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Stolb23 [73]
3 years ago
13

Give an example of a set of five positive numbers that have a median of 10 and whose mean is larger than ten?

Mathematics
1 answer:
barxatty [35]3 years ago
3 0
The answer is {1, 2, 10, 50, 75}. It has a median of 10 and a mean larger than 10
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596 divided by 3 using long divison
Basile [38]

Answer:

198.6 with the remaining 6

Step-by-step explanation:

596/3

596/3 = 198.666666667 as a decimal form

596/3 = 198.67 in 2 decimal places

596/3 = 198.7 to the nearest tenth

596/3 = 198.67 to the nearest hundredth

596/3 = 198.667 to the nearest thousandth

3 0
3 years ago
analyze the diagram below and complete the instructions that follow . find the value of z. A.11 B.32 C.50 D.68​
masha68 [24]

Answer:

50

Step-by-step explanation:

x = √30² - 18² = √576 = 24

tan A = 18/24 = 3/4

∠A = ∠B

tan B = tan A = 24 / y = 3 / 4

y = (24 x 4) / 3 = 32

Z = y + 18 = 32 + 18 = 50

3 0
3 years ago
How would i do this problem <br>slope= -2, y-intercept = 7
marin [14]
SINCE the slope is -2 it would be like this -2/1 but first plot the y-intercept onto the graph and go down 2 and over 1.
4 0
3 years ago
Read 2 more answers
A purchaser of electric relays buys from two suppliers, A and B. Supplier A supplies two of every three relays used by the compa
inessss [21]

Answer:

The probability that at most 38 of the 60 relays are from supplier A is P(X≤38)=0.3409.

Step-by-step explanation:

We can model this question with a binomial distribution random variable.

The sample size is n=60.

The probability that the relay come from supplier A is p=2/3 for any relay.

If we use a normal aproximation, we have the mean and standard deviation:

\mu=np=60*(2/3)=40\\\\\sigma=\sqrt{npq}=\sqrt{60*(2/3)*(1/3)} =3.65

The probability that at most 38 of the 60 relays are from supplier A is P(X≤38)=0.3409:

P(X\leq38)=P(X

5 0
3 years ago
A rectangular prism is filled with 360 cubes with 1/2 unit side lengths. What is the volume of the prism in cubic units?
icang [17]
180 cause half of 360 is 180 in the length
7 0
3 years ago
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