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3 years ago

What is the average height of a 13 year old boy?

Mathematics

1 answer:

olga2289 [7]3 years ago

7 0

62 inches / 175 cm, plus or minus about 2.5 inches (6.5 cm).c

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The number of seeds found in a sample of 7 different oranges were: 3, 3, 4, 5, 7, 8, and 12. What is the range of the number of

marusya05 [52]

Answer:

Step-by-step explanation:

Range is Biggest Number - Smallest Number.

12 is the biggest number and 3 is the smallest number.

So, 12 - 3 = 9

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2 years ago

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40(x-2)=2/5(x+4)<br><br><br> Please help

kherson [118]

Answer:

x = 68/33

Explanation:

Step 1: Simplify both sides of the equation.

Step 2: Subtract 2/5x from both sides.

Step 3: Add 80 to both sides.

Step 4: Multiply both sides by 5/198.

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3 years ago

The teacher still wont help

forsale [732]

Answer:

18 licks per lollipop

Step-by-step explanation:

54 licks / 3 lollipops

2. 2 lollipops

3. 90 licks

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3 years ago

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the entry ticket for a baseball game costs $5 for an adult ticket and $3 for a child.A group of 20 people paid total entry fees

sesenic [268]

There would be 15 adults and 5 children in that group.
15 adults = $75
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4 years ago

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Suppose that you were not given the sample mean and sample standard deviation and instead you were given a list of data for the

Troyanec [42]

Answer:

So, the sample mean is 31.3.

So, the sample standard deviation is 6.98.

Step-by-step explanation:

We have a list of data for the speeds (in miles per hour) of the 20 vehicles. So, N=20.

We calculate the sample mean :

$\mu=\frac{19 +19 +22 +24 +25 +27 +28+ 37 +35 +30+ 37+ 36+ 39+ 40+ 43+ 30+ 31+ 36+ 33+ 35}{20}\\\\\mu=\frac{626}{20}\\\\\mu=31.3$

So, the sample mean is 31.3.

We use the formula for a sample standard deviation:

$\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_i-\mu)^2}$

Now, we calculate the sum

$\sum_{i=1}^{20}(x_i-31.3)^2=(19-31.3)^2+(19-31.3)^2+(22-31.3)^2+(24-31.3)^2+(25-31.3)^2+(27-31.3)^2+(28-31.3)^2+(37-31.3)^2+(35-31.3)^2+(30-31.3)^2+(37-31.3)^2+(36-31.3)^2+(39-31.3)^2+(40-31.3)^2+(43-31.3)^2+(30-31.3)^2+(31-31.3)^2+(36-31.3)^2+(33-31.3)^2+(35-31.3)^2\\\\\sum_{i=1}^{20}(x_i-31.3})^2=926.2\\$

Therefore, we get

$\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_i-\mu)^2}\\\\\sigma=\sqrt{\frac{1}{19}\cdot926.2}\\\\\sigma=6.98$

So, the sample standard deviation is 6.98.

3 0

3 years ago