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3 years ago

Mean of 6 numbers is 41. Remove one number and the mean of the five numbers is 46. What is the unknown number

Mathematics

1 answer:

Alex17521 [72]3 years ago

6 0

Answer:

- 16

Step-by-step explanation:

This is very tricky. Be careful how you do it.

The total of the six numbers is 6*41 = 246

The total of five numbers = 5*46 = 230

What have you you removed to make the number average go up with a smaller sample? The answer is a negative.

230 - 246 = - 16

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True or false please help it’s due right now

Free_Kalibri [48]

Answer:

false

Step-by-step explanation:

Triangle = 5*6=30/2=15

Parallelogram = 6*5 = 30

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3 years ago

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Tina and maryann each play tennis. Maryann has won 5 more games than Tina. Is it possible for tina to have won games if the sum

Soloha48 [4]

Answer:

Tina won 5 games more than maryann, and the sum of their games is 29.. 29-5/2= 12

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3 years ago

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Estimate the answer by rounding to the nearest tenth 3.07442 + 1.352

Aliun [14]

3.10+1.40= 4.50 nearest tenth

7 0

3 years ago

Use the following image to assist in your answer. a = - 6 - 9 - 4

vfiekz [6]

Step-by-step explanation:

Pythagoras' theorem for the smallest one :

$c^2 = 4^2 + 6^2$

$c^2 = 16 + 36$

$c^{2}$ = 52

Pythagoras' theorem for the middle one :

$b^{2}$ = $6^{2}$ + $a^{2}$

Pythagoras' theorem for the biggest one :

$(4+a)^2 = c^2 + b^2$

$16 + 8a + a^2 = 52 + b^2$

Using the formula before (for $b^2$ ) it becomes :

$16 + 8a + a^2 = 52 + (6^2 + a^2)$

$16 + 8a + a^2 = 52 + 6^2 + a^2$

$16 + 8a = 52 + 6^2$

16 + 8a = 52 + 36

16+8a = 88

8a = 88-16

8a = 72

$a = \frac{72}{8}$

a = 9

Verifying :

$b^2 = 6^2 + a^2$

$b^2 = 36 + 81$

$b^2 = 117$

$b^{2}$ = 117

The biggest one :

$(4+a)^2 = c^2 + b^2$

$(4+9)^2 = 52 + 117$

$13^2 = 169$

True

8 0

2 years ago

what is perpendicular to y= 3x -9 and passes through the point (3,1)

GREYUIT [131]

$(3,1) , \ \ y= 3x -9 \\ \\ The \ slope \ is :m _{1} =3 \\ \\ If \ m_{1} \ and \ m _{2} \ are \ the \ gradients \ of \ two \ perpendicular \\ \\ lines \ we \ have \ m _{1} \cdot m _{2} = -1 \\ \\3\cdot m_{2}=-1 \ \ /:3\\ \\ m_{2}=-\frac{1}{3}$

$\Now \ your \ equation \ of \ line \ passing \ through \ (3,1) would \ be: \\ \\ y=m_{2}x+b \\ \\1=-\frac{1}{ 3} \cdot3 + b$

$1=-1+b\\ \\ b=1+1\\ \\b=2 \\ \\ y =- \frac{1}{3}x+2$

6 0

2 years ago