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lyudmila [28]
3 years ago
15

Students conducted a survey and found out that 36% of their peers on campus had tattoos. If 100 students were surveyed, can thes

e students use the Normal approximation to study the proportion of students in the population who have tattoos?
Mathematics
1 answer:
Nadusha1986 [10]3 years ago
6 0

Answer:

Normal approximation can be used.

Step-by-step explanation:

We are given the following in the question:

Proportion of student that has tattoos =

p = 36\% = 0.36

Sample size, n = 100

Condition for normal approximation of sampling distribution:

np \geq 10\\n(1-p) \geq 10

Putting values, we get,

100(0.36) = 36 \geq 10\\100(1-0.36) = 64 \geq 10

Since both the conditions are satisfied normal approximation can be used to study the proportion of students in the population who have tattoos.

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Determine the standard form of the equation of the line that passes through (-2, 0) and (-8,5)
Naya [18.7K]

Answer:

-5x-6y=10      ←   in standard form

Step-by-step explanation:

The equation of a line in  standard form  is.

Ax+By=C

were

  • A is a positive integer and
  • B, C are integers

As the equation in  point-slope form

y-y_1=m\left(x-x_1\right)

where m is the slope and \left(x_1,\:y_1\right)  is a point on the line.  

as

\mathrm{Slope\:between\:two\:points}:\quad \mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}

\left(x_1,\:y_1\right)=\left(-2,\:0\right),\:\left(x_2,\:y_2\right)=\left(-8,\:5\right)

m=\frac{5-0}{-8-\left(-2\right)}

m=-\frac{5}{6}

using   m=-\frac{5}{6}  and \left(x_1,\:y_1\right)=\left(-2,\:0\right)  then

y-0=-\frac{5}{6}\left(x-\left(-2\right)\right)

-\frac{5}{6}\left(x-\left(-2\right)\right)=y-0

6\left(-\frac{5}{6}\left(x-\left(-2\right)\right)\right)=6y

-5\left(x+2\right)=6y

-5x-10=6y

-5x-6y=10      ←   in standard form

8 0
3 years ago
Identify where the given information is suficient for graphing a linear equation. Check all that apply.
oee [108]

Answer:

  • All points are applicable apart from the second

Step-by-step explanation:

<u>One point and the slope of the line </u>

  • Yes.

<u>One point </u>

  • No

<u>One of the intercepts and the slope of the line </u>

  • Yes, it is equivalent to first option

<u>Both the intercepts </u>

  • Yes, it is equivalent to two points

<u>Two points</u>

  • Yes
5 0
3 years ago
Read 2 more answers
Devide (a2+ 7a+ 12)÷ (a+3)​
arlik [135]

Answer:

(a - 3) • (a - 4)

Step-by-step explanation:

The first term is,  a2  its coefficient is  1 .

The middle term is,  -7a  its coefficient is  -7 .

The last term, "the constant", is  +12

Multiply the coefficient of the first term by the constant   1 • 12 = 12

Find two factors of  12  whose sum equals the coefficient of the middle term, which is   -7 .

     -12    +    -1    =    -13

     -6    +    -2    =    -8

     -4    +    -3    =    -7    That's it

Rewrite the polynomial splitting the middle term using the two factors, -4  and  -3

                    a2 - 4a - 3a - 12

Add up the first 2 terms, pulling out like factors :

                   a • (a-4)

             Add up the last 2 terms, pulling out common factors :

                   3 • (a-4)

Add up the four terms

                   (a-3)  •  (a-4)

Really hope this helps :)

           

5 0
3 years ago
Someone please help me with this, is very urgent
ioda

Answer:

They are congruent

Step-by-step explanation:

Because they size are same

5 0
3 years ago
Help me! l want the correct answers °_°​
Ket [755]

Answer:

see explanation

Step-by-step explanation:

(a)

The angle on the circumference is half the central angle , that is

∠ A = ∠ BOC ÷ 2 = 50° ÷ 2 = 25°

(b)

Δ AOC is isosceles ( OA = OC , radii of the same circle ) then the base angles are congruent , that is

∠ ACO = ∠ A = 25°

(c)

Δ BOC is isosceles ( OB = OC, radii of the same circle ) then the base angles are congruent, that is

∠ BCO = \frac{180-50}{2} = \frac{130}{2} = 65°

(d)

∠ ACO + ∠ BCO = 25° + 65° = 90° ( angle in a semicircle )

8 0
3 years ago
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