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I am Lyosha [343]
2 years ago
11

Write down three integers, all less than 25, whose

Mathematics
2 answers:
MariettaO [177]2 years ago
5 0

Answer:

18 and 8

Step-by-step explanation:

Range= 18-8

Mean=(18+8)/2=13

Masja [62]2 years ago
3 0

Answer: 9 11 19

Step-by-step explanation

9 11 19

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What is the probability of an event that is certain?
horrorfan [7]

Answer:

1

Step-by-step explanation:

An event that is certain to happen has a probability of 1.

4 0
3 years ago
Solve for x in<br> 2 - 3(x + 4) = 8
AysviL [449]

Answer:

x = -6

Step-by-step explanation:

2 - 3(x + 4) = 8

=> 3(x + 4) = 2 - 8 = -6

=> x + 4 = -6/3 = -2

=> x = -2 - 4 = -6

7 0
3 years ago
Read 2 more answers
There are 2 yellow marbles and 11 blue marbles in a bag. You randomly choose one of the marbles. What is the probability of choo
Dahasolnce [82]

Answer:

The probability of choosing a yellow marble is 2/13

Step-by-step explanation:

There are 2 yellow and 11 blue marbles in the bag or a total of 13 marble

P(yellow) = number of yellow marbles / total marble

                = 2/13

The probability of choosing a yellow marble is 2/13

5 0
3 years ago
Read 2 more answers
GETT IT RIGHT AND GET BRAINLIEST
oksian1 [2.3K]

Answer:

18

Step-by-step explanation:

58 - |-40| =

58 -40=

18

7 0
4 years ago
Read 2 more answers
g In R simulate a sample of size 20 from a normal distribution with mean µ = 50 and standard deviation σ = 6. Hint: Use rnorm(20
Illusion [34]

Answer:

> a<-rnorm(20,50,6)

> a

[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905

[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501

Then we can find the mean and the standard deviation with the following formulas:

> mean(a)

[1] 50.72451

> sqrt(var(a))

[1] 7.470221

Step-by-step explanation:

For this case first we need to create the sample of size 20 for the following distribution:

X\sim N(\mu = 50, \sigma =6)

And we can use the following code: rnorm(20,50,6) and we got this output:

> a<-rnorm(20,50,6)

> a

[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905

[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501

Then we can find the mean and the standard deviation with the following formulas:

> mean(a)

[1] 50.72451

> sqrt(var(a))

[1] 7.470221

5 0
3 years ago
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