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3 years ago

Not sure what this question is asking...

Mathematics

1 answer:

zalisa [80]3 years ago

6 0

3)

$\bf ~~~~~~~~~~~~\textit{negative exponents}\\\\
a^{-n} \implies \cfrac{1}{a^n}
\qquad \qquad
\cfrac{1}{a^n}\implies a^{-n}
\qquad \qquad 
a^n\implies \cfrac{1}{a^{-n}}
\\\\
-------------------------------\\\\
\cfrac{-9}{2x^5}\implies \cfrac{-3^2}{2x^5}\implies \cfrac{-3^2}{1}\cdot \cfrac{1}{2}\cdot \cfrac{1}{x^5}\implies -\cfrac{1}{3^{-2}}\cdot 2^{-1}\cdot x^{-5}
\\\\\\
-\cfrac{2^{-1}x^{-5}}{3^{-2}}$

4)

$\bf 5^{-2}+5^0\implies \cfrac{1}{5^2}+1\implies \cfrac{1}{25}+1\implies \cfrac{1+25}{25}\implies \cfrac{26}{25}$

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3 years ago

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klasskru [66]

Answer:

<h2>B. 20% because he only got 400 back </h2>

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3 years ago

Collins is working.......

lesya [120]

Let n represent the amount Colin earned on Sunday.

On Sat. he earned n/2; on Sun. he earned n; and on Friday he earned (1/2)(n/2).

Then n/2 + n + n/4 = $70

Mult. all terms by 4 to eliminate fractions:

2n + 4n + n = $280

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Colin earned n/2, or $20, on Saturday; n, or $40, on Sunday; and n/4, or $10, on Friday.

Note that $20 and $40 and $10 add up to $70, as they must.

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3 years ago

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

vredina [299]

Answer:

c) P(270≤x≤280)=0.572

d) P(x=280)=0.091

Step-by-step explanation:

The population of bearings have a proportion p=0.90 of satisfactory thickness.

The shipments will be treated as random samples, of size n=500, taken out of the population of bearings.

As the sample size is big, we will model the amount of satisfactory bearings per shipment as a normally distributed variable (if the sample was small, a binomial distirbution would be more precise and appropiate).

The mean of this distribution will be:

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The standard deviation will be:

$\sigma_s=\sqrt{np(1-p)}=\sqrt{500*0.90*0.10}=\sqrt{45}=6.7$

We can calculate the probability that a shipment is acceptable (at least 440 bearings meet the specification) calculating the z-score for X=440 and then the probability of this z-score:

$z=(x-\mu_s)/\sigma_s=(440-450)/6.7=-10/6.7=-1.49\\\\P(z>-1.49)=0.932$

Now, we have to create a new sampling distribution for the shipments. The size is n=300 and p=0.932.

The mean of this sampling distribution is:

$\mu=np=300*0.932=279.6$

The standard deviation will be:

$\sigma=\sqrt{np(1-p)}=\sqrt{300*0.932*0.068}=\sqrt{19}=4.36$

c) The probability that between 270 and 280 out of 300 shipments are acceptable can be calculated with the z-score and using the continuity factor, as this is modeled as a continuos variable:

$P(270\leq x\leq280)=P(269.5$

d) The probability that 280 out of 300 shipments are acceptable can be calculated using again the continuity factor correction:

$P(X=280)=P(279.5$

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3 years ago

If Steve runs 4/5 mile each day, how many miles will Steve run in 5 days?

Aleksandr-060686 [28]

Answer:

4 miles

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3 years ago