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Troyanec [42]
3 years ago
8

PLZZ help ZOOM IN THE SEE IT CLEAR easy math problem (written math problem) (30 POINTS)

Mathematics
2 answers:
djverab [1.8K]3 years ago
6 0

Conner is right and Jana is wrong I hope this helps you have a great day or a great night where ever you are

Anika [276]3 years ago
3 0

Answer:

<h2>Conner's work is correct</h2>

Step-by-step explanation:

We know:

a^n\cdot a^m=a^{n+m}

Therefore:

(3^56^8)(3^96^{10})=(3^53^9)(6^86^{10})=3^{5+9}6^{8+10}=3^{14}6^{18}

Jana wrong used

(a^n)^m=a^{nm}

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(a) The usual load is not 13 credits.

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According to the Central limit theorem, if a large sample (<em>n</em> ≥ 30) is selected from an unknown population then the sampling distribution of sample mean follows a Normal distribution.

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\mu_{\bar x}=\bar x=13.65\\SE=\frac{s}{\sqrt{n}}=\frac{1.91}{\sqrt{100}}=0.191

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The (1 - <em>α</em>) % confidence interval for population mean is given by:

CI=\bar x\pm z_{\alpha/2}\times SE

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Construct the 95% confidence interval as follows:

CI=\bar x\pm z_{\alpha/2}\times SE\\=13.65\pm (1.96\times0.191)\\=13.65\pm0.3744\\=(13.2756, 14.0244)\\=(13.28, 14.02)

As the null value, <em>μ</em> = 13 is not included in the 95% confidence interval the null hypothesis will be rejected.

Thus, it can be concluded that the usual load is not 13 credits.

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Compute the probability that a a student at this college takes 16 or more credits as follows:

P(X\geq 16)=P(\frac{X-\mu}{\sigma}\geq \frac{16-13.65}{1.91})\\=P(Z>1.23)\\=1-P(Z

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