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VladimirAG [237]
3 years ago
6

Ben has a box of marbles. 30 of the marbles are red, 14 are green and 16 are blue. Ben randomly selects a marble from the box, r

ecords the color and then replaces the marble in the box. If he selects a marble 30 times, how many times do you predict he will select a blue marble?
Mathematics
1 answer:
jasenka [17]3 years ago
3 0

Answer:

he will select about 5 blue marbles

Step-by-step explanation:

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HELP QUESTION NUMBER THREE
yKpoI14uk [10]

Answer:

(-3,2) is the solution to the system of equation

Step-by-step explanation:

Here, we want to know the point that serves as a solution to the system of equation

We equate the y

-2x + 4 = -1/3x-1

Multiply through by 3

-6x + 12 = -x-3

-6x + x = -3-12

5x = 15

x = 15/5

x = 3

Substitute x into one of the y

y = -2(3) + 4

y = -6 + 4 = -2

So the solution is;

(3,-2)

8 0
3 years ago
Why does multiple of 3 add up digits trick work?
kherson [118]
You must be referring to the quick way to determine whether a given integer is divisible by 3; that is, 3 divides an integer x whenever the digits of x sum to a multiple of 3.

Suppose x has n\ge1 digits. We can expand it as a sum of the numbers in any given digits place by the corresponding power of 10. For example, if x=2148, we can write

2148=2\times10^3+1\times10^2+4\times10^1+8\times10^0

More generally, if

x=d_{n-1}d_{n-2}\ldots d_1d_0

(where d_i denotes the numeral in the 10^i-th's place), then we have the expansion

x=d_{n-1}10^{n-1}+d_{n-2}10^{n-2}+\cdots+d_110^1+d_0

Notice that for any integer k, we have

10^k-1=\underbrace{99\ldots99}_{k\text{ 9s}}

which is clearly divisible by 3. So from each power of 10 in the expansion of x, we can add and subtract 1, then rearrange the terms of the sum:

x=d_{n-1}10^{n-1}+\cdots+d_110^1+d_0
x=d_{n-1}(10^{n-1}-1+1)+\cdots+d_1(10^1-1+1)+d_0
x=d_{n-1}(10^{n-1}-1)+\cdots+d_1(10^1-1)+(d_{n-1}+\cdots+d_1+d_0)

We know 10^k-1 is divisible by 3, which means the remainder upon dividing x by 3 is just the sum of the digits of x. If this remainder is divisible by 3, then so must be the original number, x.

Back to our previous example: if x=2148, then we have the expansion

2148=2\times10^3+10^2+4\times10+8
2148=2(999+1)+(99+1)+4(9+1)+8
2148=2\times999+99+4\times9+(2+1+4+8)

Dividing through by 3, we get a remainder of 2+1+4+8=15, which is divisible by 3, and so 2148 must also be a multiple of 3.

In case for some reason you're not convinced 15 is a multiple of 3, you can apply the same trick:

15=10+5
15=9+1+5

Dividing through by 3 leaves a remainder of 1+5=6, which is also a multiple of 3, so that 15 must be, too.
7 0
3 years ago
The equations y+x^2+6x+8 and y =(x+2) (x+4) both define the same quadratic function. Without graphing, identify the x- and y-int
Goryan [66]

Answer:

we get the  

x-intercepts setting y=0 and the in the equation

y-intercept setting x=0 in the equation.

Step-by-step explanation:

x-intercepts( y=0)

0=x^2+6x+8=(x+2)(x+4)

Hence x=-2 and x=-4 ( We got two x-intercepts)

y-intercept( x=0)

y=0^2+6(0)+8=8.

the y intercept is y=8 ( only is possible to find no more thant one).

6 0
3 years ago
Read 2 more answers
How are these triangles similar
MAXImum [283]

Answer:

they both have right angles

6 0
3 years ago
Read 2 more answers
Tyra's family is spending the afternoon in Millersville. They plan to see a movie and then explore the town. The movie will cost
12345 [234]
Answer: B. $4 hope this helps(:
3 0
3 years ago
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