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2 years ago

Which data value has the highest frequency? 1 1/6 3 1/6 3/8 5/8

Mathematics

1 answer:

Marat540 [252]2 years ago

8 0

Answer:

3 1/6

Step-by-step explanation:

To find the highest value, we have to evaluate each term individually,

1) 1 1/6 = 7/6 = 1.167

2) 3 1/6 = 19/6 = 3.16

3) 3/8 = 0.375

4) 5/8 = 0.625

Therefore, second data value (3 1/6) is highest.

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Is d= -12 a solution to the equation 8d+6= -8d-186

kupik [55]

Answer:

It is a solution

Step-by-step explanation:

8d+6= -8d-186

Add 8d to each side

8d+8d+6= -8d+8d-186

16d +6 = -186

Subtract 6 from each side

16d +6-6 = -186 -6

16d = -192

Divide each side by 16

16d/16 = -192/16

d = -12

8 0

2 years ago

Can you that special son plz help me out mad let it be correct!!!

yarga [219]

Answer:Yes

Hope this helps!

Step-by-step explanation:

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2 years ago

Read 2 more answers

HELP ME PLEASE I'll give brainless.

Lana71 [14]

Answer:

∠FGI=40°

Step-by-step explanation:

CE║FH ⇒ ∠CDG=∠FGI=40°

3 0

2 years ago

The average cost of a home in a certain neighborhood ranges from $225,000 to $385,000. Write an absolute value inequality to mod

nataly862011 [7]

Answer:

D. |C -305,000| ≤ 80,000

Step-by-step explanation:

Such an equation will say that the difference from the average of these values is at most half of their difference.

| C -(225000+385000)/2 | ≤ (385000 -225000)/2

|C -305,000| ≤ 80,000 . . . . . matches choice D

8 0

3 years ago

Given the following right triangle, find cose, sine, tane sececsce, and cote. Do not

IgorLugansk [536]

Step-by-step explanation:

Use the Pythagorean theorem:

$leg^2+leg^2=hypotenuse^2$

We have

$leg=4,\ hypotenuse=10$

Substitute:

$4^2+leg^2=10^2$

$16+leg^2=100$ <em>subtract 16 from both sides</em>

$leg^2=84\to leg=\sqrt{84}\\\\leg=\sqrt{(4)(21)}\\\\leg=\sqrt4\cdot\sqrt{21}\\\\leg=2\sqrt{21}$

$sine=\dfrac{opposite}{hypotenuse}\\\\cosine=\dfrac{adjacent}{hypotenuse}\\\\tangent=\dfrac{opposite}{adjacent}\\\\cotangent=\dfrac{adjacent}{opposite}\\\\secant=\dfrac{hypotenuse}{adjacent}\\\\cosecant=\dfrac{hypotenuse}{opposite}$

Substitute:

$hypotenuse=10,\ opposite=4,\ adjacent=2\sqrt{21}$

$\sin\theta=\dfrac{4}{10}=\dfrac{2}{5}\\\\\cos\theta=\dfrac{2\sqrt{21}}{10}=\dfrac{\sqrt{21}}{5}\\\\\tan\theta=\dfrac{4}{2\sqrt{21}}=\dfrac{2}{\sqrt{21}}\cdot\dfrac{\sqrt{21}}{\sqrt{21}}=\dfrac{2\sqrt{21}}{21}\\\\\cot\theta=\dfrac{2\sqrt{21}}{4}=\dfrac{\sqrt{21}}{2}\\\\\sec\theta=\dfrac{10}{2\sqrt{21}}=\dfrac{5}{\sqrt{21}}\cdot\dfrac{\sqrt{21}}{\sqrt{21}}=\dfrac{5\sqrt{21}}{21}\\\\\csc\theta=\dfrac{10}{4}=\dfrac{5}{2}$

5 0

3 years ago