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A statistics content developer at Aplia wanted to know whether study skills are related to memory quality. She invited student v

nekit [7.7K]

Answer:

$\begin{array}{cccc}{Score\ Interval} & {f} & {Proportion} & {Percentage} & {9-10} & {29} &{0.19} & {19\%} & {7-8} & {53} & {0.34} & {34\%} &{5-6}& {50} & {0.32} & {32\%} &{3-4} & {22} & {0.14} & {14\%} & {1-2} & {1} & {0.01} & {1\%} \ \end{array}$

Step-by-step explanation:

Given

$\begin{array}{cccc}{Score\ Interval} & {f} & {Proportion} & {Percentage} & {9-10} & {29} &{0.19} & {19\%} & {7-8} & {53} & {} & {} &{5-6}& {50} & {} & {} &{3-4} & {22} & {0.14} & {14\%} & {1-2} & {1} & {0.01} & {1\%} \ \end{array}$

Required

Fill in the gaps

First, we calculate the total frequency:

$Total = \sum f$

$Total = 29+53+50+22+1$

$Total = 155$

The proportion (p) is calculated by:

$p = \frac{f}{Total}$

The percentage (P) is calculated by:

$P = p * 100\%$

For interval 7 - 8, we have:

$p = \frac{53}{155} = 0.34$

$P = 0.34 * 100\% = 34\%$

For interval 5 - 6, we have:

$p = \frac{50}{155} = 0.32$

$P = 0.32 * 100\% = 32\%$

So, the complete table is:

$\begin{array}{cccc}{Score\ Interval} & {f} & {Proportion} & {Percentage} & {9-10} & {29} &{0.19} & {19\%} & {7-8} & {53} & {0.34} & {34\%} &{5-6}& {50} & {0.32} & {32\%} &{3-4} & {22} & {0.14} & {14\%} & {1-2} & {1} & {0.01} & {1\%} \ \end{array}$

6 0

3 years ago

Express as a trinomial.<br> (x + 4)(x - 1)

katovenus [111]

Answer:

x^2+3x−4

Step-by-step explanation:

8 0

3 years ago

Find the retail price of a $24 hat with a markup of 40%

Amanda [17]

Retail price for a 24 dollar hat with a markup of 40% =

24 * .40 = 9.6

24 + 9.6 = $33.60 dollars

5 0

3 years ago

You are testing the claim that the mean GPA of night students is different from the mean GPA of day students. You sample 30 nigh

kirill [66]

Answer:

Step-by-step explanation:

This is a test of 2 independent groups. The population standard deviations are not known. Let μ1 be the mean GPA of night students and μ2 be the mean GPA of day students.

The random variable is μ1 - μ2 = difference in the mean GPA of night students and the mean GPA of day students.

We would set up the hypothesis.

The null hypothesis is

H0 : μ1 = μ2 H0 : μ1 - μ2 = 0

The alternative hypothesis is

H1 : μ1 ≠ μ2 H1 : μ1 - μ2 ≠ 0

This is a two tailed test.

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(x1 - x2)/√(s1²/n1 + s2²/n2)

From the information given,

μ1 = 2.35

μ2 = 2.58

s1 = 0.46

s2 = 0.47

n1 = 30

n2 = 25

t = (2.35 - 2.58)/√(0.46²/30 + 0.47²/25)

t = - 1.8246

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [0.46²/30 + 0.47²/25]²/[(1/30 - 1)(0.46²/30)² + (1/25 - 1)(0.47²/25)²] = 0.00025247091/0.00000496862

df = 51

We would determine the probability value from the t test calculator. It becomes

p value = 0.0746

Since alpha, 0.05 < than the p value, 0.0746, then we would fail to reject the null hypothesis.

3 0

3 years ago

3x = 12 isolates the variable on one side?

lorasvet [3.4K]

First, Divide by 3 on both sides.
So 3x=12 will turn to 3x/3=12/3
Which isolates the variable x and the answer: x= 4

3 0

3 years ago