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3 years ago

I need some help with this question, I got a different answer.

Mathematics

1 answer:

N76 [4]3 years ago

5 0

Answer:

The median is the middle number.

Step-by-step explanation:

You have to order the numbers from least to greatest, or greatest to least, and choose the number in the middle. Since there is an even number of values, you have to take the mean of the two middle numbers. I hope this helps.

You might be interested in

I am thinking of a number when I double my number and then subtract the resulting from I get -1 what is my number

igor_vitrenko [27]

Answer: 1

Step-by-step explanation:

-1 + 2 = 1 ????

4 0

3 years ago

Janie and Jasmine are playing three games at an arcade. Each of the games requires either 2, 3, or 4 tokens. The girls plan to p

Brrunno [24]

Answer:

m+2n+2p=16 tokens or m+2(n+p)= 16 tokens

Step-by-step explanation:

First, we must understand how many tokens they have.

Janie played the 3-token game four times, so she used 12 tokens (3n x 4 = 12). Jasmine played the 4-token game five times, so she used 20 tokens (4p x 5 = 20).

Added up, the girls have 32 tokens (12+20 = 32 tokens (t)).

In order to play a round of the three games, 9 tokens are required (m+n+p = 9 / 2+3+4 = 9) by each girl. (32t-18t=14) 14 tokens are left after taking a ride in every game each. 14 left divided by 2 will leave 7 more tokens for each girl. You can split these 7 tokens into an n+p = 7 tokens

Thus,

Each girl can ride as follows:

m+2n+2p=16 tokens or m+2(n+p)= 16 tokens

4 0

3 years ago

Please help fast will give brainleist

egoroff_w [7]

Answer:

The picture wont load im sorry

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Is a < 2 ?

Vladimir79 [104]

Answer:

yes to both

Step-by-step explanation:

(1) For y=1, and x=a, ...

a² + 1² ≤ 3

-√2 ≤ a ≤ √2 . . . . . subtract 1 and take the square root

All values of "a" are less than 2 inside the circle.

(2) For y=4 and x=a, ...

2·4 +4·a = 10

4a = 2 . . . . . . . subtract 8

a = 1/2 . . . . . . .divide by 4

The value of "a" for point (a, 4) on the line is less than 2.

4 0

3 years ago

The probability that a lab specimen contains high levels of contamination is 0.10. Five samples are checked, and the samples are

raketka [301]

Answer:

(a) 0.59049 (b) 0.32805 (c) 0.40951

Step-by-step explanation:

Let's define

$A_{i}$ : the lab specimen number i contains high levels of contamination for i = 1, 2, 3, 4, 5, so,

$P(A_{i})=0.1$ for i = 1, 2, 3, 4, 5

The complement for $A_{i}$ is given by

$A_{i}^{$c$}$ : the lab specimen number i does not contains high levels of contamination for i = 1, 2, 3, 4, 5, so

P( $A_{i}^{$c$}$ )=0.9 for i = 1, 2, 3, 4, 5

(a) The probability that none contain high levels of contamination is given by

P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )= $(0.9)^{5}$ =0.59049 because we have independent events.

(b) The probability that exactly one contains high levels of contamination is given by

P( $A_{1}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}$ )=5×(0.1)× $(0.9)^{4}$ =0.32805

because we have independent events.

P( $A_{1}$ ∪ $A_{2}$ ∪ $A_{3}$ ∪ $A_{4}$ ∪ $A_{5}$ )=1-P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )=1-0.59049=0.40951

6 0

3 years ago