A) X - 12x + 32 b) x = 14x + 48 c) x

The equation a = $\frac{1}{2}(b_1 + b_2 )h$ can be used to determine the area , a, of a trapezoid with height , h, and base lengths, $b_1$ and $b_2$ . Which are equivalent equations?

(a) $\frac{2a}{h} - b_2 = b_1$

(b) $\frac{a}{2h} - b_2 = b_1$

(c) $\frac{2a - b_2}{h}$ = $b_1$

(d) $\frac{2a}{b_1 + b_2} = h$

(e) $\frac{a}{2(b_1 + b_2)}$ = h

Answer: (a) $\frac{2a}{h} - b_2 = b_1$ ; (d) $\frac{2a}{b_1 + b_2} = h$ ;

Step-by-step explanation: To determine $b_1$ :

a = $\frac{1}{2}(b_1 + b_2 )h$

2a = ( $b_1 + b_2$ )h

$\frac{2a}{h} = b_1 + b_2$

$\frac{2a}{h} - b_2 = b_1$

To determine h:

a = $\frac{1}{2}(b_1 + b_2 )h$

2a = $(b_1 + b_2)h$

$\frac{2a}{(b_1 + b_2)}$ = h

To determine $b_2$

a = $\frac{1}{2}(b_1 + b_2 )h$

2a = $(b_1 + b_2)h$

$\frac{2a}{h} = (b_1 + b_2)$

$\frac{2a}{h} - b_1 = b_2$

Checking the alternatives, you have that $\frac{2a}{h} - b_2 = b_1$ and $\frac{2a}{(b_1 + b_2)}$ = h, so alternatives A and D are correct.

4 0

3 years ago

Slices of pizza for a certain brand of pizza have a mass that is approximately normally distributed with a mean of 67.7 grams an

Alexxx [7]

Answer:

a. s.e. = 0.338

b. The probability of finding a random slice of pizza with a mass of less than 67.2 grams is 0.5871 or 58.71%

c. The probability of finding a 20 random slices of pizza with a mean mass of less than 67.2 grams is 0.0694 or 6.94%

d. The sample mean of 62.75 would represent the 15th percentile

Step-by-step explanation:

1. Let's review the information provided to us to answer the question properly:

μ of the mass of slices of pizza for a certain brand = 67.7 grams

σ of the mass of slices of pizza for a certain brand = 2.28 grams

2. For samples of size 20 pizza slices, what is the standard deviation for the sampling distribution of the sample mean?

Let's recall that the standard deviation of the sampling distribution of the mean is called the standard error of the mean and its formula is:

μσs.e.= √σ/n, where n is the sample size.

s.e. = √2.28/20

s.e. = 0.338

3. What is the probability of finding a random slice of pizza with a mass of less than 67.2 grams?

Let's find the z-score for X = 67.2, this way:

z-score = (X - μ)/σ

z-score = (67.2 - 67.7)/2.28

z-score = 0.5/2.28

z-score = 0.22 (rounding to the next hundredth)

Now, using the z-table, let's find p, the probability:

p (z = 0.22) = 0.5871

The probability of finding a random slice of pizza with a mass of less than 67.2 grams is 58.71%

4. What is the probability of finding a 20 random slices of pizza with a mean mass of less than 67.2 grams?

Let's use the central limit theorem to find the z-score, this way:

z-score = X - μ/s.e

z-score = 67.2 - 67.7/0.338

z-score = -0.5/0.338

z-score = - 1.48

Now, using the z-table, let's find p, the probability:

p (z = -1.48) = 0.0694

5. What sample mean (for a sample of size 20) would represent the bottom 15% (the 15th percentile)?

For p = 0.150, let's find the z-score:

z-score = - 2.17

z-score = X - μ/s.e

- 2.17 = (X - 67.7)/2.28

- 4.95 = X - 67.7

X = 67.7 - 4.95

X = 62.75 (rounding to the next hundredth)

The sample mean of 62.75 would represent the 15th percentile

7 0

3 years ago

A) X - 12x + 32b) x = 14x + 48c) x - 3x + 2​

A) X - 12x + 32b) x = 14x + 48c) x - 3x + 2