All categories

2 years ago

What is the mode of these numbers? 10, 12, 8, 11, 10, 15, 7

Mathematics

2 answers:

lana66690 [7]2 years ago

5 0

Answer:

The mode is 10

Step-by-step explanation:

Westkost [7]2 years ago

5 0

10 because mode is the number that occurs the most

You might be interested in

Tim has $20 to buy snacks for 12 people in an office. Each person will get one snack. Tim is buying bags of pretzels that cost $

Greeley [361]

Answer:

8 bags of pretzels

Step-by-step explanation:

Let x represent the number of bags of pretzels Tim buys. We assume he spends exactly $20 on exactly 12 bags of snack food. Then his purchase is ...

1.50x + 2.00(12-x) = 20.00

-0.50x +24.00 = 20.00 . . . eliminate parentheses, collect terms

-0.50x = -4.00 . . . . . . . . . . .subtract 24

x = 8 . . . . . . . . . . divide by -0.50

Tim will buy 8 bags of pretzels.

4 0

3 years ago

ANSWER PLSSS!! Brian puts $2000 in the bank. He leaves it there for 2 years. The bank uses an interest rate of 4%. How much mone

irinina [24]

Answer:

the answer is $1920

Step-by-step explanation:

4% of $2000 is $80 so if they take 4% its just $2000 - $80 = $1920

Hope this helped! <3

3 0

3 years ago

There are 90 girls and 60 boys Participating in a race. The coach separates them into two equal groups made of all girls or or b

Lady_Fox [76]

90/2 is 45.. 60/2 is 30.. 30+45=75... 75 students would be able to be in each of the 2 groups, each containing 45 girls and 30 boys ... Hope I helped (:

7 0

3 years ago

The area of each square is 16 square units find the perimeter of a figure

vivado [14]

area=length*width

4*4=16

4+4+4+4=16

the perimeter is also 16

6 0

2 years ago

Solve the following equation by factoring:9x^2-3x-2=0

olya-2409 [2.1K]

Answer:

The two roots of the quadratic equation are

$x_1= - \frac{1}{3} \text{ and } x_2= \frac{2}{3}$

Step-by-step explanation:

Original quadratic equation is $9x^{2}-3x-2=0$

Divide both sides by 9:

$x^{2} - \frac{x}{3} - \frac{2}{9}=0$

Add $\frac{2}{9}$ to both sides to get rid of the constant on the LHS

$x^{2} - \frac{x}{3} - \frac{2}{9}+\frac{2}{9}=\frac{2}{9}$ ==> $x^{2} - \frac{x}{3}=\frac{2}{9}$

Add $\frac{1}{36}$ to both sides

$x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{2}{9} +\frac{1}{36}$

This simplifies to

$x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{1}{4}$

Noting that (a + b)² = a² + 2ab + b²

If we set a = x and b = $\frac{1}{6}\right)$ we can see that

$\left(x - \frac{1}{6}\right)^2$ = $x^2 - 2.x. (-\frac{1}{6}) + \frac{1}{36} = x^{2} - \frac{x}{3}+\frac{1}{36}$

$\left(x - \frac{1}{6}\right)^2=\frac{1}{4}$

Taking square roots on both sides

$\left(x - \frac{1}{6}\right)^2= \pm\frac{1}{4}$

So the two roots or solutions of the equation are

$x - \frac{1}{6}=-\sqrt{\frac{1}{4}}$ and $x - \frac{1}{6}=\sqrt{\frac{1}{4}}$

$\sqrt{\frac{1}{4}} = \frac{1}{2}$

So the two roots are

$x_1=\frac{1}{6} - \frac{1}{2} = -\frac{1}{3}$

and

$x_2=\frac{1}{6} + \frac{1}{2} = \frac{2}{3}$

7 0

1 year ago