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4 years ago

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:

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

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

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andreev551 [17]

Answer:

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Step-by-step explanation:

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3 years ago

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Does someone know the answer

nata0808 [166]

Answer:

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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

-1a-2=0

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

a=-2

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3 years ago

Read 2 more answers

) 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