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Ostrovityanka [42]
3 years ago
9

The probability that an egg on a production line is cracked is 0.01. Two eggs are selected at random from the production line. F

ind the probability that at least one egg is cracked. Write the entire decimal answer.
Mathematics
1 answer:
musickatia [10]3 years ago
5 0

Answer:

P(X \geq 1)=1-P(X

P(X=0)=(2C0)(0.01)^0 (1-0.01)^{2-0}=0.9801

And replacing we got:

P(X \geq 1) = 1-0.9801 = 0.0199

Step-by-step explanation:

Let X the random variable of interest "number of craked eggs", on this case we now that:

X \sim Binom(n=2, p=0.01)

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 \geq 1)=1-P(X

And we can find the probability:

P(X=0)=(2C0)(0.01)^0 (1-0.01)^{2-0}=0.9801

And replacing we got:

P(X \geq 1) = 1-0.9801 = 0.0199

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faltersainse [42]

Answer:

x=15

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Step-by-step explanation:

<em><u>Converse:</u></em>

Let's start with the converse, so if line x and y are parallel then angles on the same line are supplementary angles. I'm not sure if that's the converse your teacher wants but my teacher taught me this.

<em><u>Finding x:</u></em>

Now we know that the angles are supplementary angles, that means if w add them together we get 180°. So our equation will be 10x-22+3x+7=180. Now we just solve it. We start by combining like terms, then adding the 15 to both sides, and lastly we divide the 13 to both sides and get...

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7 0
3 years ago
A team averaging 110 points is likely to do very well during the regular season. The coach of your team has hypothesized that yo
aev [14]

Answer:

1. M=108

2. μ=110

3. In the explanation.

4. Test statistic t = -1.05

5. P-value = 0.1597

Step-by-step explanation:

The question is incomplete: to solve this problem, we need the sample information: size, mean and standard deviation.

We will assume a sample size of 10 matches, a sample mean of 108 points and a sample standard deviation of 6 points.

1. The mean points is the sample points and has a value of 108 points.

2. The null hypothesis is H0: μ=110, meaning that the mean score is not significantly less from 110 points.

3. This is a hypothesis test for the population mean.

The claim is that the mean score is significantly less than 110.

Then, the null and alternative hypothesis are:

H_0: \mu=110\\\\H_a:\mu< 110

The significance level is 0.05.

The sample has a size n=10.

The sample mean is M=108.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=6.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{6}{\sqrt{10}}=1.9

4. Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{108-110}{1.9}=\dfrac{-2}{1.9}=-1.05

The degrees of freedom for this sample size are:

df=n-1=10-1=9

5. This test is a left-tailed test, with 9 degrees of freedom and t=-1.05, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=P(t

As the P-value (0.1597) is bigger than the significance level (0.05), the effect is  not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the mean score is significantly less than 110.

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Answer:

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3 is the rising number 4 is the run

meaning it goes up 3 and 4 to the side.

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