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3 years ago

What is the probability that when we roll 5 fair 6-sided dice, at most 4 of them will show a 1?

1 answer:

4 0

If you're rolling 5 dice, and all dice have a 1 on them, then you still have a 4/5 chance that four of the five dice will show 1.

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Help me please!! I need someone to explain this for me. I need to find the area of this trapezoid!!

OverLord2011 [107]

You would multiply five and nine to get 45. Hope this helps:)

6 0

3 years ago

Read 2 more answers

Write the equation of a parabola, in standard form, that goes through these points: (0, 3) (1, 4) (-1, -6)

igomit [66]

$f(x)=ax^2+bx+c\\\\(0;\ 3)\to x=0;\ y=3\to f(0)=3\\therefore\\3=a\cdot0^2+b\cdot0+c\\\boxed{c=3}\\\\(1;\ 4)\to x=1;\ y=4\to f(1)=4\\therefore\\4=1^2\cdot a+1\cdot b+3\\4=a+b+3\ \ \ |subtract\ 3\ from\ both\ sides\\(1)\boxed{a+b=1}\\\\(-1;-6)\to x=-1;\ y=-6\to f(-1)=-6\\therefore\\-6=(-1)^2a+(-1)b+3\\-6=a-b+3\ \ \ \ |subtract\ 3\ from\ both\ sides\\(2)\boxed{a-b=-9}$

We have (1) & (2):

$\underline{+\left\{\begin{array}{ccc}a+b=1\\a-b=-9\end{array}\right}\ \ \ \ |add\ both\ sides\\.\ \ \ \ \ \ 2a=-8\ \ \ \ |divide\ both\ sides\ by\ 2\\.\ \ \ \ \ \ \ \boxed{a=-4}\\\\subtitute\ the\ value\ of\ "a"\ to\ the\ first\ equation\\\\-4+b=1\ \ \ \ |add\ 4\ to\ both\ sides\\\boxed{b=5}\\\\Answer{\boxed{\boxed{f(x)=-4x^2+5x+3}}$

8 0

3 years ago

Please help for brainliest

weeeeeb [17]

Answer:

84 and 144

Step-by-step explanation:

If the two numbers are a and b and their LCM is x and their HCF is y then there is a rule that is a×b=x×y

in this case:

a×b=x×y

a×b=(504)×(24)

a×b=12096

when finding factors of 12096 that are between 60 and 160, we can see that 84 and 144 is the best pair.

hope this helps! :D

have a miraculous day!! <3

4 0

2 years ago

Please help, thanks if you do :)

Basile [38]

Your answer should be, 3x= 22y to the tenth power...

6 0

3 years ago

The graph depicts the growth and costs using of line segment use the midpoint formula to approximate the cost during the year 19

marysya [2.9K]

Answer:

The approximate cost during the year 1987 was $7,533

Step-by-step explanation:

The complete question in the attached figure

we have

y -----> is the cost in dollars

x -----> is the year

Looking at the graph we have the points

(1983,5,047) and (1991, 10,019)

To determine the cost in the year 1987, find out the midpoint between the two given points

The formula to calculate the midpoint between two points is equal to

$M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

substitute the values

$M=(\frac{1983+1991}{2},\frac{5,047+10,019}{2})$

$M=(1987,7,533)$

therefore

The approximate cost during the year 1987 was $7,533

7 0

4 years ago