What is the GCF of 22,88,10,50?

If the mean, median, and mode are all the same for 4, 9, 7, 8, and x, what is the value of x?

seropon [69]

<h3>Answer: x = 7</h3>

===========================================================

Explanation:

Since we have an odd number of values, this tells us that the median is part of the data set. It's the middle most item after we sort the values.

Recall that the mode is the most frequent item. Since the mode and median are the same, this must mean x can only be equal to one of the following

4, 9, 7 or 8

We can only pick one of those values.

----------------------

If x = 4, then the set {4,9,7,8,x} updates to {4,9,7,8,4} which sorts to {4,4,7,8,9}

The middle most item is in slot 3, which would be 7. So the median is 7.

The median 7 does not match with the mode 4.

So we cross x = 4 off the list.

-----------------------

If x = 7, then we have {4,7,7,8,9}

The mode is 7 and the median is 7. So far, so good.

Now let's compute the mean. Add up the values and divide by 5 because there are 5 items.

(4+7+7+8+9)/5 = 35/5 = 7

We've shown that the set {4,7,7,8,9} has mean 7.

Overall, that set has the same mean, median and mode. So the answer is confirmed.

I'll let you check the cases when x = 8 and x = 9.

8 0

3 years ago

Which graph does NOT represent a function?

masha68 [24]

Answer:

the answer is D because it does not have a function

7 0

2 years ago

Read 2 more answers

2 times 2 times 13 what is the missing number

atroni [7]

The answer is 52 if you are looking for the answer to 2x2x13=? then it is 52. Otherwise idk what you are asking.

5 0

3 years ago

A curve is given by y=(x-a)√(x-b) for x≥b, where a and b are constants, cuts the x axis at A where x=b+1. Show that the gradient

ankoles [38]

<u>Answer:</u>

A curve is given by y=(x-a)√(x-b) for x≥b. The gradient of the curve at A is 1.

<u>Solution:</u>

We need to show that the gradient of the curve at A is 1

Here given that ,

$y=(x-a) \sqrt{(x-b)}$ --- equation 1

Also, according to question at point A (b+1,0)

So curve at point A will, put the value of x and y

$0=(b+1-a) \sqrt{(b+1-b)}$

0=b+1-c --- equation 2

According to multiple rule of Differentiation,

$y^{\prime}=u^{\prime} y+y^{\prime} u$

so, we get

${u}^{\prime}=1$

$v^{\prime}=\frac{1}{2} \sqrt{(x-b)}$

$y^{\prime}=1 \times \sqrt{(x-b)}+(x-a) \times \frac{1}{2} \sqrt{(x-b)}$

By putting value of point A and putting value of eq 2 we get

$y^{\prime}=\sqrt{(b+1-b)}+(b+1-a) \times \frac{1}{2} \sqrt{(b+1-b)}$

$y^{\prime}=\frac{d y}{d x}=1$

Hence proved that the gradient of the curve at A is 1.

7 0

2 years ago

Are<br> Which pair of angles<br> angles?<br> corresponding

-BARSIC- [3]

Answer:

D. <3 and <7

Step-by-step explanation:

Corresponding angles are angles that are on matching sides and that are cut by a transversal.

Hope this helps!

4 0

2 years ago