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2 years ago

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A car’s odometer measures distance to the nearest 0.1 miles. Which is the most appropriate way to report the distance driven usi

ng the car’s odometer?
A. 182 miles

B. 200 miles

C. 181.7 miles

D. 181.77 miles

1 answer:

olga55 [171]2 years ago

6 0

Answer:

C. 181.7 miles

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While standing on a highway overpass, Jennifer wonders what proportion of the vehicles that pass on the highway below are trucks

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Step-by-step explanation:

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Prove this (sinx-tanx)(cosx-cotx)=(sinx-1)(cosx-1)

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1 year ago

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Find the square root of 15129 by division method

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$\underline{\ \ \ \ \ \ 123}\\1\ \ \ |15129\\\underline{\ \ \ \ \|1}\\22\ |\ 51\\\underline{\ \ \ \ |\ 44}\\243|\ \ 729\\\underline{\ \ \ \ \ |\ 729}\\.\qquad\ \ \ 0$

$\sqrt{15129}=123$

$\begin{array}{c|c}15129&3\\5043&3\\1681&41\\41&41\\1\end{array}15129=3\cdot3\cdot41\cdot41=3^2\cdot41^2\\\\\sqrt{15129}=\sqrt{3^2\cdot41^2}=\sqrt3^2}\cdot\sqrt{41^2}=3\cdot41=123$

$Used:\\\\\sqrt{a\cdot b}=\sqrt{a}\cdot\sqrt{b}\\\\\sqrt{a^2}=a\ for\ a\geq0$

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2 years ago

answer A radio station located 120 miles due east of Collinsville has a listening radius of 100 miles. A straight road joins Col

melamori03 [73]

Answer:

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Step-by-step explanation:

One way to solve this problem is by using an equation that describes the listening radius of the station, and another for the road, then the points where this two-equation intersect each other will represent when the driver starts and stops listening to the station, and the distance between the points is the miles that the driver will receive the signal.

The equation for the listening radius (the radio station is at (0,0)):

$x^2+y^2=100^2$

The equation for the road that past through the points (-120,0) and (80,100) (Collinsville and Harmony respectively):

$m=\frac{y_2-y_1}{x_2-x_1} =\frac{100-0}{80-(-120)}=\frac{100}{200}=\frac{1}{2}$

$y-y_1=m(x-x_1)\\y-0=\frac{1}{2}(x-(-120))\\ y=\frac{1}{2}x+60$

Substitutes the value of y in the equation of the circle:

$x^2+(\frac{1}{2}x+60)^2=100^2\\x^2+\frac{1}{4} x^2+60x+3600=10000\\\frac{5}{4} x^2+60x+3600=10000\\\frac{5}{4} x^2+60x+3600-10000=0\\\frac{5}{4} x^2+60x-6400=0\\5 x^2+240x-25600=0\\x^2+48x-5120=0\\$

The formula to solve second-degree equations:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac} }{2a} \\x_{1,2}=\frac{-48\pm\sqrt{48^2-4(1)(-5120)} }{2(1)}\\x_{1,2}=\frac{-48\pm\sqrt{2304+20480} }{2}\\x_{1,2}=\frac{-48\pm\sqrt{22784} }{2}\\x_{1,2}=\frac{-48\pm16\sqrt{89} }{2}\\x_{1,2}=-24\pm8\sqrt{89} \\x_1=-24+8\sqrt{89}\approx51.4718\\x_2=-24-8\sqrt{89}\approx-99.4718\\$

Using the values in x to find the values in y:

$y_1=\frac{1}{2}x_1+60\\y_1=\frac{1}{2}(-24+8\sqrt{89} )+60\\y_1=-12+4\sqrt{89}+60\\ y_1=48+4\sqrt{89}\approx85.7359$

$y_2=\frac{1}{2}x_2+60\\y_2=\frac{1}{2}(-24-8\sqrt{89} )+60\\y_1=-12-4\sqrt{89}+60\\ y_1=48-4\sqrt{89}\approx10.2641$

The distance between the points (51.4718,85.7359) and (-99.4718,10.2641) :

$d=\sqrt{(x_1 -x_2 )^2+(y_1 -y_2)^2} \\d=\sqrt{(-24+8\sqrt{89} -(-24-8\sqrt{89}) )^2+(48+4\sqrt{89} -(48-4\sqrt{89}) )^2}\\d=\sqrt{(-24+8\sqrt{89} +24+8\sqrt{89} )^2+(48+4\sqrt{89} -48+4\sqrt{89} )^2}\\d=\sqrt{(16\sqrt{89} )^2+(8\sqrt{89} )^2}\\d=\sqrt{22784+5696}\\d=\sqrt{28480}\\d=8\sqrt{445}\approx168.7602miles$

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2 years ago

Suppose that birth weights are normally distributed with a mean of 3466 grams and a standard deviation of 546 grams. Babies weig

Anon25 [30]

Answer:

3.84% probability that it has a low birth weight

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

$\mu = 3466, \sigma = 546$

If we randomly select a baby, what is the probability that it has a low birth weight?

This is the pvalue of Z when X = 2500. So

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

$Z = \frac{2500 - 3466}{546}$

$Z = -1.77$

$Z = -1.77$ has a pvalue of 0.0384

3.84% probability that it has a low birth weight

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2 years ago