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Hitman42 [59]
4 years ago
9

What is the value of this expression A) -29 B) -26 C) 3 D) 4

Mathematics
2 answers:
netineya [11]4 years ago
7 0

Answer:

4

Step-by-step explanation:

(6-4) ^3 ÷2

PEMDAS says parentheses first

(6-4) ^3 ÷2

2^3 ÷2

The we do exponents

8 ÷2

Finally we divide

4

QveST [7]4 years ago
3 0

Answer:

D. 4

Step-by-step explanation:

(6 - 4)³ / 2

2³/ 2

(2 * 2 * 2) / 2

8 / 2

4

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Step-by-step explanation:

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nata0808 [166]

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Step-by-step explanation:

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3 years ago
Solve for x: 2(2a+3)-6=5a+2
coldgirl [10]

Answer:

a=-2

We move all terms to the left:

2(2a+3)-6-(5a+2)=0

We multiply parentheses

4a-(5a+2)+6-6=0

We get rid of parentheses

4a-5a-2+6-6=0

We add all the numbers together, and all the variables

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We move all terms containing a to the left, all other terms to the right

-a=2

a=2/-1

a=-2

Step-by-step explanation:

We move all terms to the left:

2(2a+3)-6-(5a+2)=0

We multiply parentheses

4a-(5a+2)+6-6=0

We get rid of parentheses

4a-5a-2+6-6=0

We add all the numbers together, and all the variables

-1a-2=0

We move all terms containing a to the left, all other terms to the right

-a=2

a=2/-1

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3 0
3 years ago
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) Determine the probability that a bit string of length 10 contains exactly 4 or 5 ones.
yanalaym [24]

Answer: 0.4512

Step-by-step explanation:

A bit string is sequence of bits (it only contains 0 and 1).

We assume that the  0 and 1 area equally likely to any place.

i.e. P(0)= P(1)= \dfrac{1}{2}

The length of bits : n = 10

Let X = Number of getting ones.

Then , X \sim Bin(n=10,\ p=\dfrac{1}{2})

Binomial distribution formula : P(X=x)=^nC_x p^x q^{n-x} , where p= probability of getting success in each event and q= probability of getting failure in each event.

Here , p=q=\dfrac{1}{2}

Then ,The probability that a bit string of length 10 contains exactly 4 or 5 ones.

P(X= 4\ or\ 5)=P(x=4)+P(x=5)\\\\=^{10}C_4(\dfrac{1}{2})^{10}+^{10}C_4(\dfrac{1}{2})^{10}

=\dfrac{10!}{4!6!}(\dfrac{1}{2})^{10}+\dfrac{10!}{5!5!}(\dfrac{1}{2})^{10}

=(\dfrac{1}{2})^{10}(\dfrac{10!}{4!6!}+\dfrac{10!}{5!5!})

=(\dfrac{1}{2})^{10}(210+252)

=(0.0009765625)(462)

=0.451171875\approx0.4512

Hence, the  probability that a bit string of length 10 contains exactly 4 or 5 ones is 0.4512.

3 0
4 years ago
Find the slope of the line that passes through (4,2) and (4,5)
Hunter-Best [27]

Answer: Undefined

Step-by-step explanation:

(slope = m)

m=\frac{y_2-y_1}{x_2-x_1}=\frac{5-2}{4-4}=\frac{3}{0}

(3 divided by 0 is undefined, so the slope is undefined)

4 0
3 years ago
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