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2 years ago

A set of elementary school student heights are normally distributed with a mean of 105 centimeters and a

Mathematics

1 answer:

Nana76 [90]2 years ago

7 0

Answer:

7.5

Step-by-step explanation:

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Taseem’s loose change consists of dimes and quarters. When you ask her how many of each type of coin she has, she gives you the

sweet-ann [11.9K]

The statement first, and the statement second are correct because the number of dimes is 34, and the number of quarters is 66

What is a linear equation?

It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.

If in the linear equation, one variable is present, then the equation is known as the linear equation in one variable.

We have:

0.1x + 0.25y = 19.9 and

x + y = 100

Let x = number of dimes

y = number of quarters

After solving the above system of equations by substitution method.

$\rm \dfrac{398-5y}{2}+y=100$

y = 66

$\rm x=\dfrac{398-5\cdot \:66}{2}$

x = 34

Thus, the statement first, and the statement second are correct because the number of dimes is 34, and the number of quarters is 66

Learn more about the linear equation here:

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2 years ago

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3 years ago

What is the perimeter of a 6-inch square? 20 inches 24 inches 18 inches 14 inches

Dmitry [639]

The perimeter of a square with a side measurement of 6 inches is equal to 24 inches.

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1 year ago

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Step-by-step explanation:

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2 years ago

Triangle ABC has vertices at A(−1 , 2), B(4, 2), and C(3, −1). Classify the triangle according to the side lengths.

Orlov [11]

Answer:

B) Isosceles

Step-by-step explanation:

The given triangle has vertices at A(-1,2), B(4,2) and C(3,-1).

We must first determine the length of the sides of the triangle, before we can classify it.

We apply the distance formula to find length of the sides.

$|AB|=\sqrt{(4--1)^2+(2-2)^2}$

$\Rightarrow |AB|=\sqrt{(4+1)^2+(2-2)^2}$

$\Rightarrow |AB|=\sqrt{5^2+(0)^2}$

$\Rightarrow |AB|=\sqrt{25}$

$\Rightarrow |AB|=5 units$ .

The length of side BC

$|BC|=\sqrt{(3-4)^2+(-1-2)^2}$

$\Rightarrow |BC|=\sqrt{(-1)^2+(-3)^2}$

$\Rightarrow |BC|=\sqrt{1+9}$

$\Rightarrow |BC|=\sqrt{10}$

The length of side AC

$|AC|=\sqrt{(3--1)^2+(-1-2)^2}$

We simplify to obtain;

$|AC|=\sqrt{(3+1)^2+(-3)^2}$

$\Rightarrow |AC|=\sqrt{(4)^2+(-3)^2}$

$|AC|=\sqrt{16+9}$

$|AC|=\sqrt{25}$

$|AC|=5\:units$

Since $|AC|=5\:units=|AB|$ , the given triangle is an isosceles triangle.

The correct answer is

6 0

3 years ago

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