last digit, e (it must be even).

Let's assume that the first digit at which we assign its value is this, e, the last one.

For this digit we have 9 options, {1, 2, 3, 4, 5, 6, 7, 8, 9}

4 of these options are even numbers.

So there are 4 out of 9 cases where the <u>last digit is even</u>, then the probability that the 5-digit number will be even is just:

P = 4/9

Now the other four digits can be anything (even or odd), so we only care for the probability above.

If you want to learn more about probability, you can read:

brainly.com/question/23044118

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I need help please with step by step it would help me a lot
pav-90 [236]

Answer:

Domain: (-infinity, +infinity) since you can pick any x values.

Range: [0, +infinity) since it does not go below the x axis.

Step-by-step explanation:

The graph is a parabola given by y=x^2

lets pick a few x values:

x = 1 gives us y = 1^2, which = 1

x = -1 gives us y = (-1)^2, which = 1

The parabola's domain is any x value as it extends to infinity.

For its range, you can see that it does not go below the x axis at x = 0. Therefore, the range of the parabola is from [0, infinity]

4 0
3 years ago
A credit reporting agency claims that the mean credit card debt in a town is greater than $3500. A random sample of the credit c
Lera25 [3.4K]

Answer:

Claim is false

Step-by-step explanation:

Claim : A credit reporting agency claims that the mean credit card debt in a town is greater than $3500.

H_0:\mu \leq 3500\\H_a:\mu = 3500

n = 20

Since n <30

So we will use t test

Formula : t =\frac{x-\mu}{\frac{s}{\sqrt{n}}}

s = standard deviation = 391

x = 3600

n = 20

t =\frac{3600-3500}{\frac{391}{\sqrt{20}}}

t =1.14

Degree of freedom = n-1 = 20-1 = 19

α=0.10

So, using t table

t_({\frac{\alpha}{2},d.f.}) = 1.72

t critical > t calculated

So we accept the null hypothesis

Hence we reject the claim that the mean credit card debt in a town is greater than $3500.

7 0
3 years ago
A data set includes 103 body temperatures of healthy adult humans having a mean of 98.3degreesF and a standard deviation of 0.73
faust18 [17]

Answer:

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher

Step-by-step explanation:

Data provided in the question:

sample size, n = 103

Mean temperature, μ = 98.3

°

Standard deviation, σ = 0.73

Degrees of freedom, df = n - 1 = 102

Now,

For Confidence level of 99%, and df = 102, the t-value = 2.62      [from the standard t table]

Therefore,

CI = (Mean - \frac{t\times\sigma}{\sqrt{n}},Mean + \frac{t\times\sigma}{\sqrt{n}})

Thus,

Lower limit of CI =  (Mean - \frac{t\times\sigma}{\sqrt{n}})

or

Lower limit of CI =  (98.3 - \frac{2.62\times0.73}{\sqrt{103}})

or

Lower limit of CI = 98.11

and,

Upper limit of CI =  (Mean + \frac{t\times\sigma}{\sqrt{n}})

or

Upper limit of CI =  (98.3 + \frac{2.62\times0.73}{\sqrt{103}})

or

Upper limit of CI = 98.49

Hence,

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher and  the mean temperature could very possibly be 98.6°F

7 0
3 years ago
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