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Gelneren [198K]

3 years ago

7

If a student answer 67 exam questions correctly out of 100,what fraction and what personage of question did the students answers

incorrectly.
A. 67/100
B.67%
C. 0.67
D. 30/100
E. 33 %
F. 33/100

2 answers:

katovenus [111]3 years ago

6 0

I think it’s A and B

VikaD [51]3 years ago

5 0

Answer:

A and D

Step-by-step explanation:

67 right out of 100 for the fraction and 67/100 = 67%

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3 years ago

What percent of 2b is: 0.04b; 0.2b; 0.56b; 1.8b; 2.5b; 3b; By what percent are they each larger or smaller than 2b?

Marina86 [1]

Answer:

1. 0.04b. <u>0.04b is 2% of 2b. It is smaller than 2b by 98%.</u>

2. 0.2b. <u>0.2b is 10% of 2b. It is smaller than 2b by 90%.</u>

3. 0.56b. <u>0.56b is 28% of 2b. It is smaller than 2b by 72%.</u>

<u>4. </u>1.8b. <u>1.8b is 90% of 2b. It is smaller than 2b by 10%.</u>

<u>5. </u>2.5b. <u>2.5b is 125% of 2b. It is larger than 2b by 25%.</u>

<u>6. 3b. 3b is 150% of 2b. It is larger than 2b by 50%.</u>

Step-by-step explanation:

1. 0.04b

<u>For all the cases, we will use Rule of Three, this way.</u>

Percentage * 2 = 100 * 0.04

Percentage = (100 * 0.04)/2 (Dividing by 2 at both sides)

Percentage = 4/2 = 2%

<u>0.04b is 2% of 2b. It is 98% smaller than 2b.</u>

<u>For all cases, we will use (100 - x) for calculating how smaller or larger the expression is related to 2b, where x is the percentage of the expression itself). In this case, 100 - 2 = 98.</u>

2. 0.2b

Again, we will use Rule of Three, this way:

Percentage * 2 = 100 * 0.2

Percentage = (100 * 0.2)/2 (Dividing by 2 at both sides)

Percentage = 20/2 = 10%

<u>0.2b is 10% of 2b. It is 90% smaller than 2b.</u>

3. 0.56b

Again, we will use Rule of Three, this way:

Percentage * 2 = 100 * 0.56

Percentage = (100 * 0.56)/2 (Dividing by 2 at both sides)

Percentage = 56/2 = 28%

<u>0.56b is 28% of 2b. It is 72% smaller than 2b.</u>

<u>4. </u>1.8b

Again, we will use Rule of Three, this way:

Percentage * 2 = 100 * 1.8

Percentage = (100 * 1.8)/2 (Dividing by 2 at both sides)

Percentage = 180/2 = 90%

<u>1.8b is 90% of 2b. It is 10% smaller than 2b.</u>

5. 2.5b

Again, we will use Rule of Three, this way:

Percentage * 2 = 100 * 2.5

Percentage = (100 * 2.5)/2 (Dividing by 2 at both sides)

Percentage = 250/2 = 125%

<u>2.5b is 125% of 2b. It is 25% larger than 2b.</u>

<u>In this case, 100 - 125 = -25. It means the second expression is larger than 2b by 25%.</u>

<u>6. </u>3b.

Again, we will use Rule of Three, this way:

Percentage * 2 = 100 * 3

Percentage = (100 * 3)/2 (Dividing by 2 at both sides)

Percentage = 300/2 = 150%

<u>3b is 150% of 2b. It is 50% larger than 2b.</u>

<u>In this case, 100 - 150 = -505. It means the second expression is larger than 2b by 50%.</u>

6 0

3 years ago

Location is known to affect the number, of a particular item, sold by an auto parts facility. Two different locations, A and B,

Mama L [17]

We have two samples, A and B, so we need to construct a 2 Samp T Int using this formula:

$\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$

In order to use t*, we need to check conditions for using a t-distribution first.

Random for both samples -- NOT STATED in the problem ∴ <u><em>proceed with caution</em></u>!
Independence for both samples: 130 < all items sold at Location A; 180 < all items sold at Location B -- we can reasonably assume this is true
Normality: CLT is not met; <u>n < 30</u> for both locations A and B ∴ <u><em>proceed with caution</em></u>!

<u>Since 2/3 conditions aren't met, we can still proceed with the problem but keep in mind that the results will not be as accurate until more data is collected or more information is given in the problem.</u>

<u>Solve for t*:</u>

<u></u>

We need the <u>tail area </u>first.

$\displaystyle \frac{1-.9}{2}= .05$

Next we need the <u>degree of freedom</u>.

The degree of freedom can be found by subtracting the degree of freedom for A and B.

The general formula is df = n - 1.

df for A: 13 - 1 = 12
df for B: 18 - 1 = 17
df for A - B: |12 - 17| = 5

Use a calculator or a t-table to find the corresponding <u>t-score for df = 5 and tail area = .05</u>.

t* = -2.015

Now we can use the formula at the very top to construct a confidence interval for two sample means.

$\overline {x}_A=39$
$s_A=8$
$n_A=13$
$\overline {x}_B = 55$
$s_B=2$
$n_B=18$
$t^{*}=-2.015$

Substitute the variables into the formula: $\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$ .

$39-55 \ \pm \ -2.015 \big{(}\sqrt{\frac{(8)^2}{13} +\frac{(2)^2}{18} } } \ \big{)}$

Simplify this expression.

$-16 \ \pm \ -2.015 (\sqrt{5.1453} \ )$
$-16 \ \pm \ 3.73139$

Adding and subtracting 3.73139 to and from -16 gives us a confidence interval of:

$(-20.5707,-11.4293)$

If we want to <u>interpret</u> the confidence interval of (-20.5707, -11.4293), we can say...

<u><em>We are 90% confident that the interval from -20.5707 to -11.4293 holds the true mean of items sold at locations A and B.</em></u>

5 0

1 year ago