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Solnce55 [7]
3 years ago
12

Mike put 210 baseballs into 7 trash bins. He put the same number of baseballs in each bin. He took 2 trash bins of baseballs to

the
baseball field. How many baseballs did Mike take?
A. 30
B. 896
C. 90
D. 60
Mathematics
1 answer:
OLEGan [10]3 years ago
4 0

Answer:

D. 60

Step-by-step explanation:

210 baseballs divided evenly between 7 trash bins = 30 balls per bin

2 trash bins multiplied by 30 balls per bin = 60 balls in two bins

Hope this helps!

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grin007 [14]

Answer:

a) z_1 =\frac{20-19}{4} =0.25

z_2 =\frac{15-19}{4} =-1

And we can use the complement rule and we got:

P(z>0.25)=1-P(z

P(z>-1)=1-P(z

And replacing we got:

P(X >20| X>15)= \frac{0.401}{0.841}= 0.477

b) z_1 =\frac{20-19}{4} =0.25

z_2 =\frac{18-19}{4} =-0.25

And we can use the complement rule and we got:

P(z>0.25)=1-P(z

P(z>-0.25)=1-P(z

And replacing we got:

P(X >20| X>18)= \frac{0.401}{0.599}= 0.669

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the length of time waiting to be seated of a population, and for this case we know the distribution for X is given by:

X \sim N(19,4.0)  

Where \mu=19 and \sigma=4.0

Part a

For this cae we want to find this probability:

P(X >20| X>15)

And if we use the conditional probability formula we got:

P(X >20| X>15)= \frac{P(X >20 \cap X>15)}{P(X>15)}=\frac{P(X>20)}{P(X>15)}

We can solve the problem using the z score formula given by:

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

z_1 =\frac{20-19}{4} =0.25

z_2 =\frac{15-19}{4} =-1

And we can use the complement rule and we got:

P(z>0.25)=1-P(z

P(z>-1)=1-P(z

And replacing we got:

P(X >20| X>15)= \frac{0.401}{0.841}= 0.477

Part b

P(X >20| X>18)= \frac{P(X >20 \cap X>18)}{P(X>18)}=\frac{P(X>20)}{P(X>18)}

We can solve the problem using the z score formula given by:

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

z_1 =\frac{20-19}{4} =0.25

z_2 =\frac{18-19}{4} =-0.25

And we can use the complement rule and we got:

P(z>0.25)=1-P(z

P(z>-0.25)=1-P(z

And replacing we got:

P(X >20| X>18)= \frac{0.401}{0.599}= 0.669

5 0
4 years ago
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