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4 years ago

What is the value of this expression? Select all that apply.

Mathematics

1 answer:

beks73 [17]4 years ago

3 0

Answer: A, D.

Step-by-step explanation:

(8.59 x 10^4) x (3.2 x 10^3) = 82,700

8.27 x 10^4 = 82,700

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Find the solutions <br> x^2=16x-28

ioda

Minus 16x both sides and add 28 to both sides
x^2-16x+28=0
factor
hmm, what 2 numbers multiply to 28 and add to -16
hmm
1 and 28? nope
2 and 14? yep
since middle term is negative and last tem is positive, both of the factors are negative

(x-14)(x-2)=0
set each to zero
x-14=0
x=14

x-2=0
x=2

x=2 and 14

7 0

3 years ago

A common factor for two numbers is one that is a divisor of each of

oee [108]

Answer:

The answer is 12.

Step-by-step explanation:

Please mark as Brainliest, this is the right answer

5 0

3 years ago

A coin is flipped eight times where each flip comes up either heads or tails. How many possible outcomes a) are there in total?

attashe74 [19]

Answer:

There are 256 ways in total.

There are 56 possible outcomes contain exactly three heads.

The possible outcomes contain at least three heads is 219.

The possible outcomes contain the same number of heads and tails are 70.

Step-by-step explanation:

Consider the provided information.

A coin is flipped eight times where each flip comes up either heads or tails.

Part (a) How many possible outcomes are there in total?

Each time we flip a coin it comes up either heads or tail.

Therefore the total number of ways are:

$2\times 2\times 2\times 2\times 2\times 2\times 2\times 2=2^8=256$

Hence, there are 256 ways in total.

Part (b) contain exactly three heads?

We want exactly 3 heads, therefore,

n=8 and r=3

According to the definition of combination: $\binom{n}{r}=\frac{n!}{r!(n-r)!}$

$\binom{8}{3}=\frac{8!}{3!(5)!}=56$

Hence, there are 56 possible outcomes contain exactly three heads.

Part (c) contain at least three heads?

For 3 heads: $\binom{8}{3}=\frac{8!}{3!(5)!}=56$

For 4 heads: $\binom{8}{4}=\frac{8!}{4!(4)!}=70$

For 5 heads: $\binom{8}{5}=\frac{8!}{5!(3)!}=56$

For 6 heads: $\binom{8}{6}=\frac{8!}{6!(2)!}=28$

For 7 heads: $\binom{8}{7}=\frac{8!}{7!(1)!}=8$

For 8 heads: $\binom{8}{8}=1$

Now add them as shown:

56+70+56+28+8+1=219

Hence, the possible outcomes contain at least three heads is 219.

Part (d) contain the same number of heads and tails?

Same number of heads and tails means that the value of r=4.

Therefore,

$\binom{8}{4}=\frac{8!}{4!(4)!}=70$

Hence, the possible outcomes contain the same number of heads and tails are 70.

6 0

3 years ago

Write -3i+(3/4+2i)-(9/3+3i) as a complex number in standard form.

spayn [35]

-3i+3/4+2i-9/3-3i=(3/4-9/3) +(-3i+2i-3i)=(3/4-3)+(-4i)=(3/4-12/4)-4i=
=-9/4-4i

8 0

3 years ago

Read 2 more answers

-3x-ly = -16<br> 2x +1y = 11

Aleksandr [31]

Answer: (-2/15, -53/5)

Step-by-step explanation:

multiply, simplify, thn rimate the one variable by adding the equations then divide sides by -165 then substitute

5 0

3 years ago

Read 2 more answers

What is the value of this expression? Select all that apply.​

What is the value of this expression? Select all that apply.