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1 year ago

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What is the value of x innthe equation 1.5(x+4)-3=4.5(x-2)

1 answer:

daser333 [38]1 year ago

3 0

$~~~~~~1.5(x+4)-3 = 4.5(x-2)\\\\\implies 15(x+4) -30 =45(x-2)~~~~~~~~;[\text{Multiply both sides by 10]}\\\\\implies 3(x+4)-6=9(x-2)~~~~~~~~~~;[\text{Divide both sides by 5]}\\\\\implies 3x+12-6=9x-18\\\\\implies 3x+6=9x-18\\\\\implies 9x-3x = 18+6\\\\\implies 6x = 24\\\\\implies x = \dfrac{24}6\\\\\implies x =4$

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mezya [45]

Answer:

The confidence interval is (24116.3878,24883.6122).

Step-by-step explanation:

We are given the following information in the question:

Population mean, $\mu$ = $23,500

Sample mean, $\bar{x}$ = $24,500

Sample standard deviation,s = $2,800

Sample size, n = 146

Confidence interval:

$\bar{x} \pm t_{critical}\frac{s}{\sqrt{n}}$

Putting the values, we get,

$t_{critical}\text{ at}~\alpha_{0.05} = \pm 1.655436$

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The confidence interval is (24116.3878,24883.6122).

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2 years ago

Deshawn has two bags of marbles. The first bag has 2 blue, 3 orange, and 5 red. The second bag has 4 pink, 10 blue, and 6 brown.

madreJ [45]

Answer: 0.1

Step-by-step explanation:

Given

Bag-I has 2 blue,3 orange, 5 red

Bag-II has 4 Pink,10 blue, 6 brown

No of ways of choosing a blue marble from bag-I

$\Rightarrow ^2C_1$

Total no of ways of choosing a marble from bag-I

$\Rightarrow ^{10}C_1$

No of ways of choosing a blue marble from bag-II

$\Rightarrow ^{10}C_1$

Total no of ways of choosing a marble from bag-II

$\Rightarrow ^{20}C_1$

The probability that he will pull out a blue marble from each bag is

$\Rightarrow P=\text{Probability of pulling a blue marble from bag-I}\times \text{Probability of pulling a blue out bag-II}$

$\Rightarrow P=\dfrac{^2C_1}{^{10}C_1}\times \dfrac{^{10}C_1}{^{20}C_1}\\\\\Rightarrow P=\dfrac{2}{10}\times \dfrac{10}{20}=\dfrac{1}{10}$

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Answer:

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Step-by-step explanation:

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