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1 year ago

Stem and left test score help (B help needed only)

Mathematics

1 answer:

marusya05 [52]1 year ago

4 0

Given:

6 | 1 means a score of 61.

The scores of the students from the stem-and-leave are

61, 62, 63, 71, 73, 75, 77, 78, 78, 79, 82, 82, 84, 85, 89, 95, 96, and 99.

Required:

We need to find the lowest score in the 80s, the highest score overall, and the number of students who scored in the 70s.

Explanation:

The scores in the 80s are

82, 82, 84, 85, and 89.

The smallest score is 82.

The lowest score in the 80s is 82.

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The frequency table was made using a box containing slips of paper. Each slip of paper was numbered 0, 1, 2, 3, or 4.

inna [77]

In order to make a frequency plot first we need to find the proportion of each outcome.

Total number of results = 15+20+5+5+5 = 50

Frequency of 0 = 15
Proportion of 0 = 15/50 = 0.3

Frequency of 1 = 20
Proportion of 0 = 20/50 = 0.4

Frequency of 2 = 5
Proportion of 2 = 5/50 = 0.1

Frequency of 3 = 5
Proportion of 3 = 5/50 = 0.1

Frequency of 4 = 5
Proportion of 4 = 5/50 = 0.1

Now we need to plot the data on a frequency plot. The x-axis shows the outcomes from 0 to 4 and y-axis shows the frequency of each outcomes. The frequency plot is shown in the figure attached with.

4 0

3 years ago

Read 2 more answers

Plz help me so confused

Alex Ar [27]

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4 0

3 years ago

g A trial jury of 6 people is selected from 20 people: 8 women and 12 men. What is the probability that the jury will have an od

cricket20 [7]

Answer: P(odd) = 0.499

Step-by-step explanation:

Given:

Total number of people = 20

Number of men = 12

Number of women = 8

Number of jury to be selected = 6

For the jury to have an odd number of women. it must have either of the three.

1. 1 woman , 5 men

2. 3 women, 3 men

3. 5 women, 1 man

The total possible ways of selecting the 6 people jury is;

N = 20C6 = 20!/6!(20-6)!

N = 38760

The possible ways of selecting;

Case 1 : 1 woman, 5 men

N1 = 8C1 × 12C5

N1 = 8 × 792 = 6336

Case 2 : 3 women , 3 men

N2 = 8C3 × 12C3

N2 = 12320

Case 3 : 5 women, 1 man

N3 = 8C5 × 12C1

N3 = 672

P(Odd) = (N1+N2+N3)/N

P(odd) = (6336+12320+672)/38760

P(odd) = 19328/38760

P(odd) = 0.499

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2 years ago

No links or wrong answers please!! 55 points

juin [17]

The areas of the figures are 4(x + 1), 7(d + 4) and y(y + 3)

<h3>How to determine the total areas?</h3>

<u>The figure 1</u>

In this figure, we have

Length = x + 1

Width = 4

The area is calculated as:

Area = Length * Width

So, we have

Area = 4(x + 1)

<u>The figure 2</u>

In this figure, we have

Length = d + 4

Width = 7

The area is calculated as:

Area = Length * Width

So, we have

Area = 7(d + 4)

<u>The figure 3</u>

In this figure, we have

Length = y + 3

Width = y

The area is calculated as:

Area = Length * Width

So, we have

Area = y(y + 3)

Hence, the areas of the figures are 4(x + 1), 7(d + 4) and y(y + 3)