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Ahat [919]
3 years ago
10

This is 3rd grade math. Also here is the question,

Mathematics
2 answers:
nika2105 [10]3 years ago
6 0

Answer:

The exact amount of people is 5,173 and the estimate amount is 5,000

Step-by-step explanation:

How I got 5,173 is simple. You take the amount of people that chose each location, and add both of those totals together. You would then get 5,173 if you wanted the exact answer. If you wanted to get the estimate amount here’s what you do. The first step is to take the first place, which is the beach, or you can choose the other place, the amusement park. Next is you estimate the amount of the total number. To estimate 2,862 (beach) or 2,311 (amusement park) you look at the first number which is 2 in this scenario. you then look at the next number which is either 8 (beach) or 3 (amusement park) then you tell yourself “is it greater then 5 or exactly 5?” if the number solves that answer then increase the first number which is 2. You then increase the 2 to 3. Now you do this for the next number. And the final step is after you estimate both numbers you add the estimates together. After you get the total (to know whether you did it correctly the answer has to end in 0’s) that is your estimate answer. (SORRY THIS WAS REALLY LONG, I JUST HAD ALOT OF INFORMATION TO GIVE. lol)

Aneli [31]3 years ago
3 0

Answer:

The exact answer is 5,173. The rounded estimate (rounded to the nearest tenth) is 5,200.

Step-by-step explanation:

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Find the area under the standard normal probability distribution between the following pairs of​ z-scores. a. z=0 and z=3.00 e.
prohojiy [21]

Answer:

a. P(0 < z < 3.00) =  0.4987

b. P(0 < z < 1.00) =  0.3414

c. P(0 < z < 2.00) = 0.4773

d. P(0 < z < 0.79) = 0.2852

e. P(-3.00 < z < 0) = 0.4987

f. P(-1.00 < z < 0) = 0.3414

g. P(-1.58 < z < 0) = 0.4429

h. P(-0.79 < z < 0) = 0.2852

Step-by-step explanation:

Find the area under the standard normal probability distribution between the following pairs of​ z-scores.

a. z=0 and z=3.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 3.00) = 0.9987

Thus;

P(0 < z < 3.00) = 0.9987 - 0.5

P(0 < z < 3.00) =  0.4987

b. b. z=0 and z=1.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 1.00) = 0.8414

Thus;

P(0 < z < 1.00) = 0.8414 - 0.5

P(0 < z < 1.00) =  0.3414

c. z=0 and z=2.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 2.00) = 0.9773

Thus;

P(0 < z < 2.00) = 0.9773 - 0.5

P(0 < z < 2.00) = 0.4773

d.  z=0 and z=0.79

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 0.79) = 0.7852

Thus;

P(0 < z < 0.79) = 0.7852- 0.5

P(0 < z < 0.79) = 0.2852

e. z=−3.00 and z=0

From the standard normal distribution tables,

P(Z< -3.00) = 0.0014  and P(Z< 0) = 0.5

Thus;

P(-3.00 < z < 0 ) = 0.5 - 0.0013

P(-3.00 < z < 0) = 0.4987

f. z=−1.00 and z=0

From the standard normal distribution tables,

P(Z< -1.00) = 0.1587  and P(Z< 0) = 0.5

Thus;

P(-1.00 < z < 0 ) = 0.5 -  0.1586

P(-1.00 < z < 0) = 0.3414

g. z=−1.58 and z=0

From the standard normal distribution tables,

P(Z< -1.58) = 0.0571  and P(Z< 0) = 0.5

Thus;

P(-1.58 < z < 0 ) = 0.5 -  0.0571

P(-1.58 < z < 0) = 0.4429

h. z=−0.79 and z=0

From the standard normal distribution tables,

P(Z< -0.79) = 0.2148  and P(Z< 0) = 0.5

Thus;

P(-0.79 < z < 0 ) = 0.5 -  0.2148

P(-0.79 < z < 0) = 0.2852

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A multiple must be greater than or equal to the largest starting number. True or False
podryga [215]

I'll use multiples of 2 and 4 as an example:

Multiples of 2: 2, 4, 6, 8...

Multiples of 4: 4, 8, 12, 16...

The least common multiple in this case is 4. The LCM is always ≥ the largest starting number, which is 4 for this example. Therefore, the statement is true.

<em>Hope this helps! :)</em>

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