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Lisa [10]
3 years ago
5

Solve the given equation for h B=V/h Solve the given equation for V B=V/h

Mathematics
1 answer:
sergeinik [125]3 years ago
8 0

Answer:

Step-by-step explanation:

B = V/h

V/h = B

V = Bh

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What is the median of the data shown in the box plot below?<br>A:23<br>B:28<br>C:30<br>D:35​
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It has to be 25 it is the middle number

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This talented group of dogs has 25 yards of ribbon. They want to cut the ribbon to make bows and awards. If each piece of ribbon
lbvjy [14]

Answer:

18.75 bows

Step-by-step explanation:

This talented group of dogs has 25 yards of ribbon. They want to cut the ribbon to make bows and awards. If each piece of ribbon needs to be 3/4 of a yard, how many bows will they be able to make?

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3/4 × 25 yards

= 18.75 bows.

8 0
3 years ago
Verify the following identity. <br><br> <img src="https://tex.z-dn.net/?f=-%5Cfrac%7Bcos%283x%29%2Bcos%285x%29%7D%7Bsin%283x%29-
kogti [31]

Recall the angle sum identities:

cos(a + b) = cos(a) cos(b) - sin(a) sin(b)

cos(a - b) = cos(a) cos(b) + sin(a) sin(b)

sin(a + b) = sin(a) cos(b) + sin(b) cos(a)

sin(a - b) = sin(a) cos(b) - sin(b) cos(a)

Notice that adding the first two together, and subtract the last from the third, we get two more identities:

cos(a + b) + cos(a - b) = 2 cos(a) cos(b)

sin(a + b) + sin(a - b) = 2 sin(b) cos(a)

Let a = 4x and b = x. Then

cos(5x) + cos(3x) = 2 cos(4x) cos(x)

sin(5x) - sin(3x) = 2 sin(x) cos(4x)

Now,

-\dfrac{\cos(3x)+\cos(5x)}{\sin(3x)-\sin(5x)}=\dfrac{\cos(5x)+\cos(3x)}{\sin(5x)-\sin(3x)}=\dfrac{2\cos(4x)\cos x}{2\sin x\cos(4x)}=\dfrac{\cos x}{\sin x}=\cot x

as required.

4 0
4 years ago
Many smartphones, especially those of the LTE-enabled persuasion, have earned a bad rap for exceptionally bad battery life. Batt
BartSMP [9]

Answer:

Null hypothesis: the variance in hours of usage for talking is not greater than the the variance in hours of usage for internet.

H_o: \sigma _1 ^2 \leq \sigma _2^2

Alternative hypothesis: the variance  in hours of usage for talking is  greater than the the variance in hours of usage for internet.

H_a : \sigma_1^2 > \sigma_2^2

\mathbf{s_ 1 =16.11}

\mathbf{s_2 = 7.98}

Step-by-step explanation:

Let x_1 and x_2 be the two variables that represents the battery life in hours for talking usage and battery life in hours for internet usage respectively.

The hypothesis can be formulated as:

Null hypothesis: the variance in hours of usage for talking is not greater than the the variance in hours of usage for internet.

H_o: \sigma _1 ^2 \leq \sigma _2^2

Alternative hypothesis: the variance  in hours of usage for talking is  greater than the the variance in hours of usage for internet.

H_a : \sigma_1^2 > \sigma_2^2

The standard deviation for the battery usage for talking is :

\bar x_1 = \dfrac{1}{n_1} \sum x_i  \\ \\ \bar x_1  = \dfrac{1}{12}(35.8 +22.4+...+35.5) \\ \\ \bar x_1 = \dfrac{241.2}{12}  \\ \\ \bar x_1 =20.1

The standard deviation Is:

s_ 1 = \sqrt{\dfrac{1}{n_1-1}\sum (x{_1i}-\bar x_i)^2}

s_ 1 = \sqrt{\dfrac{1}{12-1}\sum (35.8- 20.1)^2+ (35.5-20.1)^2}

s_ 1 = \sqrt{259.568}

\mathbf{s_ 1 =16.11}

The standard deviation for the battery life usage for the internet is :

\bar x_2 = \dfrac{1}{n_2} \sum x_{2i}

\bar x_2 = \dfrac{1}{10} (24.0+12.5+36.4+...+4.7})

\bar x_2 = \dfrac{115}{10}

\bar x_2 = 11.5

Thus; the standard deviation is:

s_2 = \sqrt{\dfrac{1}{n_2-1}(x_{2i}- \bar x_2)^2}

s_2 = \sqrt{\dfrac{1}{10-1}(24-11.5)^2+(4.7-11.5)^2}

s_2 = \sqrt{63.60}

\mathbf{s_2 = 7.98}

4 0
3 years ago
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