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3 years ago

5

If I roll a standard number cube (labeled 1 to 6), and I roll it twice, what is the probability that I roll an odd number each t

ime?

2 answers:

suter [353]3 years ago

7 0

Answer:

It is a 50% chance that you will roll and odd number each time.

Step-by-step explanation:

Brut [27]3 years ago

3 0

Answer:

Step-by-step explanation:

5 VESES

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I need to know the answer to number one

Law Incorporation [45]

Supplementary angles are two angles whose measures sum to a 180 degrees and complementary are the sum have to add up to 90 degrees.So 180÷2=90;90÷2=45;180÷6=30 and 90÷6=15.Altough the answer is A.

5 0

3 years ago

What is the numerical probability of selecting a jury with 10 men and 2 women?

VladimirAG [237]

Answer:

Let E denote the event of choosing a jury with 10 men and 2 women.

The sample space for selecting 12 members from the pool contains 55C12 elements.

The number of ways of selecting 10 men and 2 women is 26C10 29C2.

The probability of event E =

The probability of selecting a jury with 10 men and 2 women is 0.005 (0.5%).

Step-by-step explanation:

see the image :

3 0

2 years ago

What is the distance between point C and point D?

e-lub [12.9K]

Answer:

B. 13 units

Step-by-step explanation:

let

(x1 , y1) = (-4 , -6)

(x2 , y2) = (1 , 6)

distance between CD = root (x2 - x1)^3 + (y2 - y1)

= root {1 - ( -4)}^2 + {6 - ( -6)}^2

= root {1 +4}^2 + {6+6}^2

= root 5^2 + 12^2

= root 25 + 144

= root 169

= 13 units ans

6 0

2 years ago

Read 2 more answers

What is the volume of a box with a width 2 1/2 inches, length 1 3/4 inches, and height 3 1/4 inches?

enot [183]

Answer:

The volume of the box is 41.21 inches cube.

Step-by-step explanation:

Given,

Width of the box $2\frac{1}{2}$ inches

$=\frac{5}{2}$ inches

Length of the box $1\frac{3}{4}$ inches

$=\frac{7}{4}$ inches

Height of the box $3\frac{1}{4}$ inches

$= \frac{13}{4}$ inches

The box is in cuboid form.

The volume of cuboid is given by (length × breadth × height ).

So,

volume of the box is $( \frac{5}{2}$ × $\frac{7}{4}$ × $\frac{13}{4}$ $)$

= $( \frac{455}{32} )$

$= 14.21$ inches cube

The volume of the box is 41.21 inches cube.

8 0

3 years ago

Solve the following equations.<br> log2(x^2 − 16) − log^2(x − 4) = 1

Alenkasestr [34]

Answer:

$x=\frac{4*(2+e)}{e-2}$

Step-by-step explanation:

Let's rewrite the left side keeping in mind the next propierties:

$log(\frac{1}{x} )=-log(x)$

$log(x*y)=log(x)+log(y)$

Therefore:

$log(2*(x^{2} -16))+log(\frac{1}{(x-4)^{2} })=1\\ log(\frac{2*(x^{2} -16)}{(x-4)^{2}})=1$

Now, cancel logarithms by taking exp of both sides:

$e^{log(\frac{2*(x^{2} -16)}{(x-4)^{2}})} =e^{1} \\\frac{2*(x^{2} -16)}{(x-4)^{2}}=e$

Multiply both sides by $(x-4)^{2}$ and using distributive propierty:

$2x^{2} -32=16e-8ex+ex^{2}$

Substract $16e-8ex+ex^{2}$ from both sides and factoring:

$-(x-4)*(-8-4e-2x+ex)=0$

Multiply both sides by -1:

$(x-4)*(-8-4e-2x+ex)=0$

Split into two equations:

$x-4=0\hspace{3}or\hspace{3}-8-4e-2x+ex=0$

Solving for $x-4=0$

Add 4 to both sides:

$x=4$

Solving for $-8-4e-2x+ex=0$

Collect in terms of x and add $4e+8$ to both sides:

$x(e-2)=4e+8$

Divide both sides by e-2:

$x=\frac{4*(2+e)}{e-2}$

The solutions are:

$x=4\hspace{3}or\hspace{3}x=\frac{4*(2+e)}{e-2}$

If we evaluate x=4 in the original equation:

$log(0)-log(0)=1$

This is an absurd because log (x) is undefined for $x\leq 0$

If we evaluate $x=\frac{4*(2+e)}{e-2}$ in the original equation:

$log(2*((\frac{4e+8}{e-2})^2-16))-log((\frac{4e+8}{e-2}-4)^2)=1$

Which is correct, therefore the solution is:

$x=\frac{4*(2+e)}{e-2}$

6 0

3 years ago