Please help and thank you

Translate the sentence into an equation: the sum of a number divided by 7 and 9 equals 3

ludmilkaskok [199]

X/7 + 9 = 3
a number: x
a number divided by 7: x/7
sum of number divided by 7 and 9: x/7 + 9 (and means plus)
equals 3: x/7 + 9 = 3

3 0

3 years ago

Simplify (6 +i)(8 – 3i) 51 – 10i 48 – 32i 45 – 10i 48 – 10i – 3i^2 ...?

mote1985 [20]

= 48 - 18i + 8i - 3i²
= 48 – 10i – 3i²

4 0

3 years ago

There are 30 homes in Neighborhood A. Each year, the number of homes increases by 20%. Just down the road, Neighborhood B has 45

nignag [31]

Part A

f(x)=30∗0.2xandf(x)=45+3x
___________________________________________________________

Part B

f(x)=30+0.2(5)f(x)=45+3(5)
Neighborhood A and neighborhood B both have 60 houses after 5 years

___________________________________________________________

Part C

1 year
A=36
B=48

2 years
A=42
B=51

3 years
A=48
B=54

4 years
A = 54
B = 57

5 years
A = 60
B= 60

5 0

3 years ago

Read 2 more answers

Write the series 1-0.5-2+ ... for 12 terms using summation notation

lions [1.4K]

Answer:

The series

1 -0.5-2-3.5 -5-6.5-8-9.5-11-12.5-14-15.5 upto 12 terms

Step-by-step explanation:

Step(i):-

Given series 1 -0.5 -2 +.........

we know that the sum of the sequence is called a series

The sequence is 1, -0.5, -2, -3.5,.....is in AP

a = 1 and the difference between the two terms is equal 'd' = -1.5

Step(ii):-

adding 1.5 value of each term

The series

1 -0.5-2-3.5 -5-6.5-8-9.5-11-12.5-14-15.5 upto 12 terms

5 0

2 years ago

If a random sample of size nequals=6 is taken from a population, what is required in order to say that the sampling distributio

goldenfox [79]

Answer:

$\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})$

In order to satisfy this distribution we need that each observation on this case comes from a normal distribution, because since the sample size is not large enough we can't apply the central limit theorem.

Step-by-step explanation:

For this case we have that the sample size is n =6

The sample man is defined as :

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And we want a normal distribution for the sample mean

$\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})$

In order to satisfy this distribution we need that each observation on this case comes from a normal distribution, because since the sample size is not large enough we can't apply the central limit theorem.

So for this case we need to satisfy the following condition:

$X_i \sim N(\mu , \sigma), i=1,2,...,n$