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3 years ago

2*10⁴-7³ please help

Mathematics

1 answer:

melisa1 [442]3 years ago

7 0

Answer:

19657

Step-by-step explanation:

By PEMDAS, we know we have to calculate the exponents first, then multiply the 2, and then subtract the 7³

2*10⁴ = 20000

20000 - 7³ = 19657

This is the third time I've seen this question. Are you posting this question again and again?

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Complete the equation of the line through (−8,8), and (1,-10 ). Use exact numbers. Help me please my dog just died and I want to

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Answer: (Mayadc821 wrote this anwser check theyre account for more)

Since you know the x and y values, you just have to plug these into the linear slope-intercept equation:

y = mx + b

-10 = m(1) + b

m + b = -10

b = -10 - m

Now that we have a value for b, we can plug this into the other equation:

8 = m(-8) + b

8 = -8m + (-10 - m)

8 = -9m -10

18 = -9m

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Now that we know what m is equal to, we can plug this into our first equation to get an answer for b:

b = -10 - m

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3 years ago

A box contains 5 tickets numbered 1,2,3,4, and 5. Two tickets are drawn at random from the box. Find the chance that the numbers

adelina 88 [10]

Answer:

(a) The probability that the numbers on the two tickets differ by two or more if the draws are made with replacement is 0.48.

(b) The probability that the numbers on the two tickets differ by two or more if the draws are made without replacement is 0.60.

Step-by-step explanation:

The tickets are drawn such that the difference between the two numbers is at least 2.

That is, 1st number - 2nd number ≥ 2.

The sample space such that this condition is satisfied is:

S = {(1, 3), (1, 4), (1, 5), (2, 4), (2, 5) and (3, 5)} = 6 possible pairs.

But there is also case where the numbers can be drawn in reverse order, i.e we can draw (3, 1) instead of (1, 3).

This makes the total number of possible pairs as, 6 × 2 = 12 pairs.

(a) With Replacement

In case the tickets are selected with replacement, then the probability of selecting the 1st ticket is same as for the 2nd ticket is:

$P(Difference\geq 2)=P(1st\ ticket)\times P(2nd\ ticket)\times No.\ of\ Possible\ pairs$

$=\frac{1}{5}\times \frac{1}{5}\times12\\ =0.48$

Thus, the probability that the numbers on the two tickets differ by two or more if the draws are made with replacement is 0.48.

(b) Without Replacement:

In case the tickets are selected without replacement, then the probability of selecting the 1st ticket is same as for the 2nd ticket is:

$P(Difference\geq 2)=P(1st\ ticket)\times P(2nd\ ticket)\times No.\ of\ Possible\ pairs$

$=\frac{1}{5}\times \frac{1}{4}\times12\\ =0.60$

Thus, the probability that the numbers on the two tickets differ by two or more if the draws are made without replacement is 0.60.

7 0

3 years ago

2*10⁴-7³ please help​

2*10⁴-7³ please help