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ankoles [38]
3 years ago
6

What is 329,625 rounded to the place value of the underlined digit?

Mathematics
2 answers:
dexar [7]3 years ago
4 0

Answer:

330,000

Step-by-step explanation:

if the underlined digit is 9, then the six rounds the 9 up to a 10, making the rounded number amount to 330,000.

Ulleksa [173]3 years ago
3 0
The answer is 330,000
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Let U = {1, 2, 3, 4, 5, 6, 7}, A= {1, 3, 4, 6}, and B= {3, 5, 6}. Find the set A’ U B’
Art [367]

Answer:

Step-by-step explanation:

A'={2,5,7}

B'={1,2,4,7}

A'UB'={1,2,4,5,7}

8 0
3 years ago
If there are four teams in a league, how many games will have to be played so that each team plays every other team once?
Sedaia [141]

Answer:

6 matches

Step-by-step explanation:

Let’s call the teams A B C D

A will play B C D = 3 matches

B will play only C and D as it already played A, making 2 matches

C will play D, making 1 match

D has already played all

Total number of matches is thus 3 + 2 + 1 = 6 matches

3 0
3 years ago
Order the following decimals from least to greatest. 9.036, 9.982, 9.535
Leona [35]
The order is: 9.036, 9.535, and 9.982.
8 0
3 years ago
Suppose that 40 percent of the drivers stopped at State Police checkpoints in Storrs on Spring Weekend show evidence of driving
lesantik [10]

Answer:

a) 0.778

b) 0.9222

c) 0.6826

d) 0.3174

e) 2 drivers

Step-by-step explanation:

Given:

Sample size, n = 5

P = 40% = 0.4

a) Probability that none of the drivers shows evidence of intoxication.

P(x=0) = ^nC_x P^x (1-P)^n^-^x

P(x=0) = ^5C_0  (0.4)^0 (1-0.4)^5^-^0

P(x=0) = ^5C_0 (0.4)^0 (0.60)^5

P(x=0) = 0.778

b) Probability that at least one of the drivers shows evidence of intoxication would be:

P(X ≥ 1) = 1 - P(X < 1)

= 1 - P(X = 0)

= 1 - ^5C_0 (0.4)^0 * (0.6)^5

= 1 - 0.0778

= 0.9222

c) The probability that at most two of the drivers show evidence of intoxication.

P(x≤2) = P(X = 0) + P(X = 1) + P(X = 2)

^5C_0  (0.4)^0  (0.6)^5 + ^5C_1  (0.4)^1  (0.6)^4 + ^5C_2  (0.4)^2  (0.6)^3

= 0.6826

d) Probability that more than two of the drivers show evidence of intoxication.

P(x>2) = 1 - P(X ≤ 2)

= 1 - [^5C_0  (0.4)^0  (0.6)^5 + ^5C_1  (0.4)^1  (0.6)^4 + ^5C_2 * (0.4)^2  (0.6)^3]

= 1 - 0.6826

= 0.3174

e) Expected number of intoxicated drivers.

To find this, use:

Sample size multiplied by sample proportion

n * p

= 5 * 0.40

= 2

Expected number of intoxicated drivers would be 2

7 0
3 years ago
In one game, the ratio of three-point baskets to three-point tries for one team was 3:8. If the team scored 27 points from three
nasty-shy [4]

Since we’ve been given a ratio, 3:8, we can convert that into a fraction and form an equation from it:

\frac{3}{8} = \frac{27/3}{?}

The reason the numerator on the second fraction is 27/3 is because the team scored 27 points, not 27 baskets.

Simplify (Also replace the ? with an unknown variable):

\frac{3}{8} = \frac{9}{n}

Cross multiply:

3*n = 9*8

Simplify:

3n = 72

Divide both sides by 3:

n = 24

The team had 33 three point tries, 9 successes, and 24 fails.

-T.B.

7 0
3 years ago
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