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lara [203]
2 years ago
11

Is x = 4 a solution to 2x - 6 + 4 = 14? Yes or no

Mathematics
1 answer:
astra-53 [7]2 years ago
4 0

Answer:

no

Step-by-step explanation:

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What is HCF and LCM ? # confused
GaryK [48]
HCF= highest common factor
LCM= least common multiple
7 0
3 years ago
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Help it time and i have no answer left on branly​
Marina CMI [18]

Answer:

4/3

10/3

24/3

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The weights (in pounds) of 18 preschool children are
ivanzaharov [21]

Answer:

22.8 and 37.5

Step-by-step explanation:

Given the

The weights (in pounds) of 18 preschool children are

32, 20, 25, 27, 31, 22, 30, 23, 44, 37, 33, 45, 41, 24, 34, 21, 39, 29

To find its 20th percentile and 75th percentile.

In ascending order we get like this

Position  X (Asc. Order)

1  20

2  21

3  22

4  23

5  24

6  25

7  27

8  29

9  30

10  31

11  32

12  33

13  34

14  37

15  39

16  41

17  44

18  45

Percentile position = (no of entries +1)20/100 = 19/5 = 3.8

Since posiiton is not integer we use interpolation method.

The value of 3.8 - 3 = 0.8 corresponds to the proportion of the distance between 22 and 23 where the percentile we are looking for is located at.

Hence 20th percentile = 22+0.8(23-22)=22.8

So answer is 22.8

----------------------

75th percentile

Percentile posiiton = 19(75)/100 = 14.25

75th percentile= 37+0.25(39-37)=37.5

6 0
3 years ago
How many students must be randomly selected to estimate the mean weekly earnings of students at one college? We want 95% confide
kari74 [83]

Answer:

The sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:

CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}

The margin of error of a (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:

MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}

The information provided is:

<em>σ</em> = $60

<em>MOE</em> = $2

The critical value of <em>z</em> for 95% confidence level is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Compute the sample size as follows:

MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}

       n=[\frac{z_{\alpha/2}\times \sigma }{MOE}]^{2}

          =[\frac{1.96\times 60}{2}]^{2}

          =3457.44\\\approx 3458

Thus, the sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.

8 0
3 years ago
On the first day of ticket sales the school sold 1 senior citizen ticket and 5 child tickets for a total of $60. The school took
Stells [14]

Answer:

Step-by-step explanation:

Let's start off with the first equation

5x + y = 60

We need to find what X and Y are.

Let's try subtracting a few numbers from 60.

Let's do 60 - 55 = 5

11x5 = 55

11 = Children Ticket Cost

5 = Seniors Ticket Cost

Without them being the same price!

So using the numbers we found, we can now solve the second one

14x5 = 70

11x11 = 121

70 + 121 = 191

There we go!

Hope this helped! Please give brainiest if you can :)

3 0
3 years ago
Read 2 more answers
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