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olga55 [171]
3 years ago
13

Ln (1/x)+ln (2x^3)=ln3

Mathematics
1 answer:
Lapatulllka [165]3 years ago
3 0
\ln \frac{1}{x} +\ln2x^3=\ln3
\ln \frac{2x^3}{x} = \ln3
\ln2x^2=\ln3
2x^2=3
x=\sqrt{ \frac{3}{2} }
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The sum of marshall and williams ages is 27 . William is 7 years older than marshall . What is marshall age
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Answer:

10

Step-by-step explanation: 17+10=27, 17-10=7

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2 years ago
Roxanne bought a 40-inch television that measures 24 inches in height. What is the width of the television?
Natali5045456 [20]

Answer:

The width is 32 inches

Step-by-step explanation:


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3 years ago
Let u = <-3, 4>, v = <8, 2>. Find u + v.
Liono4ka [1.6K]

Answer:

WHAT SUBJECT IS THAT? TELL ME WHAT SUBJECT AND I'LL ANSWER IT PROPERLY

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A nutritionist has developed a diet that she claims will help people lose weight. Twelve people were randomly selected to try th
deff fn [24]

The diet which is developed by the nutritionist is effective at the 0.05 level of significance. Claim is agreed.

<h3>When do we use two-sample t-test?</h3>

The two-sample t-test is used to determine if two population means are equal.

A nutritionist has developed a diet that she claims will help people lose weight. In this,

  • Twelve people were randomly selected to try the diet.
  • Their weights were recorded prior to beginning the diet and again after 6 months.

Here are the original weights, in pounds, with the weight after 6 months in parentheses.

  • Before 192 212 171 215 180 207 165 168 190 184 200 196
  • After    183 196  174 211 160 191   162 175 190  179  189 195

The mean of the weights before 6 moths is,

\overline X_1=\dfrac{192+ 212 +171 +215 +180 +207 +165 +168 +190 +184 +200 +196 }{12}\\\overline X_1=190

The mean of the weights after 6 months is,

\overline X_2=\dfrac{ 183 +196  +174 +211 +160 +191   +162 +175 +190  +179  +189 +195  }{12}\\\overline X_1=183.75

Standard deviation of both the data is 16.9 and 14.7.

1. Null and Alternative Hypotheses.

The following null and alternative hypotheses need to be tested:

\begin{array}{ccl} H_0: \mu_1 & = & \mu_2 \\\\ \\\\ H_a: \mu_1 & > & \mu_2 \end{array}

This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

  • (2) Rejection Region

Based on the information provided, the significance level is α=0.05 and the degrees of freedom are df = 22. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

df_{Total} = df_1 + df_2 = 11 + 11 = 22

Hence, it is found that the critical value for this right-tailed test is

t_c=1.717, for α=0.05 and df=22

The rejection region for this right-tailed test is,

R = \{t: t > 1.717\}

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

t = \displaystyle \frac{\bar X_1 - \bar X_2}{\sqrt{ \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}(\frac{1}{n_1}+\frac{1}{n_2}) } }

\displaystyle \frac{ 190 - 183.75}{\sqrt{ \frac{(12-1)16.9^2 + (12-1)14.7^2}{ 12+12-2}(\frac{1}{ 12}+\frac{1}{ 12}) } } = 0.967

  • (4) Decision about the null hypothesis

Since it is observed that t=0.967≤tc=1.717 it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.1721 and since p=0.1721≥0.05p = 0.1721  it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis H₀ is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1​ is greater than μ2​, at the α=0.05 significance level.

Confidence Interval

The 95% confidence interval is −7.16<μ<19.66

Thus, the diet which is developed by the nutritionist is effective at the 0.05 level of significance. Claim is agreed.

Learn more about the two-sample t-test here;

brainly.com/question/27198724

#SPJ1

8 0
2 years ago
Select the graph for the solution of the open sentence. Click until the correct graph appears. 5|x| + 3 &lt; 18
netineya [11]

Answer:

see below

Step-by-step explanation:

5|x| + 3 < 18

Subtract 3 from each side

5|x| + 3-3 < 18-3

5|x| < 15

Divide by 5

5|x| /5 < 15/5

|x|  < 3

There are  two solutions for an absolute value, one positive and one negative ( the negative flips the inequality)

x <3  and  x > -3

5 0
3 years ago
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