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tiny-mole [99]
2 years ago
15

The penny size of a nail indicates the length of the nail. The penny size d is given by the literal equation d=4n−2, where n is

the length (in inches) of the nail. a. Solve the equation for n. n= Question 2 b. Use the equation from part (a) to find the lengths of nails with the following penny sizes: 3, 6, and 10. A 3-penny nail is inches. A 6-penny nail is inches. A 10-penny nail is inches.
Mathematics
2 answers:
sleet_krkn [62]2 years ago
6 0
D=4n-2, d+2=4n-2+2, d+2=4n
(d+2)/4=(4n)/4, 1/4d+1/2=n
n=1/4n+1/2

size 3=
n=1/4*3/1=3/4, 3/4+1/2=5/4=1 1/4
size 3=1 1/4

size 6=
n=1/4*6/1=6/4, 6/4+1/2=8/4=2
size 6=2

size 10=
n=1/4*10/1=10/4, 10/4+1/2=12/4=3
size 10=3
LekaFEV [45]2 years ago
5 0

Answer:

The penny size d is given by the literal equation d=4n-2

Where n is the length (in inches) of the nail.

We can deduce the value of n from equation above:

(d+2)/4=n

Now we have to find n with various values of d.

1. when d = 3 inches

n=(3+2)/4

=>n=5/4

n = 1.25 inches

2. when d = 6 inches

n=(6+2)/4

=>n=8/4

n = 2 inches

3. when d = 10 inches

n=(10+2)/4

=>n=12/4

n = 3 inches

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Isosceles △ABC (AC=BC) is inscribed in the circle k(O). Prove that the tangent to the circle at point C is parallel to AB .
timofeeve [1]

Explanation:

Let M be the midpoint of AB. Then CM is the perpendicular bisector of AB. As such, center O is on CM, and OC is a radius (and CM). The tangent is perpendicular to that radius (and CM), so is parallel to AB, which is also perpendicular to CM.

If you need to go any further, you can show that triangles CMA and CMB are congruent, so (linear) angles CMA and CMB are congruent, hence both 90°.

5 0
2 years ago
• A researcher claims that less than 40% of U.S. cell phone owners use their phone for most of their online browsing. In a rando
antiseptic1488 [7]

Answer:

We failed to reject H₀

Z > -1.645

-1.84 > -1.645

We failed to reject H₀

p > α

0.03 > 0.01

We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.

Step-by-step explanation:

Set up hypotheses:

Null hypotheses = H₀: p = 0.40

Alternate hypotheses = H₁: p < 0.40

Determine the level of significance and Z-score:

Given level of significance = 1% = 0.01

Since it is a lower tailed test,

Z-score = -2.33 (lower tailed)

Determine type of test:

Since the alternate hypothesis states that less than 40% of U.S. cell phone owners use their phone for most of their online browsing, therefore we will use a lower tailed test.

Select the test statistic:  

Since the sample size is quite large (n > 30) therefore, we will use Z-distribution.

Set up decision rule:

Since it is a lower tailed test, using a Z statistic at a significance level of 1%

We Reject H₀ if Z < -1.645

We Reject H₀ if p ≤ α

Compute the test statistic:

$ Z =  \frac{\hat{p} - p}{ \sqrt{\frac{p(1-p)}{n} }}  $

$ Z =  \frac{0.31 - 0.40}{ \sqrt{\frac{0.40(1-0.40)}{100} }}  $

$ Z =  \frac{- 0.09}{ 0.048989 }  $

Z = - 1.84

From the z-table, the p-value corresponding to the test statistic -1.84 is

p = 0.03288

Conclusion:

We failed to reject H₀

Z > -1.645

-1.84 > -1.645

We failed to reject H₀

p >  α

0.03 > 0.01

We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.

8 0
2 years ago
3.30 Survey response rate. Pew Research reported in 2012 that the typical response rate to their surveys is only 9%. If for a pa
Artist 52 [7]

Answer:

0% probability that at least 1,500 will agree to respond

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 15000, p = 0.09

So

\mu = E(X) = np = 15000*0.09 = 1350

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{15000*0.09*0.91} = 35.05

What is the probability that at least 1,500 will agree to respond

This is 1 subtracted by the pvalue of Z when X = 1500-1 = 1499. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{1499 - 1350}{35.05}

Z = 4.25

Z = 4.25 has a pvalue of 1.

1 - 1 = 0

0% probability that at least 1,500 will agree to respond

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Researchers are assessing a new screening tool (blood-based assay) for early identification of prostate cancer. The test is run
natima [27]

Answer: B. Accurate but not reliable

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An experiment is said to be " accurate "

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