Can someone solve? i’ll mark brainliest!

Mathematics

1 answer:

Vlad1618 [11]3 years ago

5 0

If you are looking for x the answer would be -3 because 2•-3=-6. -6+12=6
Hope this helps!

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You can copy the data into Excel by highlighting the data, right-clicking and selecting Copy, then opening Excel, clicking on a

Iteru [2.4K]

Answer:

Low income group :

Mean = 22.100

Median = 22.000

Standard deviation = 4.717

High Income group :

Mean = 19.771

Median = 20.000

Standard deviation = 3.255

Step-by-step explanation:

Low income : 19,20,13,28,21,27,16,21,19, 19,34,24, 24,14,24,25,23,22,20,21,14,23,17,26,24,24,20,24,27,16,22,17,29,20,25,22,20,34,24,22

High Income : 23 19 19 18 20 18 17 25 13 24 25 17 21 19 17 22 22 15 19 23 19 23 22 20 20 20 13 21 16 13 21 24 23 21 20

Using Excel :

Mean of low income, Σx / n = 884/40 = 22.100

Median of low income group : 1/2(n+1)th term = Median of low income = 22.000

Standard deviation of low income group :

√(Σ(x - mean)²/n) = 4.717

High Income group :

Mean of high income, Σx / n = 692/35 = 19.771

Median of high income group : 1/2(n+1)th term = Median of high income = 20.000

Standard deviation of low income group :

√(Σ(x - mean)²/n) = 3.255

5 0

3 years ago

I need help don’t know how to do

ivann1987 [24]

Let x be the unknown angle.
We know that x+50=120 because the value of an exterior angle of a triangle is equal to the sum of the other 2 interior angles.

x+50 = 120
Subtract both sides by 50
x = 70

The value of the unknown angle is 70 degrees.

Have an awesome day! :)

5 0

3 years ago

Read 2 more answers

Suppose that a study of elementary school students reports that the mean age at which children begin reading is 5.4 years with a

Tpy6a [65]

Answer:

By the Central Limit Theorem, the mean of the sampling distribution of sample means would be 5.4 years.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .