Please help me with this statistical question if you can, it would mean alot im really struggling right now

StartFraction 12 minus 3 y Over 2 EndFraction plus y StartFraction 2 y minus 4 Over y EndFraction for y = 3.

Neko [114]

Answer:
Step-by-step explanation:
The point slope form of the equation is expressed as
y - 7/2 = 1/2(x - 4)
Comparing with the point slope form of an equation which is expressed as
y - y1 = m(x - x1)
m represents slope
Therefore
m = 1/2
y1 = 7/2
x1 = 4
The slope intercept form of the equation of a straight line is expressed as
y = mx + c
Where c represents the intercept
Substituting y = 7/2, m = 1/2 and x = 4 into y = mx + c, it becomes
7/2 = 1/2 × 4 + c
7/2 = 2 + c
c = 2 - 7/2 = (4 - 7)/2 = - 3/2
Therefore, the y intercept is - 3/2

7 0

4 years ago

Read 2 more answers

A baker bakes 50 muffins. 1/5 of the muffins are chocolate chip. 1/2 of the muffins are blueberry. The rest are cinnamon. How ma

vitfil [10]

Answer:

15 muffins are cinnamon.

Step-by-step explanation:

Given that:

Number of muffins baked = 50 muffins

Chocolate chip muffins = 1/5 of 50 = $\frac{1}{5}*50$

Chocolate chip muffins = 10 muffins

Blueberry muffins = 1/2 of 50 = $\frac{1}{2}*50$

Blueberry muffins = 25

Cinnamon muffins = Muffins baked - chocolate chip muffins - blueberry muffins

Cinnamon muffins = 50 - 10 - 25

Cinnamon muffins = 50 - 35 = 15

Hence,

15 muffins are cinnamon.

5 0

3 years ago

A sphere and a cylinder have the same radius and height. The volume of the cylinder is 21m. what is the volume of the sphere.

Goshia [24]

Answer:

The volume of the sphere is 14m³

Step-by-step explanation:

Given

Volume of the cylinder = $21m^3$

Required

Volume of the sphere

Given that the volume of the cylinder is 21, the first step is to solve for the radius of the cylinder;

<em>Using the volume formula of a cylinder</em>

The formula goes thus

$V = \pi r^2h$

Substitute 21 for V; this gives

$21 = \pi r^2h$

Divide both sides by h

$\frac{21}{h} = \frac{\pi r^2h}{h}$

$\frac{21}{h} = \pi r^2$

The next step is to solve for the volume of the sphere using the following formula;

$V = \frac{4}{3}\pi r^3$

Divide both sides by r

$\frac{V}{r} = \frac{4}{3r}\pi r^3$

Expand Expression

$\frac{V}{r} = \frac{4}{3}\pi r^2$

Substitute $\frac{21}{h} = \pi r^2$

$\frac{V}{r} = \frac{4}{3} * \frac{21}{h}$

$\frac{V}{r} = \frac{84}{3h}$

$\frac{V}{r} = \frac{28}{h}$

Multiply both sided by r

$r * \frac{V}{r} = \frac{28}{h} * r$

$V = \frac{28r}{h}$ ------ equation 1

From the question, we were given that the height of the cylinder and the sphere have equal value;

This implies that the height of the cylinder equals the diameter of the sphere. In other words

$h = D$ , where D represents diameter of the sphere

Recall that $D = 2r$

So, $h = D = 2r$

$h = 2r$

Substitute 2r for h in equation 1

$V = \frac{28r}{2r}$

$V = \frac{28}{2}$

$V = 14$

Hence, the volume of the sphere is 14m³

4 0

3 years ago

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Find the remainder when the function is <br>

prisoha [69]

according to the question

3x-1=0

3x=1

x=⅓

so

f(x)=18x³+x-1

f(⅓)=18.(⅓)³+⅓-1

f(⅓)=18.⅓.⅓.⅓+⅓-1

f(⅓)=6.⅑+⅓-1

f(⅓)=⅔+½-1

f(⅓)=0

<h3>therefore</h3><h3> the remainder is 0</h3>

4 0

3 years ago

How many equivalence relations are there on the set 1, 2, 3]?

Alex787 [66]

Answer:

We need to find how many number of equivalence relations are on the set {1,2,3}

A relation is an equivalence relation if it is reflexive, transitive and symmetric.

equivalence relation R on {1,2,3}

1.For reflexive, it must contain (1,1),(2,2),(3,3)

2.For transitive, it must satisfy: if (x,y)∈R then (y,x)∈R

3. For symmetric, it must satisfy: if (x,y)∈R,(y,z)∈R then (x,z)∈R

Since (1,1),(2,2),(3,3) must be there is R, (1,2),(2,1),(2,3),(3,2),(1,3),(3,1). By symmetry,

we just need to count the number of ways in which we can use the pairs (1,2),(2,3),(1,3) to construct equivalence relations.

This is because if (1,2) is in the relation then (2,1) must be there in the relation.

the relation will be an equivalence relation if we use none of these pairs (1,2),(2,3),(1,3) . There is only one such relation: {(1,1),(2,2),(3,3)}

we can have three possible equivalence relations:

{(1,1),(2,2),(3,3),(1,2),(2,1)}

{(1,1),(2,2),(3,3),(1,3),(3,1)}

{(1,1),(2,2),(3,3),(2,3),(3,2)}

6 0

4 years ago