Han divided 3 10 by 6. His work is shown below. Is han's answer correct? 3 10 ÷ 6 = 6 10 ÷ 6 = 1 10

Can someone help fill in the gaps in the table and calculate an estimate for the mean

Mumz [18]

Answer:

Step-by-step explanation:

5x3= 15

8x9=72

12x11=132

14x17=138

9x17=153

15+72+132+138+153=510

510/50=10.2= mean

7 0

3 years ago

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

The Blair family was one of the first to come to the original 13 colonies (now part of the USA). They had 4 children. Assuming t

Mumz [18]

Answer:

31.25% probability that the Blair family had at least 3 girls

Step-by-step explanation:

For each children, there are only two possible outcomes. Either it was a girl, or it was not. The probability of a child being a girl is independent of any other children. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

They had 4 children.

This means that $n = 4$

The probability of a child being a girl is 0.5

This means that $p = 0.5$

Probability of at least 3 children:

$P(X \geq 3) = P(X = 3) + P(X = 4)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{4,3}.(0.5)^{3}.(0.5)^{1} = 0.25$

$P(X = 4) = C_{4,4}.(0.5)^{4}.(0.5)^{0} = 0.0625$

$P(X \geq 3) = P(X = 3) + P(X = 4) = 0.25 + 0.0625 = 0.3125$

31.25% probability that the Blair family had at least 3 girls

3 0

3 years ago

What is 2 times 2 times 2 times 2?

kolezko [41]

The answer 16 (sixteen)

7 0

3 years ago

Read 2 more answers

Solve for X:

saveliy_v [14]

$- \frac{6}{5} x - 4 = - 46 \\ - \frac{6}{5} x = - 42 \\ - 6x = - 210 \\ x = 35$
Translate the words into math and solve for x. Any questions let me know.

3 0

3 years ago