All categories

3 years ago

How many times someone would land on a number less than or equal to 5 if they rolled a die 90 times.

Mathematics

1 answer:

galina1969 [7]3 years ago

5 0

Answer:

75 times

Step-by-step explanation:

If they rolled each number 15 times, they would roll 5 or lower 75 times.

You might be interested in

Solve and write solution in interval notation 4(x+1)+3>x-5

soldi70 [24.7K]

Answer:

$\large\boxed{x\in\left(-\dfrac{13}{3},\ \infty\right)}$

Step-by-step explanation:

$4(x+1)+4>x-5\qquad\text{use the distributive property}\\\\4x+4+4>x-5\\\\4x+8>x-5\qquad\text{subtract 8 from both sides}\\\\4x>x-13\qquad\text{subtract}\ x\ \text{from both sides}\\\\3x>-13\qquad\text{divide both sides by 3}\\\\x>-\dfrac{13}{3}\to x\in\left(-\dfrac{13}{3},\ \infty\right)$

3 0

3 years ago

Can you guys figure it out ??

Studentka2010 [4]

I'm not sure but I think it's 15

7 0

4 years ago

The price of a gallon of milk is 3.70 5 dollars what is the price that is rounded to the nearest dollar

ira [324]

4.00....................

7 0

3 years ago

Read 2 more answers

Consider the function on the interval (0, 2π). f(x) = 7 sin2(x) + 7 sin(x) (a) Find the open interval(s) on which the function i

-BARSIC- [3]

Answer with Step-by-step explanation:

Given

$f(x)=7sin(2x)+7sin(x)$

Differentiating both sides by 'x' we get

$14cos(2x)+7cos(x)=f'(x)$

Now we know that for an increasing function we have

$f'(x)>0\\\\14cos(2x)+7cos(x)>0\\\\2cos(2x)+cos(x)>0\\\\2(2cos^{2}(x)-1)+cos(x)>0\\\\4cos^{2}(x)+cos(x)-2>0\\\\(2cos(x)+\frac{1}{2})^2-2-\frac{1}{4}>0\\\\(2cos(x)+\frac{1}{2})^2>\frac{9}{4}\\\\2cos(x)>\frac{3}{2}-\frac{1}{2}\\\\\therefore cos(x)>\frac{1}{4}\\\\\therefore x=[0,cos^{-1}(1/4)]\cup [2\pi-cos^{-1}(1/4),2\pi ]$

Similarly for decreasing function we have

$[tex]f'(x)$

Part b)

To find the extreme points we equate the derivative with 0

$f'(x)=0\\\\cos(x)=\frac{1}{4}\\\\x=cos^{-1}(\frac{1}{4})$

Thus point of extrema is only 1.

4 0

4 years ago

3 < 2x + 1 < 9<br> How do I solve this?

Vitek1552 [10]

Answer:

(1,4) or 1 < x < 4

Step-by-step explanation:

3 = less than 2x + 1 and 2x + 1 is less than 9

we can split the two parts of the equation

3< 2x + 1 and 2x + 1 < 9

for 3< 2x + 1 x has to be greater than 1

for 2x+ 1 < 9 x has to be less than 4 so the answer is (1,4) or 1 < x < 4

5 0

3 years ago