What is the value represented by the letter E on the box plot of the data?

The relative locations of a swing set, a garden, and a sandbox in Gina's backyard are shown in the diagram.

Archy [21]

Let us suppose the distance between the sandbox and garden is x feet.

Apply the sine law in the given triangle, we have

$\frac{x}{\sin 65^{\circ}}=\frac{36}{\sin 104^{\circ}}\\
\\
\text{On coss multiplying, we get}\\
x\sin 104^{\circ}=36\sin 65^{\circ}\\
\\
x=\frac{36\sin 65^{\circ}}{\sin 104^{\circ}}\\
\\
x\approx 33.6$

Therefore, in the nearest foot, the distance between the sandbox and garden is 34 foot.

8 0

4 years ago

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HELP!!!!!!!!!!!!!!!!!!!!!!!

skad [1K]

Answer:

A D 1 Vertical

Step-by-step explanation:

x always represents 1

5 0

3 years ago

Which number products an irrational number when multiplied by 2 over 5

statuscvo [17]

Answer: π

Ur answer is pie

4 0

3 years ago

You draw one card at random from a standard deck of 52 playing cards. Find the probability that (a) the card is an even-numbered

DerKrebs [107]

Answer:

$P(even )=\frac{20}{52} =\frac{5}{13}$

$P(heart or diamond)=\frac{13}{52} +\frac{13}{52} =\frac{1}{2}$

$P(nine or face card)=\frac{4}{52} +\frac{12}{52} =\frac{4}{13}$

Step-by-step explanation:

You draw one card at random from a standard deck of 52 playing cards

Total cards = 52

5 even numbers in each suit. 4 times 5 = 20 even number cards

$P(even )=\frac{20}{52} =\frac{5}{13}$

(b) the card is a heart or a diamond

there are 13 heart and 13 diamonds

$P(heart or diamond)=\frac{13}{52} +\frac{13}{52} =\frac{1}{2}$

(c) the card is a nine or a face card

There are 4 nine and 12 face cards

$P(nine or face card)=\frac{4}{52} +\frac{12}{52} =\frac{4}{13}$

8 0

3 years ago

A normal distribution curve, where x = 70 and σ = 15, was created by a teacher using her students’ grades. What information abou

mash [69]

Answer:

The median and mode of the students grade is 70.

Most of the students scored between 40 and 100.

Step-by-step explanation:

From the provided information it can be seen that the mean of the distribution is, <em>μ</em> = 70 and the standard deviation is, <em>σ</em> = 15.

For a Normal distributed data the mean, median and mode are the same.

So, the median and mode of the students grade is 70.

The standard deviation of the data represents the spread of the observation, i.e. how dispersed the values are along the curve.

In statistics, the 68–95–99.7 rule, also recognized as the empirical rule, is a shortcut used to recall that 68.27%, 95.45% and 99.73% of the values of a Normally distributed data lie within one, two and three standard deviations of the mean, respectively.

$P(\mu-\sigma$

$P(\mu-2\sigma$

$P(\mu-3\sigma$

Assuming that maximum marks of the exam is 100, it can be said that most of the students scored between 40 and 100.

3 0

3 years ago

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