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klemol [59]
2 years ago
4

A random variable X counts the number of successes in 20 independent trials. The probability that any one trial is unsuccessful

is 0.42. What is the probability of exactly eight successful trials
Mathematics
1 answer:
frutty [35]2 years ago
7 0

Answer:

P(X=8)

And using the probability mass function we got:

P(X=8)=(20C8)(0.58)^8 (1-0.58)^{20-8}=0.0486

Step-by-step explanation:

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=20, p=1-0.42=0.58)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

And we want to find this probability:

P(X=8)

And using the probability mass function we got:

P(X=8)=(20C8)(0.58)^8 (1-0.58)^{20-8}=0.0486

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A principal of $2000 is placed in a savings account at 3% per annum compounded annually. How much is in the account after one ye
natita [175]

Answer:

Step-by-step explanation:

After one year

A=p(1+r/n)^nt

=2000(1+0.03/12)^12*1

=2000(1+0.0025)^12

=2000(1.0025)^12

=2000(1.0304)

=$2060.8

After two-years

A=p(1+r/n)^nt

=2060.8(1+0.03/12)^12*2

=2060.8(1+0.0025)^24

=2060.8(1.0025)^24

=2060.8(1.0618)

=$2188.157

After three years

A=p(1+r/n)^nt

=2188.157(1+0.03/12)^12*3

=2188.157(1+0.0025)^36

=2188.157(1.0025)^36

=2188.157(1.0941)

=$2394.063

8 0
2 years ago
Pllllllllzzzzzzz answer this!!!♥️♥️
Umnica [9.8K]

Answer:

Use the regression calculator to compare the teams’ number of runs with their number of wins.

A 2-column table with 9 rows. Column 1 is labeled R with entries 808, 768, 655, 684, 637, 619, 613, 609, 563. Column 2 is labeled W with entries 93, 94, 66, 81, 86, 75, 61, 69, 55.

What is the y-intercept of the trend line, to the nearest hundredth?

Step-by-step explanation:

6 0
2 years ago
The XO Group Inc. conducted a survey of 13,000 brides and grooms married in the United States and found that the average cost of
slavikrds [6]

Answer:

a) 0.0392

b) 0.4688

c) At least $39,070 to be among the 5% most expensive.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 29858, \sigma = 5600

a. What is the probability that a wedding costs less than $20,000 (to 4 decimals)?

This is the pvalue of Z when X = 20000. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{20000 - 29858}{5600}

Z = -1.76

Z = -1.76 has a pvalue of 0.0392.

So this probability is 0.0392.

b. What is the probability that a wedding costs between $20,000 and $30,000 (to 4 decimals)?

This is the pvalue of Z when X = 30000 subtracted by the pvalue of Z when X = 20000.

X = 30000

Z = \frac{X - \mu}{\sigma}

Z = \frac{30000 - 29858}{5600}

Z = 0.02

Z = 0.02 has a pvalue of 0.5080.

X = 20000

Z = \frac{X - \mu}{\sigma}

Z = \frac{20000 - 29858}{5600}

Z = -1.76

Z = -1.76 has a pvalue of 0.0392.

So this probability is 0.5080 - 0.0392 = 0.4688

c. For a wedding to be among the 5% most expensive, how much would it have to cost (to the nearest whole number)?

This is the value of X when Z has a pvalue of 0.95. So this is X when Z = 1.645.

Z = \frac{X - \mu}{\sigma}

1.645 = \frac{X - 29858}{5600}

X - 29858 = 5600*1.645

X = 39070

The wedding would have to cost at least $39,070 to be among the 5% most expensive.

5 0
2 years ago
The equation that represents the canned goods order is 24x + 64y = 384, where x = number of minutes spent producing fruit cans,
Lubov Fominskaja [6]
So wee need to find x

4 0
2 years ago
Read 2 more answers
Merrill Lynch Securities and Health Care Retirement Inc. are two large employers in downtown Toledo, Ohio. They are considering
ratelena [41]

Answer:

Step-by-step explanation:

We would determine the mean an standard deviation first

Mean = (107 + 92 + 97 + 95 + 105 + 101 + 91 + 99 + 95 + 104)/10 = 98.6

Standard deviation = √(summation(x - mean)/n

n = 10

Summation(x - mean) = (107 - 98.6)^2 + (92 - 98.6)^2 + (97 - 96.6)^2 + (95 - 98.6)^2 + (105 - 98.6)^2 + (101 - 98.6)^2 + (91 - 98.6)^2 + (99 - 98.6)^2 + (95 - 98.6)^2 + (104 - 98.6)^2 = 274

Standard deviation = √(274/10) = 5.23

The confidence interval is used to determine the range of values that could possibly contain a population parameter (population mean)

Confidence interval is written in the form,

(Point estimate - margin of error, Point estimate + margin of error)

The sample mean, x is the point estimate for the population mean.

We will use the t distribution because the sample size is small and the population standard deviation is not given.

Degree of freedom, df = 10 - 1 = 9

α = 1 - 0.99 = 0.01

z α/2 = 0.01/2 = 0.005

The area to the right of z 0.005 is 0.005 and the area to the left of z0.005 is 1 - 0.005 = 0.995. Using the t distribution table,

z = 3.25

Margin of error = z × s/√n

Where

s = sample standard Deviation = 5.23

Margin of error = 3.25 × 5.23/√10 = 5.38

Confidence interval = sample mean(point estimate) ± z × σ/√n

The lower boundary of the confidence interval is

98.6 - 5.38 = 93.22

The upper boundary of the confidence interval is

98.6 + 5.38 = 103.98

We estimate with 99% confidence that the mean weekly child care cost of their employees is between $93.22 and $103.98

Also,

99% of the confidence intervals constructed in this way would contain the true value for the population mean of mean weekly child care cost of their employees.

6 0
2 years ago
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