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yKpoI14uk [10]
3 years ago
9

Solve 7x - 2 = 5x + 14

Mathematics
1 answer:
Alexxx [7]3 years ago
4 0

Answer:

8

Step-by-step explanation:

7x - 5x = 14 + 2

2x = 16

x = 16/2

x = 8

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PLEASE HELP I WILL GIVE BRAINLIEST
marysya [2.9K]

Answer:

54

Step-by-step explanation:

Multiply:

<em>37x37=72</em>

<em>Subtract:</em>

<em>180-72=108</em>

<em>Divide:</em>

<em>108/2=54</em>

6 0
2 years ago
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Solve the proportion -4/7 = 6/2y+5
Veseljchak [2.6K]

Hope this is helpful :)

4 0
2 years ago
Fifty people were surveyed about their preference between chocolate and vanilla cake. The following two
GuDViN [60]

Answer:

TOTAL

Male

18

6

4

28

Female

15

5

2

22

TOTAL

33

November

6

50

In This sample, are the events "female" and "prefers chocolate" mutually exclusive

Step-by-step explanation:

7 0
3 years ago
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Construct a​ 99% confidence interval for the population​ mean, mu. Assume the population has a normal distribution. A group of 1
Zarrin [17]

Answer:

99% confidence interval for the population​ mean is [19.891 , 24.909].

Step-by-step explanation:

We are given that a group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.

Assuming the population has a normal distribution.

Firstly, the pivotal quantity for 99% confidence interval for the population​ mean is given by;

         P.Q. = \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, \bar X = sample mean age of selected students = 22.4 years

             s = sample standard deviation = 3.8 years

             n = sample of students = 19

             \mu = population mean

<em>Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.</em>

So, 99% confidence interval for the population​ mean, \mu is ;

P(-2.878 < t_1_8 < 2.878) = 0.99  {As the critical value of t at 18 degree of

                                                freedom are -2.878 & 2.878 with P = 0.5%}

P(-2.878 < \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } < 2.878) = 0.99

P( -2.878 \times {\frac{s}{\sqrt{n} } } < {\bar X - \mu} < 2.878 \times {\frac{s}{\sqrt{n} } } ) = 0.99

P( \bar X -2.878 \times {\frac{s}{\sqrt{n} } < \mu < \bar X +2.878 \times {\frac{s}{\sqrt{n} } ) = 0.99

<u>99% confidence interval for</u> \mu = [ \bar X -2.878 \times {\frac{s}{\sqrt{n} } , \bar X +2.878 \times {\frac{s}{\sqrt{n} } ]

                                                 = [ 22.4 -2.878 \times {\frac{3.8}{\sqrt{19} } , 22.4 +2.878 \times {\frac{3.8}{\sqrt{19} } ]

                                                 = [19.891 , 24.909]

Therefore, 99% confidence interval for the population​ mean is [19.891 , 24.909].

6 0
3 years ago
The diagram shows the entrance to a tunnel.
gladu [14]

Step-by-step explanation:

it filled up half the circle (up to the center point) - if we had a full circle. but a little bit is cut off (below AB).

what we see is that the shaded area is the sum of the area of the triangle AOB and 2 equally sized circle segment areas left and right of AOB.

since we are dealing with a half-circle, we have 180° in total. 120° are taken by AOB, so, that leaves us with 180-120 = 60° for both circle segments (so, one has an angle of 30°).

and 2×30° = 1×60°, so we can calculate the area of one 60° segment instead of two 30° segments.

AOB is an isoceles triangle (the legs are equally long, and therefore also the 2 side angles are equal).

the area of this triangle AOB is

1/2 × a × b × sin(C) = 1/2 × 3 × 3 × sin(120) =

= 3.897114317... m²

a circle segment area of 60° is 60/360 = 1/6 of the full circle area (as a full circle = 360°).

so, it's area is

pi×r² × 1/6 = pi×3²/6 = pi×3/2 = 4.71238898... m²

so, the total area of the shaded area is

3.897114317... m² + 4.71238898... m² =

= 8.609503297... m²

5 0
2 years ago
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