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Sholpan [36]
3 years ago
14

The probability of event A is 0. 48, the probability of event A and B is 0. 21, and the probability of events A or B is 0. 89. W

hat is the probability of event B? 0. 09 0. 20 0. 27 0. 62.
Mathematics
1 answer:
Mice21 [21]3 years ago
7 0

Probabilities are used to determine the chances of an event

The probability of event B is 0.62

The probabilities are given as:

\mathbf{P(A) = 0.48}

\mathbf{P(A\ and\ B) = 0.21}

\mathbf{P(A\ or\ B) = 0.89}

To calculate the probability of event B, we make use of the following formula

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

So, we have:

\mathbf{0.21 = 0.48 + P(B) - 0.89}

Collect like terms

\mathbf{P(B)= -0.48  +0.21 + 0.89}

\mathbf{P(B)= 0.62}

Hence, the probability of event B is 0.62

Read more about probabilities at:

brainly.com/question/11234923

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Pls help im not smart enough ;-;
evablogger [386]
PART1:

First Combination:
Pizza ($7) + Chicken Strips ($6) + Biscuits ($3) + Grapes ($4) = $20

Second Combination:
Dog Food ($13) + Bread ($3) + Crackers ($2) + Broccoli ($2) = $20

Third Combination:
Shampoo ($4) + Tissues ($3) + Pizza ($7) + Eggs ($3) + Biscuits ($3) = $20


PART 2:

First Combination:
$7.20 + $5.70 + $2.90 + $3.70 = $19.60
No, I wouldn’t have gone over the limit

Second Combination:
$13.40 + $3.50 + $2.00 + $1.90 = $20.80
Yes, I would have gone over the limit

Third Combination:
$3.50 + $2.60 + $7.20 + $2.50 + $2.90 = $18.70
No, I wouldn’t have gone over the limit

Hope this helps!!



7 0
3 years ago
Assuming that the sample mean carapace length is greater than 3.39 inches, what is the probability that the sample mean carapace
joja [24]

Answer:

The answer is "".

Step-by-step explanation:

Please find the complete question in the attached file.

We select a sample size n from the confidence interval with the mean \muand default \sigma, then the mean take seriously given as the straight line with a z score given by the confidence interval

\mu=3.87\\\\\sigma=2.01\\\\n=110\\\\

Using formula:

z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}  

The probability that perhaps the mean shells length of the sample is over 4.03 pounds is

P(X>4.03)=P(z>\frac{4.03-3.87}{\frac{2.01}{\sqrt{110}}})=P(z>0.8349)

Now, we utilize z to get the likelihood, and we use the Excel function for a more exact distribution

=\textup{NORM.S.DIST(0.8349,TRUE)}\\\\P(z

the required probability: P(z>0.8349)=1-P(z

4 0
2 years ago
The fox population in a certain region has a continuous growth rate of 7 percent per year. It is estimated that the population i
rjkz [21]

Answer:

P(t) = 14300e^0.07t

Step-by-step explanation:

Let :

Population as a function of years, t = P(t) ;

Growth rate, r = 7%

Estimated population on year 2000 = Initial population = 14300

The given scenario can be modeled using an exponential function as the change in population is based in a certain percentage increase per period.

P(t) = Initial population*e^rt

P(t) = 14300*e^(0.07t)

P(t) = 14300e^0.07t

Where, t = number of years after year 2000.

7 0
2 years ago
What's the lowest common multiple of of 2.2 and 5
alexira [117]
Hey there, just multiply 2.2 by 5, 2.2×5=11. Therefore, the lowest common multiple of 2.2 and 5 is 11.
8 0
3 years ago
I need help please. Its 2 parts.
Bezzdna [24]
18 is 6+n. The other one is 60 because n=60 and 60+6=66. 66 represents Chang's height.
4 0
3 years ago
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