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3 years ago

Find the mode of the data set below.

Mathematics

2 answers:

Wewaii [24]3 years ago

8 0

Answer:

The mode is 88 as you can see there the 88 occurs twice.

Step-by-step explanation:

I hope this helps

kramer3 years ago

4 0

The mode is 88. The mode is a number that is most repeated in a pattern, and here it is 88.

Hope this helps!

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Marina sells candles that are in the shape of cones. she makes a yellow candle that has a diameter of 4 inches and a height of 1

Katena32 [7]

The correct statements are:

the volume of the yellow candle is about 46.05 inches cubed.
the radius of the yellow candle is 2 in
he base area of the yellow candle is about 6.28 inches squared.
the price of one yellow candle is about $16.12.

<h3>What are the correct statements?</h3>

Volume of a cone = 1/3(πr²h)

π = 3.14
r = radius = diameter / 2 = 4/2 = 2 inches
h = height

Volume of a cone = 1/3 x 2² x 11 x 3.14 = 46.05 inches cubed.

Base area = πr²

3.14 x 4 = 12.56 inches squared.

Price of one yellow candle = volume x price per cubic unit

46.05 x $0.35 = $16.12

To learn more about the volume of a cone, please check: brainly.com/question/13705125

#SPJ4

3 0

2 years ago

Solve this application problem using a system of equations: Dan and June mix two kinds of feed for pedigreed dogs. They wish to

wlad13 [49]

Answer:

50 pounds

Step-by-step explanation:

Dan and june mix two kind of feed for pedigreed dogs

Feed A worth is $0.26 per pound

Feed B worth is $0.40 per pound

Let x represent the cheaper amount of feed and y the costlier type of feed

x+y= 70..........equation 1

0.26x + 0.40y= 0.30×70

0.26x + 0.40y= 21.........equation 2

From equation 1

x + y= 70

x= 70-y

Substitutes 70-y for x in equation 2

0.26(70-y) + 0.40y= 21

18.2-0.26y+0.40y= 21

18.2+0.14y= 21

0.14y= 21-18.2

0.14y= 2.8

Divide both sides by the coefficient of y which is 0.14

0.14y/0.14= 2.8/0.14

y= 20

Substitute 20 for y in equation 1

x + y= 70

x + 20= 70

x= 70-20

x = 50

Hence Dan and june should use 50 pounds of the cheaper kind in the mix

4 0

3 years ago

I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).

Aleksandr-060686 [28]

Answer:

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

Step-by-step explanation:

$\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$

Let's first isolate the trig function.

Add 1 one on both sides:

$\sqrt{3} \tan(x-\frac{\pi}{8})=1$

Divide both sides by $\sqrt{3}$ :

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

Now recall $\tan(u)=\frac{\sin(u)}{\cos(u)}$ .

$\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$

$\frac{1}{\sqrt{3}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}$

The first ratio I have can be found using $\frac{\pi}{6}$ in the first rotation of the unit circle.

The second ratio I have can be found using $\frac{7\pi}{6}$ you can see this is on the same line as the $\frac{\pi}{6}$ so you could write $\frac{7\pi}{6}$ as $\frac{\pi}{6}+\pi$ .

So this means the following:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

is true when $x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

where $n$ is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

$x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

Add $\frac{\pi}{8}$ on both sides:

$x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi$

Find common denominator between the first two terms on the right.

That is 24.

$x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi$

$x=\frac{7\pi}{24}+n \pi$ (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval $[0,2\pi)$ .

So if $\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$ and we let $u=x-\frac{\pi}{8}$ , then solving for $x$ gives us:

$u+\frac{\pi}{8}=x$ ( I just added $\frac{\pi}{8}$ on both sides.)

So recall $0\le x$ .

Then $0 \le u+\frac{\pi}{8}$ .

Subtract $\frac{\pi}{8}$ on both sides:

$-\frac{\pi}{8}\le u$

Simplify:

$-\frac{\pi}{8}\le u$

So we want to find solutions to:

$\tan(u)=\frac{1}{\sqrt{3}}$ with the condition:

$-\frac{\pi}{8}\le u$

That's just at $\frac{\pi}{6}$ and $\frac{7\pi}{6}$

So now adding $\frac{\pi}{8}$ to both gives us the solutions to:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$ in the interval:

$0\le x$ .

The solutions we are looking for are:

$\frac{\pi}{6}+\frac{\pi}{8}$ and $\frac{7\pi}{6}+\frac{\pi}{8}$

Let's simplifying:

$(\frac{1}{6}+\frac{1}{8})\pi$ and $(\frac{7}{6}+\frac{1}{8})\pi$

$\frac{7}{24}\pi$ and $\frac{31}{24}\pi$

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

5 0

3 years ago

calculate the pH of a solution prepared by dissolving 1.75g of acetate, CH3COONa, in 60.5mL of 0.25M acetic acid, CH3COOH(aq). A

pentagon [3]

Answer:

PH= 5.00

Step-by-step explanation:

4 0

3 years ago

If one angle of a set of vertical angles measures 63°, the sum of the vertical angles is °?

algol13

Answer

63rrdddd

Step-by-step explanation:

7 0

3 years ago