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VMariaS [17]
2 years ago
14

What information can you obtain about the scores in a regular frequency distribution table that is not available from a grouped

table.
A. The number and width of the class intervals

B. The exact frequency for each category on the scale of measurement

C. The overall frequency for each class interval
Mathematics
1 answer:
Feliz [49]2 years ago
8 0

Answer: B. The exact frequency for each category on the scale of measurement

Step-by-step explanation: The main difference between a grouped table and regular frequency distribution exists in the fact that ; grouped table takes in a defined range of (X) values, that is takes more than one individual value per group and takes the frequency of all scores or values in the group, hence giving the frequency of the scores in the interval rather than the frequency of each individual score. On the other hand, the regular frequency distribution treats values or scores individually and then takes the frequency of each score. This way a regular frequency distribution enables us to know the exact number or distribution of each individual score.

You might be interested in
The heights of fully grown sugar maple trees are normally distributed, with a mean of 87.5 feet and a standard deviation of 6.25
Julli [10]

Answer:

20.33%

Step-by-step explanation:

We have that the mean (m) is equal to 87.5, the standard deviation (sd) 6.25 and the sample size (n) = 12

They ask us for P (x <86)

For this, the first thing is to calculate z, which is given by the following equation:

z = (x - m) / (sd / (n ^ 1/2))

We have all these values, replacing we have:

z = (86 - 87.5) / (6.25 / (12 ^ 1/2))

z = -0.83

With the normal distribution table (attached), we have that at that value, the probability is:

P (z <-0.83) = 0.2033

The probability is 20.33%

6 0
2 years ago
Help????????? If so....
Vinil7 [7]

Answer:

I think it equals 55

4 0
2 years ago
Read 2 more answers
Matthew manages the packaging department of his company. Yesterday, the department operated for only 4 hours due to a company pi
yKpoI14uk [10]

Alright, lets get started.

If Matthew wants to complete packages at an average rate of at least 39 packages per hour.

And they worked 4 hrs only due to picnic, yesterday, it means they have to make 39*4 = 156 packages.

But they made only 112 packages means they are short of 156 - 112 = 44 packages.

Suppose they are working today t hrs, and his department will complete 43 packages per hour today.

It means they are going to make 43 t packages today.

This 43 t packages includes those 44 too , which they are short of yesterday due to picnic.

So, average will be

\frac{43 t - 44}{t} = 39 (39 average given in question)

Cross multiplying

39 t = 43 t - 44

Adding 44 in both sides

39 t + 44 = 43 t - 44 + 44

43 t = 39 t + 44

Subtracting 39 t in both sides

43 t - 39 t = 39 t + 44 - 39 t

4t = 44

Dividing 4 in both sides

t = 11 hrs

Hence they have to woth 11 hrs today : Answer

Hope it will help :)

3 0
2 years ago
Read 2 more answers
Subtract 1.05 from a certain number. Multiply the difference by 0.8, add 2.84 to the product then divide the sum by 0.01 and get
bazaltina [42]
Answer:
The number is 6.25

Explanation:
Assume that the number we are looking for is x

First, we will set up the equations as follows:
1- Subtract 1.05 from the number (assume the result is y):
x - 1.05 = y

2- Multiply the difference by 0.8 (assume the product is z):
0.8(y) = z

3- add 2.84 to the product (assume the result is w):
z + 2.84 = w

4- divide the sum by 0.01, the quotient is 700:
w / 0.01 = 700

Now, we will work backwards as follows:
4) w / 0.01 = 700
    w = 0.01 * 700
    w = 7

3) z + 2.84 = w
    z + 2.84 = 7
    z = 7 - 2.84
    z = 4.16

2) 0.8y = z
    0.8y = 4.16
    y = 4.16 / 0.8
    y = 5.2

1) x - 1.05 = y
    x - 1.05 = 5.2
    x = 5.2 + 1.05
    x = 6.25

Hope this helps :)
8 0
2 years ago
Read 2 more answers
Jane wishes to bake an apple pie for dessert. The baking instructions say that she should bake the pie in an oven at a constant
Viktor [21]

Answer:

Therefore k= \frac{ln2 }{18}, A=184

Step-by-step explanation:

Given function is

T(t)=230 -e^{-kt}

where T(t) is the temperature in °C and t is time in minute and A and k are constants.

She noticed that after 18 minutes the temperature of the pie is 138°C

Putting T(t) =138°C and t= 18 minutes

138=230 -Ae^{-k\times 18}

\Rightarrow  -Ae^{-18k}=138-230

\Rightarrow  Ae^{-18k}=92 .....(1)

Again after 36 minutes it is 184°C

Putting T(t) =184°C and t= 36 minutes

184=230-Ae^{-k\times 36}

\Rightarrow Ae^{-36k}=230-184

\Rightarrow Ae^{-36k}=46.......(2)

Dividing (2) by (1)

\frac{Ae^{-36k}}{Ae^{-18k}}=\frac{46}{92}

\Rightarrow e^{-18k}=\frac{46}{92}

Taking ln both sides

ln e^{-18k}=ln\frac{46}{92}

\Rightarrow -18k =ln (\frac12)

\Rightarrow -18k= ln1-ln2

\Rightarrow k= \frac{ln2 }{18}

Putting the value k in equation (1)

Ae^{-18\frac{ln2}{18}}=92

\Rightarrow A e^{ln2^{-1}}=92

\Rightarrow A.2^{-1}=92

\Rightarrow \frac{A}{2}=92

\Rightarrow A= 92 \times 2

⇒A= 184.

Therefore k= \frac{ln2 }{18}, A=184

7 0
2 years ago
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