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Rudiy27
3 years ago
6

2/3=1.2/x solve for x

Mathematics
1 answer:
Anvisha [2.4K]3 years ago
5 0

First isolate x by multiplying by x on both sides.  It will look like this:

(2/3)x=1.2

Then devise by (2/3) on both sides to get:

x= 1.2/(2/3)

The easiest way to get this is plug that into a calculator and get the answer:

x=1.8

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A bouncing ball reaches a height of 54 inches at its first peak, 36 inches at its second peak, and 24 inches at its third peak.
steposvetlana [31]
54, 36, 24 are the 1st 3 element of a geometric progression with 2/3 as a common ratio: PROOF:
the 1st term is 54, (a₁= 54) the 2nd term a₂ = 24, then
(a₂ = a₁.r) or 36 = 54.r → r= 36/54 = 2/3. Same logique for the 3rd term. 
So 2/3 is common ratio. We know that :U(n) = a.(r)ⁿ⁻¹. Then if a =54 and r = x (given by the problem), then f(x) = 54.xⁿ⁻¹
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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
What is the quotient of 9.83 divided by 89.56
pav-90 [236]
Answer:

<span><span><span>34</span><span>23</span></span>=<span>98</span></span>

Explanation:

<span><span><span>34</span><span>23</span></span>=<span>?
</span></span>

4 0
3 years ago
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