Choose what the expressions below best represent within the context of the word problem.

SAT verbal scores are normally distributed with a mean of 433 and a standard deviation of 90. Use the Empirical Rule to determin

laila [671]

34% of the scores lie between 433 and 523.

Solution:

Given data:

Mean (μ) = 433

Standard deviation (σ) = 90

<u>Empirical rule to determine the percent:</u>

(1) About 68% of all the values lie within 1 standard deviation of the mean.

(2) About 95% of all the values lie within 2 standard deviations of the mean.

(3) About 99.7% of all the values lie within 3 standard deviations of the mean.

$$Z(X)=\frac{x-\mu}{\sigma}$

$$Z(433)=\frac{433-\ 433}{90}=0$

$$Z(523)=\frac{523-\ 433}{90}=1$

Z lies between o and 1.

P(433 < x < 523) = P(0 < Z < 1)

μ = 433 and μ + σ = 433 + 90 = 523

Using empirical rule, about 68% of all the values lie within 1 standard deviation of the mean.

i. e. $((\mu-\sigma) \ \text{to} \ (\mu+\sigma))=68\%$

Here μ to μ + σ = $\frac{68\%}{2} =34\%$

Hence 34% of the scores lie between 433 and 523.

8 0

3 years ago

A garden box has a perimeter of 18½ feet. If the length is 5 feet, what is the area of the garden box?

Travka [436]

Answer:

$area=\frac{85}{4}feet^{2}$

Step-by-step explanation:

1. Approach

Since it is given that the garden box is a rectangle, then the opposite sides are congruent. One can use this to their advantage, by setting up an equation that enables them to solve for the width of the rectangle. After doing so, one will multiply the width by the given length and solve for the area.

2. Solve for the width

It is given that the garden box is a rectangle. As per its definition, opposite sides in a rectangle are congruent. The problem gives the length and the perimeter of the rectangle, therefore, one can set up an equation and solve for the width.

$perimeter=18\frac{1}{2}\\\\length=5$

$2(length+width)=perimeter$

Substitute,

$2(5+width)=18\frac{1}{2}$

Conver the mixed number to an improper fraction. This can be done by multiplying the "number" part of the mixed number by the denominator of the fraction. Then add the result to the numerator.

$2(5+width)=\frac{37}{2}$

Inverse operations,

$2(5+width)=\frac{37}{2}\\/2\\\\5+width=\frac{37}{4}\\-5\\\\width=\frac{17}{4}$

3. Solve for the area

Now that one has solved for the width of the box, one must solve for the area. This can be done by multiplying the length by the width. Since the width is a fraction, one must remember, that when multiplying an integer by a fraction, one will multiply the integer by the numerator (the top of the fraction), and then simplify by reducing the fraction, if possible. Reducing the fraction is when one divides both the numerator and the denominator by the GCF (Greatest Common Factor).

$length=5\\\\width = \frac{17}{4}$