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egoroff_w [7]
3 years ago
14

Math problem. I need help plz.

Mathematics
2 answers:
mash [69]3 years ago
6 0

Answer:

Never because one of them goes 1k8 and one of them goes 24 so if they that song every time then it would be the same time diffrence even if you try a million tinmes

Step-by-step explanation:

cricket20 [7]3 years ago
6 0
It’s never, they would never play again
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Solid fats are more likely to raise blood cholesterol levels than liquid fats. Suppose a nutritionist analyzed the percentage of
Andrews [41]

Answer:

t = 31.29

Step-by-step explanation:

Given

\begin{array}{ccccccc}{Stick} & {25.8} & {26.9} & {26.2} & {25.3} & {26.7}& {26.1} \ \\ {Liquid} & {16.9} & {17.4} & {16.8} & {16.2} & {17.3}& {16.8} \ \end{array}

Required

Determine the test statistic

Let the dataset of stick be A and Liquid be B.

We start by calculating the mean of each dataset;

\bar x =\frac{\sum x}{n}

n, in both datasets in 6

For A

\bar x_A =\frac{25.8+26.9+26.2+25.3+26.7+26.1}{6}

\bar x_A =\frac{157}{6}

\bar x_A =26.17

For B

\bar x_B =\frac{16.9+17.4+16.8+16.2+17.3+16.8}{6}

\bar x_B =\frac{101.4}{6}

\bar x_B =16.9

Next, calculate the sample standard deviation

This is calculated using:

s = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}

For A

s_A = \sqrt{\frac{\sum(x - \bar x_A)^2}{n-1}}

s_A = \sqrt{\frac{(25.8-26.17)^2+(26.9-26.17)^2+(26.2-26.17)^2+(25.3-26.17)^2+(26.7-26.17)^2+(26.1-26.17)^2}{6-1}}

s_A = \sqrt{\frac{1.7134}{5}}

s_A = \sqrt{0.34268}

s_A = 0.5854  

For B

s_B = \sqrt{\frac{\sum(x - \bar x_B)^2}{n-1}}

s_B = \sqrt{\frac{(16.9 - 16.9)^2+(17.4- 16.9)^2+(16.8- 16.9)^2+(16.2- 16.9)^2+(17.3- 16.9)^2+(16.8- 16.9)^2}{6-1}}

s_B = \sqrt{\frac{0.92}{5}}

s_B = \sqrt{0.184}

s_B = 0.4290

Calculate the pooled variance

S_p^2 = \frac{(n_A - 1)*s_A^2 + (n_B - 1)*s_B^2}{(n_A+n_B-2)}

S_p^2 = \frac{(6 - 1)*0.5854^2 + (6 - 1)*0.4290^2}{(6+6-2)}

S_p^2 = \frac{2.6336708}{10}

S_p^2 = 0.2634

Lastly, calculate the test statistic using:

t = \frac{(\bar x_A - \bar x_B) - (\mu_A - \mu_B)}{\sqrt{S_p^2/n_A +S_p^2/n_B}}

We set

\mu_A = \mu_B

So, we have:

t = \frac{(\bar x_A - \bar x_B) - (\mu_A - \mu_A)}{\sqrt{S_p^2/n_A +S_p^2/n_B}}

t = \frac{(\bar x_A - \bar x_B) }{\sqrt{S_p^2/n_A +S_p^2/n_B}}

The equation becomes

t = \frac{(26.17 - 16.9) }{\sqrt{0.2634/6 +0.2634/6}}

t = \frac{9.27}{\sqrt{0.0878}}

t = \frac{9.27}{0.2963}

t = 31.29

<em>The test statistic is 31.29</em>

3 0
3 years ago
Four circular cardboard pieces of radii 7 cm are placed on a paper in such a way that each piece touches other two pieces. Find
Anastasy [175]
Required area of shaded portion

= Area of square ABCD - 4 × area of one quadrant

= 14 \times 14 - 4 \times \frac{90}{360} \times \frac{22}{7} \times (7) \times (7)

= 196 - 154 = 42 \: cm {}^{2}

3 0
3 years ago
Find the dimensions of the square <br> Area=441cm over 2
MA_775_DIABLO [31]

Answer:

each side is 55.125cm

Step-by-step explanation:

you have to divide the area by 2

then do 441/2=220.5

then divide this by 4 because a square has 4 equal sides so 220.5/4=55.125

5 0
3 years ago
PLEASE HELP (question is in picture)
HACTEHA [7]

Answer:

4*3=12

12*4=

3*3=9

9*2=18

18+48=66

Step-by-step explanation:

8 0
3 years ago
How many cubes with side lengths of \dfrac14\text{ cm}
bogdanovich [222]

Answer:

Step-by-step explanation:

Correct question

How many cubes with side lengths of ¼cm needed to fill the prism of volume 4 cubic units?

We know that,

Volume of a cube is s³

V = s³

Where 's' is length of side of a cube

Given that

The cube has a length of ¼cm, and a cube has equal length

s= ¼cm

Then, it's volume is

V = s³

V = (¼)³ = ¼ × ¼ × ¼

V = 1 / 64 cubic unit

V = 0.015625 cubic unit

Then, given that the volume of the prism to be filled is 4 cubic unit

Then,

As, we have to find the number if cubes so we will divide volume of prism by volume of one cube

Then,

n = Volume of prism / Volume of cube

n = 4 / 0.015625

n = 256

So, then required cubes to filled the prism is 256 cubes.

5 0
3 years ago
Read 2 more answers
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