Solve the equation for x 5x-(4x-1)=2 A 1/9 B -1 C -1/9 D 1

Mathematics

1 answer:

Marrrta [24]3 years ago

7 0

Answer:

Step-by-step explanation:

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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

Plz help me toofree your my only hope

grandymaker [24]

Area of a circle = πr²

Area of a circle = 3.14 x 11² = 379.94 in²

Area painted in blue = 379.94 ÷ 2 = 189.97 in²

Area painted in purple = 189.97 in²

--------------------------
Answer: 189.97 in²
--------------------------

4 0

3 years ago

Order the following from least to greatest. 1/16 .0173 1/7 2.2 -0.25

Dovator [93]

Answer:

1.-0.25 2..0173 3. .1/16 4. 1/7 5.2.2

Step-by-step explanation:

You want to put all numbers in fractions to see which numbers are smallest

5 0

3 years ago

T-2(3-2t)=2t+9 algebra2

nekit [7.7K]