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grigory [225]
3 years ago
13

The probability of a dry summer is equal to 0.3, the probability of a wet summer is equal to 0.2, and the probability of a summe

r with normal precipitation is equal to 0.5.
what is the probability of observing two dry summers?
what is the probability of observing at least 2 dry summers?
what is the probability of not observing a wet summer?
Mathematics
1 answer:
brilliants [131]3 years ago
4 0

Answer:

s

Step-by-step explanation:

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What happens because a bat applies a force on a ball?​
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Newton's 3rd law states that for every action, there is an equal and opposite reaction.

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Calculate how many days there in 32 weeks​
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2 years ago
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Which statement(s) is (are) correct?
Anna71 [15]

Answer:

<em>statements 3 and 4 are correct.</em>

Step-by-step explanation:

(1)

The probability of choosing cured pasta and bear= probability that the card is king.

Hence, The probability of choosing cured pasta and bear=\dfrac{4}{52}=\dfrac{1}{13}

Probability of choosing baked cucumber and lime mutton=probability that the card is 3.

as there are 4 cards that are '3'.

Hence Probability of choosing baked cucumber and lime mutton=\dfrac{4}{52}=\dfrac{1}{13}

as both the probabilities are equal.

Hence statement 1 is incorrect.

(2)

The probability of choosing gooseberry & passion fruit cheesecake= Probability taht the card is ace.

as there are 4 cards which are ace out of 52 cards.

Hence, The probability of choosing gooseberry & passion fruit cheesecake=\dfrac{4}{52}=\dfrac{1}{13}

probability of choosing poached fennel & lemon alligator=Probability that the card is a face card.

As there are 12 face cards out of 52 cards.

Hence, probability of choosing poached fennel & lemon alligator=\dfrac{12}{52}=\dfrac{3}{13}

Hence, the probability of choosing gooseberry and passion fruit cheesecake is smaller than the probability of choosing poached fennel & lemon alligator.

Hence statement 2 is false.

(3)

The probability of choosing a praline wafer=probability that the card is a diamond.

as there are 13 diamond cards out of 52 cards.

The probability of choosing a praline wafer=\dfrac{13}{52}=\dfrac{1}{4}

the probability of choosing poached fennel & lemon alligator=Probability that the card is a face card.

As there are 12 face cards out of 52 cards.

Hence, probability of choosing poached fennel & lemon alligator=\dfrac{12}{52}=\dfrac{3}{13}

Hence, The probability of choosing a praline wafer is greater than the probability of choosing poached fennel & lemon alligator.

Hence statement 3 is correct.

(4)

The probability of choosing pressure-cooked mushroom & garlic chicken =probability that the card is red.

As there are 26 red cards out of 52 cards.

Hence,  

The probability of choosing pressure-cooked mushroom & garlic chicken =\dfrac{26}{52}=\dfrac{1}{2}

probability of choosing an oven-baked apple & lavender calzone =probability that the card is black.

As there are 26 red cards out of 52 cards.

Hence,  probability of choosing an oven-baked apple & lavender calzone=\dfrac{26}{52}=\dfrac{1}{2}

Hence, The probability of choosing pressure-cooked mushroom & garlic chicken and the probability of choosing an oven-baked apple & lavender calzone are the same.

Hence statement 4 is true.

(5)

The probability of choosing pressure-cooked mushroom & garlic chicken =probability that the card is red.

As there are 26 red cards out of 52 cards.

Hence,  

The probability of choosing pressure-cooked mushroom & garlic chicken =\dfrac{26}{52}=\dfrac{1}{2}

the probability of choosing a praline wafer=probability that the card is a diamond.

as there are 13 diamond cards out of 52 cards.

The probability of choosing a praline wafer=\dfrac{13}{52}=\dfrac{1}{4}

Hence, the probability of choosing pressure-cooked mushroom & garlic chicken and the probability of choosing a praline wafer are not same.

Hence, statement 5 is not correct.


6 0
3 years ago
For many years businesses have struggled with the rising cost of health care. But recently, the increases have slowed due to les
kaheart [24]

Answer:

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 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

The margin of error for this case is given by:

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

And replacing we got:

ME = 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.0259

And replacing into the confidence interval formula we got:

0.52 - 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.4941

0.52 + 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.5459

And the 95% confidence interval would be given (0.4941;0.5459).

Step-by-step explanation:

Data given and notation  

n=1000 represent the random sample taken    

\hat p=0.52 estimated proportion of of U.S. employers were likely to require higher employee contributions for health care coverage

\alpha=0.05 represent the significance level (no given, but is assumed)    

Solution to the problem

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 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

The margin of error for this case is given by:

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

And replacing we got:

ME = 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.0259

And replacing into the confidence interval formula we got:

0.52 - 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.4941

0.52 + 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.5459

And the 95% confidence interval would be given (0.4941;0.5459).

5 0
3 years ago
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