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4 years ago

I need help ASAP please!

Mathematics

1 answer:

topjm [15]4 years ago

8 0

Answer:

Jan 2018: 7.83 million
Feb 2017: 8.12 million

Step-by-step explanation:

Your "Grandma formula" tells you ...

N = G(1 +x/100)

Solving for G gives ...

G = N/(1 +x/100) . . . . . . G = old value; N = new value; x = % change

For this problem, N=7.9 million.

1) x = 0.9, so G = 7.9/(1 +.9/100) ≈ 7.83 . . . million

7.83 million people were looking for work in January 2018.

2) x = -2.7, so G = 7.9/(1 -2.7/100) ≈ 8.12 . . . million

8.12 million people were looking for work in February 2017.

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<h2>Answer:</h2>

The type of reasoning which starts with a given set of rules and conditions to determine what must be true is:

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<h2>Step-by-step explanation:</h2>

Deductive Reasoning--

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The value after differentiate will be;

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Given that;

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The function is,

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⇒ $\frac{dy}{dx} = 3^{x^{3} } log_{e} 3 \frac{d}{dx} (x^3)$

⇒ $\frac{dy}{dx} = 3^{x^{3} } log_{e} 3 * 3x^2$

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⇒ $\frac{dy}{dx} = 3x^2 3^{x^{3} } log_{e} 3$

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1 year ago

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Answer:

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b) The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

c) $\alpha =1-0.98=0.02$

Step-by-step explanation:

If np' and n(1-p') are higher than 5, a confidence interval for the proportion is calculated as:

$p'-z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }\leq p\leq p'+z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }$

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Then, a 98% confidence interval for the percentage of all males who identify themselves as the primary grocery shopper can be calculated replacing p' by 0.52, n by 2400, $\alpha$ by 0.02 and $z_{\alpha/2}$ by 2.33

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The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

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