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Marianna [84]
2 years ago
14

there are 147 third grade students going to the theater. Each row at the theater has ten seats. How many rows of seats will the

third grade students fill?
Mathematics
1 answer:
givi [52]2 years ago
7 0

Answer:

157

Step-by-step explanation:

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Measure the length of the paper clip to the nearest tenth of a centimeter.
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3 cm since the end is closer to the 3 than it is to 4
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Can someone help me with this please
MrRa [10]

Answer:

Andre's Method is the only one that works

Step-by-step explanation:

Noah's method won't work because he doesn't follow the order of operations. You can pull the the 9 out first you have to distribute the 2 first.

Elena's method doesn't work because to eliminate the 18 you must use the reverse operation it must be added not subtracted.

Andre's Method is the only one that works

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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
How can I solve this?
elixir [45]

Answer:

x=8/21

Step-by-step explanation:

(3/4)/(2/7) = 1/x

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x = (2/7) / (3/4)

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Answer

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Step-by-step explanation:

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