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2 years ago

Solve for x . Help me solve them

Mathematics

1 answer:

Tresset [83]2 years ago

8 0

$\\(1).\\\\(1.7898)^{2x} = (1.7898)^6\\\\\implies \ln(1.7898)^{2x} = \ln(1.7898)^6\\\\\implies 2x \ln(1.7898) = 6 \ln(1.7898)\\\\\implies 2x =6\\\\\implies x = \dfrac 62 =3\\\\(3).\\\\2^{5x -6} = 4^{x+9}\\\\\implies 2^{5x -6} = (2^2)^{x+9}\\\\\implies 2^{5x-6} = 2^{2x+18}\\\\\implies \ln(2^{5x-6}) = \ln(2^{2x+18})\\\\\implies (5x -6) \ln 2 = (2x +18) \ln 2\\\\\implies 5x -6 = 2x +18\\\\\implies 5x -2x = 18 +6\\\\\implies 3x = 24\\\\\implies x = \dfrac{24}3 =8$

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Step-by-step explanation:

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2 years ago

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Illusion [34]

Answer: Volume = 97.6

Step-by-step explanation:

Volume of the cone first

V = 1/3 h π r²

h = 0.7

r = 2

V = 1/3 × 0.7 × 2² × 3.14

V = 1/3 × 0.7 × 4 × 3.14

V = 1/3 × 8.792

V = 2.93

Then find the volume of the cylinder if the cone was there.

V = π r² h

r = 2

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V = 3.14 × 2² × 8

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Now you subtract the volume of the cone from the cylinder.

100.48 - 2.93 = 97.55

Rounded to the nearest tenth is 97.6

Hope this was helpful, I know it is a lot to look at but I tried my best.

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1 year ago

Suppose that a company prints baseball cards. They claim that 30% of the cards feature rookies, 60% feature veterans, and 10% fe

liraira [26]

Answer:

H0: The distribution of players featured on the cards is 0.30 rookies, 0.60 veterans, and 0.10 All-Stars.

Ha: At least one of the proportions in the null hypothesis is false.

Step-by-step explanation:

On this case we need to apply a Chi squared goodness of fit test, and the correct system of hypothesis would be:

H0: The distribution of players featured on the cards is 0.30 rookies, 0.60 veterans, and 0.10 All-Stars.

Ha: At least one of the proportions in the null hypothesis is false.

And in order to test it we need to have observed and expected values. On this case we can calculate the Expected values like this

$E_{rookies}=50*0.3=15$

$E_{veterans}=50*0.6=30$

$E_{All stars}=50*0.1=5$

The observed values are not provided. The statistic on this case is given by:

$\chi^2 =\sum_{i=1}^n \frac{(O_i) -E_i}{E_i}$

And this statistic follows a chi square distribution with k-1 degrees of freedom on this case k=3, since we have 3 groups.

We can calculate the p valu like this:

$P(\chi^2 > \chi^2_{calculated})=p_v$

And if the p value it's higher than the significance level we FAIL to reject the null hypothesis. In other case we reject the null hypothesis.

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2 years ago