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3 years ago

Plz help! (due tonight!)

Mathematics

1 answer:

Vilka [71]3 years ago

5 0

Answer:

There will be 9 couples, and 3 lonely girls

Hope this helps

plz like and brainly :D

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g If a hypothesis test with a significance level of α rejects H0: μ1=μ2 in favor of Ha: μ1≠μ2, then thecorresponding (1- α)% con

Effectus [21]

Answer:

The statement, (1- α)% confidence interval for (μ₁ - μ₂) does not contain zero is TRUE.

Step-by-step explanation:

The hypothesis for a test is defined as follows:

H₀: μ₁ = μ₂ vs. Hₐ: μ₁ ≠ μ₂

It is provided that the test was rejected st the significance level α%.

If a decision is to made using the confidence interval the conditions are:

If the null hypothesis value is not included in the (1 - α)% confidence interval then the null hypothesis will be rejected and vice versa.

In this case the null hypothesis value is:

H₀: μ₁ - μ₂ = 0.

If the value 0 is not included in the (1 - α)% confidence interval for the difference between two means, then the null hypothesis will be rejected.

Thus the statement, (1- α)% confidence interval for (μ1- μ2) does not contain zero is TRUE.

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Yakvenalex [24]

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A fair coin is tossed three times and the events A, B, and C are defined as follows: A: \{ At least one head is observed \} B: \

Yanka [14]

Answer:

a) $P(A)=0.875$

b) $\text{P(A or B)}=0.875$

c) $\text{P((not A) or B or (not C))}=0.625$

Step-by-step explanation:

Given : A fair coin is tossed three times and the events A, B, and C are defined as follows: A: At least one head is observed, B: At least two heads are observed, C: The number of heads observed is odd.

To find : The following probabilities by summing the probabilities of the appropriate sample points ?

Solution :

The sample space is

S={HHH,HHT,HTT,HTH,TTT,TTH,THH,THT}

n(S)=8

A: At least one head is observed

i.e. A={HHH,HHT,HTT,HTH,TTH,TTH,THH,THT}

n(A)=7

B: At least two heads are observed

i.e. B={HHH,HTT,TTH,THT}

n(B)=4

C: The number of heads observed is odd.

i.e. C={HHH,HTT,THT,TTH}

n(c)=4

a) Probability of A, P(A)

$P(A)=\frac{n(A)}{n(S)}$

$P(A)=\frac{7}{8}$

$P(A)=0.875$

b) P(A or B)

Using formula,

$\text{P(A or B)}=P(A)+P(B)-\text{P(A and B)}$

$\text{P(A or B)}=\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}-\frac{\text{n(A and B)}}{n(S)}$

$\text{P(A or B)}=\frac{7}{8}+\frac{4}{8}-\frac{4}{8}$

$\text{P(A or B)}=\frac{7}{8}$

$\text{P(A or B)}=0.875$

(c) P((not A) or B or (not C))

A={HHH,HHT,HTT,HTH,TTH,TTH,THH,THT}

not A = {TTT} = 1

B={HHH,HTT,TTH,THT}

C={HHH,HTT,THT,TTH}

not C = {HHT,HTH,THH,TTT} = 4

So, not A or B or not C = {HHH,HHT,HTH,THH,TTT}=5

$\text{P((not A) or B or (not C))}=\frac{5}{8}$

$\text{P((not A) or B or (not C))}=0.625$

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3 years ago