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3 years ago

Help plz will give brainliest

Mathematics

1 answer:

brilliants [131]3 years ago

5 0

Answer:

2*2*2*2*2*2*2*2*2*2*2*2=4096

Step-by-step explanation:

2 to the 12th power because you multiply exponents

You might be interested in

Complete each step in the following solution.

Charra [1.4K]

Answer:

1)-6

2)-14

Which it's going to give you an answer of x=-2

Step-by-step explanation:

Hope this helps

6 0

3 years ago

Can someone help pls?

scoundrel [369]

try googling it thats what I would do. I think it would be 1/6 tho

3 0

3 years ago

A rectangular bed of a truck is 1 9/10 meters long,1 1/2 meters wide, and 1/2 meter tall. What is the volume of the truck bed?

Sophie [7]

Answer:

The volume of the truck bed is $1\frac{17}{40}\ m^{3}$

Step-by-step explanation:

we know that

The volume is equal to

$V=LWH$

substitute the values , but first convert to an improper fractions

$L=1\frac{9}{10}\ m=\frac{10*1+9}{10}=\frac{19}{10}\ m$

$W=1\frac{1}{2}\ m=\frac{1*2+1}{2}=\frac{3}{2}\ m$

$H=\frac{1}{2}\ m$

Find the volume

$V=(\frac{19}{10})(\frac{3}{2})(\frac{1}{2})=\frac{57}{40}\ m^{3}$

Convert to mixed number

$\frac{57}{40}\ m^{3}=\frac{40}{40}+\frac{17}{40}=1\frac{17}{40}\ m^{3}$

3 0

3 years ago

What is the prime factorasation of 96

dimulka [17.4K]

So, the prime factors of 96 are written as 2 x 2 x 2 x 2 x 2 x 3 or 3 x 25, where 2 and 3 are the prime numbers.

5 0

3 years ago

Read 2 more answers

You flip a coin and then roll a 6-sided number cube (a die).

expeople1 [14]

Answer:

a) No, it does not matter whether you roll the die or flip the coin first, as these two events are <u>independent</u> of each other, which means they do not affect each other.

b) Yes.

Let event 1 be flipping a coin and event 2 be rolling a die.
Let event 1 be rolling a die and event 2 be flipping a coin.

The likelihood that any outcome will occur will not change, as the events are independent.

c) see attached

d) 12 outcomes (H = head, T = tail, numbers represent the value of the die)

H 1 T 1

H 2 T 2

H 3 T 3

H 4 T 4

H 5 T 5

H 6 T 6

$\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}$

$\implies \sf P(even)=\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}=\dfrac{3}{6}=\dfrac{1}{2}$

$\implies \sf P(head)=\dfrac{1}{2}$

$\implies \sf P(even)\:and\:P(head)=\dfrac{1}{2} \times \dfrac{1}{2}=\dfrac{1}{4}$

6 0

2 years ago

Help plz will give brainliest​

Help plz will give brainliest