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Vladimir [108]
3 years ago
5

What is the volume of a sphere with a diameter of 2 5/8

Mathematics
2 answers:
densk [106]3 years ago
8 0
4.19 is the correct answer
german3 years ago
3 0
V

≈

4.19

^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Show all four symbols of multiplication using the variables a/b
OlgaM077 [116]

Answer:

Multiplication Symbols:

<u>1.</u><u> Times</u> ×

<u>2.</u><u> Dot</u>  ⋅

<u>3.</u><u> Parentheses</u> ()

<u>4.</u><u> Variables next to each other</u> <em>ab</em>

Step-by-step explanation:

The symbols can be used like this...

Times: <em>a</em> × <em>b</em> = <em>c</em>

Dot: <em>a</em> ⋅ <em>b</em> = <em>c</em>

Parentheses:<em> </em>(<em>a</em>)(<em>b</em>) = <em>c</em>

Variables next to each other: <em>ab </em>= <em>c</em>

5 0
3 years ago
Please help me! i dont know what to do... And my weekend depends on you
ludmilkaskok [199]
2) 5.3
3) -6
4) -6
5) 6
7 0
3 years ago
What is 8/10 divided by 1 5/6
yulyashka [42]
The answer is <span>0.43 that is what i got</span>
7 0
3 years ago
Read 2 more answers
What is the answer? Please
NikAS [45]
I’m not sure but is it 7/10 since out of 10 times she hit an ace 7 times already so there is a 7/10 chance she would hit another ace
6 0
3 years ago
A recent study from the University of Virginia looked at the effectiveness of an online sleep therapy program in treating insomn
Ludmilka [50]

Answer:

1. The 99% confidence interval for the difference in average is -6.47377 < μ₁ - μ₂ < -11.34623

2. The possible issues in the calculations includes;

a. The confidence level used in the confidence interval can influence the result of the confidence interval observed

b. The sample size is small

Step-by-step explanation:

1. The number of adults with insomnia in the sample = 45

The number of adults that participated in the therapy, n₁ = 22

The number of candidates that served as control group, n₂ = 23

The average score for the for the 22 participants of the program, \overline x_1 = 6.59

The standard deviation for the 22 participants of the program, s₁ = 4.10

The average score for the for the 23 subjects in the control group, \overline x_2 = 15.50

The standard deviation for the 23 subjects in the control group, s₂ = 5.34

The confidence interval for unknown standard deviation, σ, is given by the following expression;

\left (\bar{x}_{1}- \bar{x}_{2}  \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}

α = 1 - 0.99 = 0.01

α/2 = 0.005

The degrees of freedom, df = 22 - 1 = 21

t_{\alpha /2} = t_{0.005, \, 21} = 1.721

Therefore, we have;

\left (6.59- 15.5  \right )\pm1.721 \cdot \sqrt{\dfrac{4.10^{2}}{22}+\dfrac{5.34^{2}}{23}}

The 99% confidence interval for the difference in average is therefore given as follows;

-6.47377 < μ₁ - μ₂ < -11.34623

Therefore, there is considerable evidence that the participants in the survey  had lower average score than the subjects in the control group

2. The possible issues in the calculations are;

a. The confidence level used in the confidence interval can influence the result of the confidence interval observed

b. The sample size is small

5 0
2 years ago
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