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viva [34]
3 years ago
11

I need help with this math problem

Mathematics
1 answer:
Fofino [41]3 years ago
4 0

Answer:

5

Step-by-step explanation:

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Property being illustrated 73=0+73
Bad White [126]

Answer:

yes???

Step-by-step explanation:

your question either doesn't make sense or I just don't know

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3 years ago
How to spell 30,906 in word
hammer [34]
Thirty thousand nine hundred and six.
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What base-ten number is 14 tens and 4 ones?¿
MrMuchimi

Answer:144


Step-by-step explanation:


7 0
4 years ago
How many revolutions does a go kart with a wheel circumference of 62.8318 inches make in a 1609 m race? Please answer as quickly
Rudiy27

Answer: 1008 revolutions

Step-by-step explanation:

1 revolution is equal to 2\pi r

To find how many revolutions the wheel makes, you must divide the distance traveled in the race between the circumference of the wheel

You must first convert 62.8318 in to meters.

We know that 1 inch is 0.0254 meters

Then

62.8318\ in*\frac{0.0254\ m}{1\ in}= 1.5959\ m

revolutions = \frac{1609}{1.5959}\\\\revolutions=1008.21\\

Finally The wheel made 1008.21 revolutions

3 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
3 years ago
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