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2 years ago

Easy question! What is 6(4+2)?

Mathematics

2 answers:

Hunter-Best [27]2 years ago

7 0

Answer:

Step-by-step explanation:

Hoping this helps!

EleoNora [17]2 years ago

3 0

Answer:

Step-by-step explanation:

6(4+2)

6(6)

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I need this one, answer please?

Stolb23 [73]

Answer:

Pair 1.

Step-by-step explanation:

On pair 1, shape A can go into shape B with a rotation.

Pair 2 would be a translation.

Pair 3 would be a reflection across the x-axis.

Pair 4 would be a reflection across the y-axis.

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3 years ago

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Christine received a $70 gift card for a coffee store. She used it in buying some coffee that cost $7.52 per pound. After buying

Sonbull [250]

Answer:

Step-by-step explanation:

take 70 and subtract the amount left, which is $22.56 divide that by 7.52 and you get your answer of 3

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3 years ago

An airplane on a transatlantic flight took 4 hours 20 minutes to get from New York to its destination, a distance of 3,000 miles

ankoles [38]

738.

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The closest to that is 738

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2 years ago

Pls someone answer pls :)

Goshia [24]

Answer:

(-x-2) (-2x -3)=0

=>[-(x +2)] [-(2x+3)]=0

=>(x + 2)(2x +3)=0

=>x =-2 or,x=-3/2

smaller x=-2

larger x =-3/2

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3 years ago

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The probability that an American CEO can transact business in a foreign language is .20. Twelve American CEOs are chosen at rand

NNADVOKAT [17]

Answer with Step-by-step explanation:

We are given that

The probability that an American CEO can transact business in foreign language=0.20

The probability than an American CEO can not transact business in foreign language= $1-0.20=0.80$

Total number of American CEOs chosen=12

a. The probability that none can transact business in a foreign language= $12C_0(0.20)^0(0.80)^{12}$

Using binomial theorem $nC_r(1-p)^{n-r}p^r$

The probability that none can transact business in a foreign language= $\frac{12!}{0!(12-0)!}(0.8)^{12}=(0.8)^{12}$

b.The probability that at least two can transact business in a foreign language= $1-P(x=0)-p(x=1)=1-((0.8)^{12}+12C_1(0.8)^{11}(0.2))=1-((0.8)^{12}+12(0.8)^{11}}(0.2))$

c.The probability that all 12 can transact business in a foreign language= $12C_{12}(0.8)^0(0.2)^{12}$

The probability that all 12 can transact business in a foreign language= $\frac{12!}{12!}(0.2)^{12}=(0.2)^{12}$

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3 years ago

Easy question! What is 6(4+2)?​

Easy question! What is 6(4+2)?