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4 years ago

8

Algebra 2---Please help with this problem!!! Will give medals and hugs! Show that the sum of the reciprocals of three different

positive integers is greater than 6 times the reciprocal of their product

1 answer:

andrezito [222]4 years ago

5 0

1/x + 1/y + 1/y > 6(1/xyz)

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Find the area of the figure

Amanda [17]

Answer:

A ≈ 26.42

Step-by-step explanation:

<em><u>The shape presented is a circle, and the formula for this is:</u></em>

A = π $r^2$

<em><u>The "</u></em> $r$ <em><u>," is the radius (2.9), so plug this into the formula:</u></em>

A = π $(2.9)^2$

A = π8.41

A ≈ 26.42

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4 years ago

Which of the following is a true statement concerning the use of a placebo?

Inessa05 [86]

Answer:

its B

Step-by-step explanation:

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3 years ago

A quality-conscious disk manufacturer wishes to know the fraction of disks his company makes which are defective. Step 2 of 2 :

stiks02 [169]

Answer:

$0.0497 - 1.96\sqrt{\frac{0.0497(1-0.0497)}{322}}=0.026$

$0.0497 + 1.96\sqrt{\frac{0.0497(1-0.0497)}{322}}=0.073$

The 90% confidence interval would be given by (0.026;0.073)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$\hat p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

The estimated proportion for this case is:

$\hat p =\frac{322-306}{322}= 0.0497$

Represent the proportion of defectives for this case

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.1$ and $\alpha/2 =0.05$ . And the critical value would be given by:

$z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.0497 - 1.96\sqrt{\frac{0.0497(1-0.0497)}{322}}=0.026$

$0.0497 + 1.96\sqrt{\frac{0.0497(1-0.0497)}{322}}=0.073$

The 90% confidence interval would be given by (0.026;0.073)

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3 years ago

Plz help me it's due this Monday

Alexxx [7]

246 muffins at 2.25 each

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3 years ago

1. The national mean (μ) IQ score from an IQ test is 100 with a standard deviation (s) of 15. The dean of a college wants to kno

Nat2105 [25]

Answer:

We conclude that the mean IQ of her students is different from the national average.

Step-by-step explanation:

We are given that the national mean (μ) IQ score from an IQ test is 100 with a standard deviation (s) of 15.

The dean of a college want to test whether the mean IQ of her students is different from the national average. For this, she administers IQ tests to her 144 students and calculates a mean score of 113

Let, Null Hypothesis, $H_0$ : $\mu$ = 100 {means that the mean IQ of her students is same as of national average}

Alternate Hypothesis, $H_1$ : $\mu\neq$ 100 {means that the mean IQ of her students is different from the national average}

(a) The test statistics that will be used here is One sample z-test statistics;

T.S. = $\frac{Xbar-\mu}{\frac{s}{\sqrt{n} } }$ ~ N(0,1)

where, Xbar = sample mean score = 113

s = population standard deviation = 15

n = sample of students = 144

So, test statistics = $\frac{113-100}{\frac{15}{\sqrt{144} } }$

= 10.4

Now, at 0.05 significance level, the z table gives critical value of 1.96. Since our test statistics is more than the critical value of z which means our test statistics will fall in the rejection region and we have sufficient evidence to reject our null hypothesis.

Therefore, we conclude that the mean IQ of her students is different from the national average.

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4 years ago