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2 years ago

What is the probability of flipping a coin 10 times and getting heads 5 times? Round your answer to the nearest tenth percent.

1 answer:

6 0

The Probability is 1/2 or .5, .5 rounded to the nearest tenth would be irregular because .5 is already in the tenth place but according to the law of rounding since ut is five it would be rounded up giving the whole number 1

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Make a table of values for each eqution<br> y = |x| + 1

Alisiya [41]

No x-intercept/Zero
y=1

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2 years ago

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Yuki888 [10]

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2 years ago

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Bumek [7]

Answer:

Step-by-step explanation:

.07x1 is .07, therefore tax rate

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2 years ago

(k^2-6k^4)-(3k^4+k^2+2)

jeka57 [31]

Step-by-step explanation:

Vamos simplificar passo a passo.

k2-6k4-(3k4+k2+2)

Distribua o sinal negativo:

=k2-6k4+-1(3k4+k2+2)

=k2+-6k4+-1(3k4)+-1k2+(-1)(2)

=k2+-6k4+-3k4+-k2+-2

Combine os termos semelhantes:

=k2+-6k4+-3k4+-k2+-2

=(-6k4+-3k4)+(k2+-k2)+(-2)

=-9k4+-2

Responda:

=-9k4-2

7 0

3 years ago

Below are survival times (in days) of 13 guinea pigs that were injected with a bacterial infection in a medical study:

tresset_1 [31]

Outliers are data that are relatively far from other data elements.

The dataset has an outlier and the outlier is 120

The dataset is given as:

91 83 84 79 91 93 95 97 97 120 101 105 98

Sort the dataset in ascending order

79 83 84 91 91 93 95 97 97 98 101 105 120

<h3>The lower quartile (Q1)</h3>

The Q1 is then calculated as:

$Q1 = \frac{N +1}{4}th$

So, we have:

$Q1 = \frac{13 +1}{4}th$

$Q1 = \frac{14}{4}th$

$Q1 = 3.5th$

This is the average of the 3rd and the 4th element

$Q1 = \frac{1}{2} \times (84 + 91)$

$Q1 = 87.5$

<h3>The upper quartile (Q3)</h3>

The Q3 is then calculated as:

$Q3 = 3 \times \frac{N +1}{4}th$

So, we have:

$Q3 = 3 \times \frac{13 +1}{4}th$

$Q3 = 3 \times 3.5th$

$Q3 = 10.5th$

This is the average of the 10th and the 11th element.

$Q_3 =\frac12 \times (98 + 101)$

$Q_3 =99.5$

<h3>The interquartile range (IQR)</h3>

The IQR is then calculated as:

$IQR = Q_3 -Q_1$

$IQR = 99.5 - 87.5$

$IQR = 12$

Also, we have:

$IQR(1.5) = 12 \times 1.5$

$IQR(1.5) = 18$

<h3>The outlier range</h3>

The lower and the upper outlier range are calculated as follows:

$Lower = Q_1 - IQR(1.5)$

$Lower = 87.5- 18$

$Lower = 69.5$

$Upper = Q_3 + IQR(1.5)$

$Upper = 99.5 + 18$

$Upper = 117.5$

120 is greater than 117.5.

Hence, the dataset has an outlier and the outlier is 120