What is the 15th term in the sequence with the given formula? An = 7n

Hint: m=7 5m-6=_?__ Due: Today

djverab [1.8K]

Answer:

29...

Step-by-step explanation:

1) Fill in

5(7)-6

2) P.E.M.D.A.S

35 - 6

29

5 0

3 years ago

Read 2 more answers

Scores on a certain IQ test are knownto have a mean of 100. A random sample of 43 students attend a series of coaching classes b

NeX [460]

Answer:

Part 1

Type II error

Part 2

No ; is not ; true

Step-by-step explanation:

Data provided in the question

Mean = 100

The Random sample is taken = 43 students

Based on the given information, the conclusion is as follows

Part 1

Since it is mentioned that the classes are successful which is same treated as a null rejection and at the same time it also accepts the alternate hypothesis

Based on this, it is a failure to deny or reject the false null that represents type II error

Part 2

And if the classes are not successful so we can make successful by making type I error and at the same time type II error is not possible

Therefore no type II error is not possible and when the null hypothesis is true the classes are not successful

6 0

3 years ago

Which fraction is shown by point D? A. 1/3 B. 1/4 C. 3/3 D. 3/4

Volgvan

Answer:

C

Step-by-step explanation:

Hope dis helps

7 0

2 years ago

Read 2 more answers

Solve 9,159 divided by 7

aniked [119]

Answer:

(9159 / 7 = 1308.429)

Step-by-step explanation:

Simply multiply the last digit by 2 and then subtract the product from the remaining digits.

If that difference is divisible by 7, then 9159 is divisible by 7.

The last digit in 9159 is 9 and the remaining digits are 915. Thus, the math to determine if 9159 is divisible by 7 using our alternate method is:

915 - (9 x 2) = 897

Since 897 is not divisible by 7, 9159 is also not divisible by 7.

Therefore, the answer to "Is 9159 Divisible By 7?" is no.

(9159 / 7 = 1308.429)

5 0

2 years ago

Read 2 more answers

Sara is working on a Geometry problem in her Algebra class. The problem requires Sara to use the two quadrilaterals below to ans

zloy xaker [14]

Part A:

Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4

The perimeter of the square is given by 4(x + 4) = 4x + 16

The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12

For the perimeters to be the same

4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2

The value of x that makes the perimeters of the quadrilaterals the same is 2.

Part B:

The area of the square is given by

$Area=(x+4)^2=x^2+8x+16$

The area of the rectangle is given by 2(3x + 4) = 6x + 8

For the areas to be the same

$x^2+8x+16=6x+8 \\ \\ \Rightarrow x^2+8x-6x+16-8=0 \\ \\ \Rightarrow x^2+2x+8=0 \\ \\ \Rightarrow x= \frac{-2\pm\sqrt{2^2-4(8)}}{2} \\ \\ = \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{-28}}{2} \\ \\ = \frac{-2\pm2i\sqrt{7}}{2} =-1\pm i\sqrt{7}$

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.

7 0

3 years ago

What is the 15th term in the sequence with the given formula? An = 7n – 10