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kodGreya [7K]
3 years ago
12

Kim rolls a dice and flips a coin. a) calculate the probability that she gets a 2 and a head

Mathematics
1 answer:
gtnhenbr [62]3 years ago
5 0

Kim rolls a dice and flips a coin.

For coin: we have either H or T  

For dice: we have 1, 2, 3, 4, 5 or 6

On a coin, the probability of heads:  \frac{1}{2}

On a dice, the probability of getting 2 : \frac{1}{6}

Hence, the probability of getting a 2 and a head is

\frac{1}{2}\times\frac{1}{6} = \frac{1}{12}

So, the probability that Kim gets a 2 and a head is \frac{1}{12}

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Which statement must be true according to the dot plot?
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*see attachment for the dot plot being referred to

Answer:

The data is symmetric and shows that he typically sent about 6 to 8 text messages per day

Step-by-step explanation:

The distribution of the data set on a dot plot can be said to be symmetric when most of the data points in the data are located or are concentrated at the center of the dot plot.

As we can observe from the given dot plot in the attachment, it shows that 6 to 8 text messages per day have more frequencies and are just right at the center of the dot plot. This shows the data is symmetric.

This also shows that Reza dents averagely 6 - 8 text messages per day. Reza can be said to have typically sent 6 - 8 text messages per day.

The rest statements about the dot plot are untrue.

6 0
3 years ago
What is the value of this expression?
mylen [45]

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Answer:

  2

Step-by-step explanation:

I find it convenient to remember that a logarithm is an exponent.

  \log_a(a^b)=b

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  log₂(8) +log₃(1/3) = log₂(2³) +log₃(3⁻¹)

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The value of the expression is 2.

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Some calculators can compute this for you directly.

6 0
3 years ago
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Alexeev081 [22]

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3 0
3 years ago
On a test with a population mean of 75 and standard deviation equal to 16, if the scores are normally distributed, what percenta
sashaice [31]

Answer:

Percentage of scores that fall between 70 and 80 = 24.34%

Step-by-step explanation:

We are given a test with a population mean of 75 and standard deviation equal to 16.

Let X = Percentage of scores

Since, X ~ N(\mu,\sigma^{2})

The z probability is given by;

           Z = \frac{X-\mu}{\sigma} ~ N(0,1)    where, \mu = 75  and  \sigma = 16

So, P(70 < X < 80) = P(X < 80) - P(X <= 70)

P(X < 80) = P( \frac{X-\mu}{\sigma} < \frac{80-75}{16} ) = P(Z < 0.31) = 0.62172

P(X <= 70) = P( \frac{X-\mu}{\sigma} < \frac{70-75}{16} ) = P(Z < -0.31) = 1 - P(Z <= 0.31)

                                              = 1 - 0.62172 = 0.37828

Therefore, P(70 < X < 80) = 0.62172 - 0.37828 = 0.24344 or 24.34%

6 0
4 years ago
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