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4 years ago

Please, please help. Answer the question, then explain how you got your answer. a= -3

Mathematics

1 answer:

Mice21 [21]4 years ago

3 0

3.5*(-3)///////3.5*(-3)//////all of this parts same

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In a continuous series, mean(x) = 80, assumed mean(A) = 60 and<br>EF=40 then find the value of EFd.

Aneli [31]

Using the given information, the value of Σfd is 800

<h3>Calculating mean using Assumed mean </h3>

From the question, we are to determine the value of Σfd

The formula for mean, using the assumed mean method is given by

$\bar x = A + \frac{\sum fd}{\sum f}$

Where $\bar x$ is the mean

A is the assumed mean

From the given information,

$\bar x = 80$

$A = 60$

$\sum f = 40$

Putting the parameters into the equation, we get

$80 = 60 + \frac{\sum fd}{40}$

$80-60 = \frac{\sum fd}{40}$

$20 = \frac{\sum fd}{40}$

$\sum fd = 20 \times 40$

$\sum fd = 800$

Hence, the value of Σfd is 800

Learn more on Mean here: brainly.com/question/20118982

#SPJ1

8 0

2 years ago

Louise is walking 12 miles to raise funds for a local cause. If she has completed 34 of the distance, how many miles has she wal

Elden [556K]

About 4 miles hope this helps

6 0

3 years ago

−(−49) = −49 true or false?

vova2212 [387]

I hope this helps you

false

cause two negative numbers multiple must be positive

(-1).(-49)

+49

5 0

3 years ago

Mrs. Brown hosts an yearly art contest keeps a record of the number of entries each year.

olganol [36]

1. Slope = rate of change
Shown on the graph, for 2016 there are 36 entries. For 2017, there are 40
Since it’s only one year
You can subtract 36 from 40
40 - 36 = 4
Solution: the slope is 4

2. Months increase the most, the lines which are steeper and going upward (since it’s increasing)
Solution: from 2015 to 2016

4 0

3 years ago

There are eight students in a class. Only one of them has passed Exam P/1 and only one of them has passed Exam FM/2. No student

Svetach [21]

Answer: There is probability of 0.57 chances that exactly three students from a group of four students have not passed Exam P/1 or Exam FM/2.

Step-by-step explanation:

Total number of students = 8

Number of student who has passed Exam P/1 = 1

Number of student who has passed Exam FM/2 = 1

No student has passed more than one exam.

According to question, exactly three students from a randomly chose group of four students have not passed Exam P/1 or Exam FM/2.

So, Probability will be

$\frac{^6C_3\times ^2C_1}{^8C_4}\\\\=\frac{20\times 2}{70}\\\\=\frac{4}{7}\\\\=0.57$

Hence, there is probability of 0.57 chances that exactly three students from a group of four students have not passed Exam P/1 or Exam FM/2.

4 0

3 years ago