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Elenna [48]
3 years ago
13

Velma must make five bottles of lemonade for the kids playing outside. Each bottle requires 3⁄4 cup of sugar. How much sugar doe

s she need?
Mathematics
1 answer:
Ket [755]3 years ago
6 0
The total number of cups of sugar needed will be given by:
Total amount=(number of bottles)×(number of cups of sugar per box)
number of bottles to be made is 5 bottles
number of cups per box is 3/4
thus the total number of cups required will be:
Total amount=3/4×5=3 3/4 cups
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Write the standard form of the equation of the circle with center (7,1) that also contains the point (−1,−5).
Karo-lina-s [1.5K]

Answer:

(x-7)^2+(y-1)^2=100

Step-by-step explanation:

<u>1) Find the radius</u>

We can do this by using the distance equation with the centre (7,1) and the given point (-1,-5):

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} where the two points are (x_1,y_1) and (x_2,y_2)

Plug in the points (7,1) and (-1,-5)

d=\sqrt{(-1-7)^2+(-5-1)^2}\\d=\sqrt{(-8)^2+(-6)^2}\\d=\sqrt{64+36}\\d=\sqrt{100}\\d=10

Therefore, the radius of the circle is 10 units.

<u>2) Plug the data into the equation of a circle</u>

Equation of a circle (when not centred at the origin):

(x-h)^2+(y-k)^2=r^2 where the centre is (h,k) and r is the radius

Plug in the centre (7,1) as (h,k)

(x-7)^2+(y-1)^2=r^2

Plug in the radius 10

(x-7)^2+(y-1)^2=10^2\\(x-7)^2+(y-1)^2=100

I hope this helps!

4 0
3 years ago
A chi-square test for independence is being used to evaluate the relationship between two variables, one of which is classified
AysviL [449]

Answer:

<em>Degrees of freedom for independence  in chi-square statistic</em>

<em>ν = ( r-1) (s-1) =6</em>

Step-by-step explanation:

<u><em>Explanation</em></u><u>:-</u>

Given data  chi-square test for independence is being used to evaluate the relationship between two variables

<em>Given "A" is classified into 3 categories</em>

<em>Second 'B' is classified into 4 categories</em>

In this chi-square test, we test if two attributes A and B under consideration are independent or not

We will assume that

<em>Null Hypothesis : H₀</em>: The two variables are independent

<em>Degrees of freedom in chi-square test for independence </em>

<em>ν = ( r-1) (s-1)</em>

<em>Given data 'r' = 3  and  's' = 4</em>

<em>Degrees of freedom for independence </em>

<em>ν = ( r-1) (s-1) = ( 3-1) ( 4-1) = 2×3 =6</em>

<em>Test statistic</em>

<em>                        χ ²  =  ∑  </em>\frac{(O-E)^{2} }{E}<em></em>

<em></em>

4 0
3 years ago
**PLEASE HELP**
maksim [4K]

Answer:

Number 4

Step-by-step explanation:

Option B would yield more in the short term but after Option A surpasses it, it would become more profitable.

6 0
3 years ago
The measure of the vertex angle in an isosceles triangle is six less than four time the measure of the base angle. find the meas
blagie [28]
An isosceles triangle is a special type of triangle that has two equal sides. Consequently, the two base angles are also equal. Since the interior angles of a triangle should sum up to 180°, the general equation should be

180° = 2*Base angle + Vertex angle

Let the base angle be x. Thus, the vertex angle is equal to
Vertex angle = 4x - 6
Substituting,
180 = 2x + 4x - 6
x = 31°

Vertex angle = 4(31°) - 6 
Vertex angle = 118°
5 0
3 years ago
For a certain population of sea turtles, 18 percent are longer than 6.5 feet. A random sample of 90 sea turtles will be selected
nika2105 [10]

Answer:

The standard deviation of the sampling distribution of the sample proportion of sea turtles longer than 6.5 feet for samples of size 90 is of 0.0405

Step-by-step explanation:

Central Limit Theorem:

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

18 percent are longer than 6.5 feet.

This means that p = 0.18

A random sample of 90 sea turtles

This means that n = 90

What is the standard deviation of the sampling distribution of the sample proportion of sea turtles longer than 6.5 feet for samples of size 90

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.18*0.82}{90}} = 0.0405

The standard deviation of the sampling distribution of the sample proportion of sea turtles longer than 6.5 feet for samples of size 90 is of 0.0405

8 0
3 years ago
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