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2 years ago

In Ellen’s sixth grade class there are 14 boys and 11 girls write each ratio using numbers in two ways

Mathematics

1 answer:

cestrela7 [59]2 years ago

7 0

Answer: 14:11

14/11

14 to 11

Step-by-step explanation:

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Find the square roots of the following numbers. Give the steps correctly.

Novay_Z [31]

Answer:

the sqare root of 19.9 is 4.46094160464

3 0

3 years ago

Someone please help...

Viefleur [7K]

Answer:

2 1/3

1 4/3

2.333...

233.3...%

Step-by-step explanation:

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2 years ago

What are the endpoint coordinates for the midsegment of △PQR that is parallel to PQ¯¯¯¯¯?

andriy [413]

Answer:

M(x₄ ,y₄) = (-3.5 , 0.5) and

N (x₅ ,y₅) = ( -1 , -0.5 )

Step-by-step explanation:

Let the endpoint coordinates for the mid segment of △PQR that is parallel to PQ be

M(x₄ ,y₄) and N(x₅ ,y₅) such that MN || PQ

point P( x₁ , y₁) ≡ ( -3 ,3 )

point Q( x₂ , y₂) ≡ (2 , 1 )

point R( x₂ , y₂) ≡ (-4 , -2)

To Find:

M(x₄ ,y₄) = ? and

N (x₅ ,y₅) = ?

Solution:

We have Mid Point Formula as

$Mid\ point(x,y)=(\frac{x_{1}+x_{2} }{2}, \frac{y_{1}+y_{2} }{2})$

As M is the mid point of PR and N is the mid point of RQ so we will have

$Mid\ pointM(x_{4} ,y_{4})=(\frac{x_{1}+x_{3} }{2}, \frac{y_{1}+y_{3} }{2})$

$Mid\ pointN(x_{5} ,y_{5})=(\frac{x_{2}+x_{3} }{2}, \frac{y_{2}+y_{3} }{2})$

Substituting the given value in above equation we get

$Mid\ pointM(x_{4} ,y_{4})=(\frac{-3+-4 }{2}, \frac{3+-2} }{2})$

∴ $Mid\ pointM(x_{4} ,y_{4})=(\frac{-7} }{2}, \frac{1}{2})$

∴ $Mid\ pointM(x_{4} ,y_{4})=(-3.5, 0.5)$

Similarly,

$Mid\ pointN(x_{5} ,y_{5})=(\frac{2+-4 }{2}, \frac{1+-2 }{2})$

∴ $Mid\ pointN(x_{5} ,y_{5})=(\frac{-2 }{2}, \frac{-1}{2})$

∴ $Mid\ pointN(x_{5} ,y_{5})=(-1, -0.5)$

∴ M(x₄ ,y₄) = (-3.5 , 0.5) and

N (x₅ ,y₅) = ( -1 , -0.5 )

3 0

2 years ago

The proportion of households in a region that do some or all of their banking on the Internet is 0.31. In a random sample of 100

Alenkasestr [34]

Answer:

Approximate probability that the number of households that use the Internet for banking in a sample of 1000 is less than or equal to 130 is less than 0.0005% .

Step-by-step explanation:

We are given that let X be the number that do some or all of their banking on the Internet.

Also; Mean, $\mu$ = 310/1000 or 0.31 and Standard deviation, $\sigma$ = 14.63/1000 = 0.01463 .

We know that Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

Probability that the number of households that use the Internet for banking in a sample of 1000 is less than or equal to 130 is given by P(X <= 130/1000);

P(X <=0.13) = P( $\frac{X-\mu}{\sigma}$ <= $\frac{0.13-0.31}{0.01463}$ ) = P(Z <= -12.303) = P(Z > 12.303)